Mathematical Reflections 2013, Issue 5 - Problem U278

Problem:
Evaluate $$\lim_{n \to \infty} \sum_{k=0}^\infty \dfrac{1}{(kn+1)k!}.$$

Proposed by Dorin Andrica.

Solution:
Observe that for any $n,k \geq 1$, it holds $n+1 \leq kn+1 \leq n(k+1)$. Therefore,
$$1+\sum_{k=1}^\infty \dfrac{1}{n(k+1)!} \leq \sum_{k=0}^\infty \dfrac{1}{(kn+1)k!} \leq 1+\sum_{k=1}^\infty \dfrac{1}{(n+1)k!},$$ i.e. $$1+\dfrac{e-2}{n} \leq \sum_{k=0}^\infty \dfrac{1}{(kn+1)k!} \leq 1+\dfrac{e-1}{n+1}.$$
By the Squeeze Theorem, we have
$$1 \leq \lim_{n \to \infty} \sum_{k=0}^\infty \dfrac{1}{(kn+1)k!} \leq 1,$$ hence $\displaystyle \lim_{n \to \infty} \sum_{k=0}^\infty \dfrac{1}{(kn+1)k!}=1$.

Mathematical Reflections 2013, Issue 5 - Problem U277

Problem:
For $n \in \mathbb{N}$, $n \geq 2$, find the greatest integer less than $2(e^{\frac{1}{n+1}}+\ldots+e^{\frac{1}{n+n}})$.

Proposed by Marius Cavachi.

Solution:
We claim that the greatest integer less than $2(e^{\frac{1}{n+1}}+\ldots+e^{\frac{1}{n+n}})$ is $2n+1$. Indeed,
$$2(e^{\frac{1}{n+1}}+\ldots+e^{\frac{1}{n+n}}) \geq 2ne^{\frac{1}{2n}} \geq 2n\left(1+\dfrac{1}{2n}\right)=2n+1,$$
but $$2(e^{\frac{1}{n+1}}+\ldots+e^{\frac{1}{n+n}}) \leq 2ne^{\frac{1}{n+1}} < 2n+2,$$
where the last inequality can be obtained observing that $$\dfrac{1}{n+1}=\int_{1}^{1+\frac{1}{n}} \dfrac{1}{1+\frac{1}{n}} \ dx <\int_{1}^{1+\frac{1}{n}} \dfrac{1}{x} \ dx=\log\left(1+\dfrac{1}{n}\right).$$

Mathematical Reflections 2013, Issue 5 - Problem S281

Problem:
Let $f(n)$ be the sum of the digits of $n^2 + 1$. Define the sequence $(a_n)_{n \geq 0}$, with $a_0$ an
arbitrary positive integer and $a_{n+1} = f(a_n)$, $n \geq 0$. Prove that $(a_n)_{n \geq 0}$ is $3$-periodic.

Proposed by Roberto Bosch Cabrera.

Solution:
Since $f(5)=8, f(8)=11$ and $f(11)=5$, it suffices to prove that for every positive integer $a_0$ there exists some $n \in \mathbb{N}$ such that $a_n \in \{5,8,11\}$. Let $m$ be the number of digits of $a_0$. We prove the statement by induction on $m$. For $m \leq 2$ we proceed by a direct check. If $a_0 \in \{5,8,11\}$ there is nothing to prove. If $a_0$ is a two-digit number, then $a^2_0 \leq 10000$, so $a_1 \leq 37$ and we reduce to analyze the cases for $a_0 \leq 37$.

(i) If $a_0 \in \{2,7,20\}$, then $a_1=5$. If $a_0 \in \{1,10,26,28\}$, then $a_1 \in \{2,20\}$, so $a_2=5$. Finally, if $a_0 \in \{3,6,9,12,15,18,27,30,33\}$, then $a_3=5$.

(ii) If $a_0 \in \{4,13,23,32\}$, then $a_1=8$.

(iii) If $a_0 \in \{17,19,21,35,37\}$, then $a_1=11$. If $a_0 \in \{14,22,24,31,36\}$, then $a_1 \in \{17,19\}$, so $a_2=11$. Finally, if $a_0 \in \{16,25,29,34\}$, then $a_3=11$.

Thus, we have proved that if $a_0$ is a one or two-digit number, then $a_n \in \{5,8,11\}$ for some $n \in \mathbb{N}$, i.e. the sequence is $3$-periodic. Let $m \geq 2$ and suppose that the statement is true for all $k \leq m$. Let $a_0$ be an $(m+1)$-digit number. Then, $10^m \leq a_0 < 10^{m+1}$ which implies $10^{2m} \leq a^2_0 < 10^{2(m+1)}$. Hence, $$a_1=f(a_0) \leq 9\cdot2(m+1)+1<10^m,$$ and by induction hypothesis, the sequence $(a_n)_{n \geq 1}$ is $3$-periodic, which implies that the sequence $(a_n)_{n \geq 0}$ is $3$-periodic, as we wanted to prove.


Note: the problem was later changed.

Mathematical Reflections 2013, Issue 5 - Problem S279

Problem:
Solve in integers the equation $$(2x+y)(2y+x)=9\min(x,y).$$

Proposed by Titu Andreescu.

Solution:
Assume without loss of generality that $x \leq y$. Then, we have to find the integer solutions of the equation
$$2y^2+5xy+(2x^2-9x)=0.$$ To this aim, the discriminant of this equation in $y$ must be a perfect square, so there exists $t \in \mathbb{Z}$ such that
$$\Delta_y=9x(x+8)=t^2 \implies x(x+8)-\dfrac{t^2}{9}=0,$$ i.e. $$\left(x+4-\dfrac{t}{3}\right)\left(x+4+\dfrac{t}{3}\right)=16.$$
If both the two factors are equal to $\pm 4$, it must be $t=0$, so $x=0$ or $x=-8$, which give $y=0$ and $y=10$ respectively. If the two factors are distinct, since they have the same parity, it must be $$\begin{array}{lll}x+4-t/3&=&\pm 2 \\ x+4+t/3&=&\pm 8, \end{array} \qquad \begin{array}{lll}x+4-t/3&=&\pm 8 \\ x+4+t/3&=&\pm 2. \end{array}$$
Since $x(x+8) \geq 0$, an easy check shows that $x=1$, and for this value we get $y=1$. Therefore, all the integer solutions are $$(0,0), (1,1), (-8,10), (10,-8).$$

Mathematical Reflections 2013, Issue 5 - Problem J281

Problem:
Solve the equation $$x+\sqrt{(x+1)(x+2)}+\sqrt{(x+2)(x+3)}+\sqrt{(x+3)(x+1)}=4.$$

Proposed by Titu Andreescu.

Solution:
We rewrite the equation in the form $$\sqrt{(x+1)(x+2)}+\sqrt{(x+2)(x+3)}=4-x-\sqrt{(x+3)(x+1)}.$$
Since $4-x \geq \sqrt{(x+3)(x+1)}$, it follows that $x \leq 13/12$. Squaring both sides and reordering, we obtain
$$12\sqrt{(x+3)(x+1)}=-12x+11,$$ from which $x \leq 11/12$. Squaring both sides again, we get
$$144x^2+576x+432=144x^2-264x+121,$$ which gives $x=-\dfrac{311}{840}$.

Mathematical Reflections 2013, Issue 5 - Problem J280

Problem:
Let $a, b, c, d$ be positive real numbers. Prove that
$$2(ab + cd)(ac + bd)(ad + bc) \geq (abc+bcd+cda+dab)^2.$$

Proposed by Ivan Borsenco.

Solution:
The given inequality is equivalent to the inequality $$\sum_{cyc} (abc)^2 \geq 2abcd\left(\sum_{cyc} (ab-a^2)+ac+bd \right).$$
Using the Rearrangement Inequality and the AM-GM Inequality, we get
$$\begin{array}{lll} \displaystyle 2abcd\left(\sum_{cyc} (ab-a^2)+ac+bd \right) & \leq & 2abcd(ac+bd)\\&=&(ac)^2 \cdot 2bd+ (bd)^2\cdot 2ac \\ & \leq & \displaystyle \sum_{cyc} (abc)^2, \end{array}$$ and the conclusion follows.

Mathematical Reflections 2013, Issue 5 - Problem J279

Problem:
Find all triples $(p, q, r)$ of primes such that $pqr = p + q + r + 2000$.

Proposed by Titu Andreescu.

Solution:
Assume without loss of generality that $p \leq q \leq r$. The given equality can be rewritten as
$$\begin{equation}\label{first-eq} (rq-1)(p-1)+(r-1)(q-1)=2002.                 (1) \end{equation}$$
If $p$ is an odd prime, then $q$ and $r$ are odd prime also, but this means that the LHS is divisible by $4$ and the RHS is not divisible by $4$, a contradiction. Thus, $p=2$ and equation $(1)$ becomes $$(2q-1)(2r-1)=4005=3^2\cdot5\cdot89.$$ Since $2q-1 \leq 2r-1$, then $(2q-1)^2 \leq 4005$, i.e. $2q-1 \leq 63$. This means that $2q-1 \in \{1,3,5,9,15,45\}$. Clearly, $2q-1 \neq 1,15$, therefore we have the four systems of equations
$$\begin{array}{lll} 2q-1&=&3 \\ 2r-1&=&1335, \end{array} \qquad \begin{array}{lll} 2q-1&=&5 \\ 2r-1&=&801, \end{array} \qquad \begin{array}{lll} 2q-1&=&9 \\ 2r-1&=&445, \end{array} \qquad \begin{array}{lll} 2q-1&=&45 \\ 2r-1&=&89. \end{array}$$
It's easy to see that the first and the last system have no solution in primes, and the other two systems give $q=3, r=401$ and $q=5, r=223$. Therefore, $(2,3,401)$ and $(2,5,223)$ are two solutions to the given problem and by symmetry all the solutions are $$(2,3,401),(2,401,3),(3,2,401),(3,401,2),(401,2,3),(401,3,2),$$ $$(2,5,223),(2,223,5),(5,2,223),(5,223,2),(223,2,5),(223,5,2).$$

Mathematical Reflections 2013, Issue 5 - Problem J278

Problem:
Find all positive integers $n$ for which $$\{\sqrt[3]{n}\} \leq \dfrac{1}{n},$$ where $\{x\}$ denotes the fractional part of $x$.

Proposed by Ivan Borsenco.

Solution:
Clearly, every perfect cube satisfies the condition. Now, let $m \in \mathbb{Z}^+$ such that $m^3<n<(m+1)^3$, i.e. $n=m^3+k$ with $1 \leq k \leq (m+1)^3-1$. Then, $$\{\sqrt[3]{m^3+k}\} \geq \{\sqrt[3]{m^3+1}\}=\sqrt[3]{m^3+1}-m>\dfrac{1}{m^3+1},$$ for all $m>2$. Therefore, it's enough to find the integers which satisfy the condition in $(1,8) \cup (8,27)$. An easy check shows that the required integers are $n=2,9$. In conclusion, $n=2,9$ or $n=m^3$ for some $m \in \mathbb{Z}^+$.

Mathematical Reflections 2013, Issue 5 - Problem J277

Problem:
Is there an integer $n$ such that $4^{5^n}+ 5^{4^n}$ is a prime?

Proposed by Titu Andreescu.

Solution:
The answer is no. Clearly, for $n<0$ the given expression is not an integer. If $n=0$ we get $4+5=9$, which is not a prime. If $n>0$, set $x=5^{4^{n-1}}$ and $y=4^{\frac{5^n-1}{4}}$. It is easy to see that $x$ and $y$ are both integers. Hence $$4^{5^n}+5^{4^n}=x^4+4y^4=(x^2+2y^2+2xy)(x^2+2y^2-2xy),$$ which is the product of two positive integers greater than $1$.

Mathematical Reflections 2013, Issue 4 - Problem S273

Problem:
Let $a,b,c$ be positive integers such that $a \geq b \geq c$ and $\dfrac{a-c}{2}$ is a prime. Prove that if $$a^2+b^2+c^2-2(ab+bc+ca)=b,$$ then $b$ is either a prime or a perfect square.

Proposed by Titu Andreescu.

Solution:

From the given conditions, we have $a=2p+c$, where $p$ is a prime number. Then, the given equation can be written as $$(2p-b)^2=b(4c+1).$$ Expanding $(2p-b)^2$, we see that $b|4p^2$, therefore $b \in \{1,2,4,p,2p,4p,p^2,2p^2,4p^2\}$. We have three cases.

(i) $b=2^{\alpha}$, where $\alpha \in \{0,1,2\}$. If $\alpha=0$, then $b=c=1$, but this implies $(2p-1)^2=5$, a contradiction. If $\alpha>0$, then $2^{2-\alpha}(p-2^{\alpha-1})^2=(4c+1)$. If $\alpha=1$ we have a contradiciton. If $\alpha=2$, then $(2p-1)^2=4c+1$ and since $c \in \{0,1,2,3\}$, we obtain $c=2$ and $p=2$, which gives $a=6$. Therefore, $$a=6, \qquad b=4, \qquad c=2.$$

(ii) $b=2^{\alpha}p$, where $\alpha \in \{0,1,2\}$. If $\alpha=0$, then $p=4c+1$, and for all primes of the form $4k+1$, with $k \in \mathbb{Z}^+$, we have $c=(p-1)/4$. Therefore, $$a=(9p-1)/4, \qquad b=p, \qquad c=(p-1)/4.$$ If $\alpha=1$ we get an easy contradiction, and if $\alpha=2$, we get $c=(p-1)/4$, but this implies $a=(9p-1)/4<4p=b$, contradiction.

(iii) $b=2^{\alpha}p^2$, where $\alpha \in \{0,1,2\}$. If $\alpha=0$, then $(2-p)^2=4c+1$, which gives $c=\dfrac{(p-1)(p-3)}{4}$ (note that this is an integer for all primes $p>3$) and $a=\dfrac{(p+1)(p+3)}{4}<p^2=b$ a contradiction. If $\alpha=1$ we get an easy contradiction, and if $\alpha=2$, we get $c=p(p-1)$ and $a=p(p+1)<4p^2=b$, a contradiction.

In conclusion, $b \in \{4,p\}$, hence $b$ is either a prime or a perfect square.  

Mathematical Reflections 2013, Issue 4 - Problem J276

Problem:
Find all positive integers $m$ and $n$ such that $$10^n-6^m=4n^2$$

Proposed by Tigran Akopyan.

Solution:
It is easy to see that if $n=1$, then $m=1$ and $(1,1)$ is a solution to the given equation. We prove that this is the only solution. Assume that $n>1$ and $n$ odd. Then, $10^n-6^m \equiv -6^m \pmod{8}$ and $4n^2 \equiv 4 \pmod{8}$ and it is clear that $-6^m \equiv 4 \pmod{8}$ if and only if $m=2$. But $10^n>36+4n^2$ for all positive integers $n>1$, therefore there are no solutions when $n$ is odd. Let $n$ be an even number, i.e. $n=2k$ for some $k \in \mathbb{Z}^+$. Hence,
$$(10^k-4k)(10^k+4k)=6^m.$$ Since there are no solutions for $k=1,2$, let us assume that $k>2$. Clearly $m \geq 4$ and simplifying by $2$, we get
$$\begin{equation} (2^{k-2}\cdot5^k - k)(2^{k-2}\cdot5^k + k)=2^{m-4}\cdot3^m.                          (1) \end{equation}$$
If $k$ is odd, then $m=4$. But $2^{k-2}\cdot5^k + k \geq 2\cdot5^3+3 > 3^4$, contradiction. Therefore, $k$ must be even. Assume that $k=2^{\alpha}h$, where $\alpha,h \in \mathbb{Z}^+$ and $h$ is odd. If $\alpha \geq k-2$, then $2^{k-2} | k$, which implies that $2^{k-2} \leq k$. This gives $k=3,4$ and an easy check shows that there are no solutions for this values. If $\alpha < k-2$, then equation $(1)$ becomes $$2^{2\alpha}(2^{k-2-\alpha}\cdot5^k - h)(2^{k-2-\alpha}\cdot5^k + h)=2^{m-4}\cdot3^m,$$ and by unique factorizaton we have $2\alpha=m-4$. Therefore, from the inequality $k>\alpha+2$, we obtain $n>2\alpha+4=m$. But this implies that $10^n=6^m+4n^2<6^n+4n^2$, which is false for all integers $n>1$. Hence, there are no solutions for $n$ even and the statement follows.

Mathematical Reflections 2013, Issue 4 - Problem J274

Problem:
Let $p$ be a prime and let $k$ be a nonnegative integer. Find all positive integer solutions $(x, y, z)$ to the
equation $$x^k(y-z)+y^k(z-x)+z^k(x-y)=p.$$

Proposed by Alessandro Ventullo.

Solution:

It's easy to see that if $k=0,1$, the equation has no solution. Suppose that $k \geq 2$ and put $f(x)=x^k(y-z)+y^k(z-x)+z^k(x-y)$. Since $f(y)=0$, then $(x-y)|f(x)$ and since the expression is cyclic, we obtain that also $(y-z)$ and $(z-x)$ divide $f(x)$, so $$x^k(y-z)+y^k(z-x)+z^k(x-y)=(x-y)(y-z)(z-x)h(x,y,z), \qquad h \in \mathbb{Z}[x,y,z].$$
Put $x-y=a,y-z=b,z-x=c$. Let $p>2$. Then $a,b,c \in \{\pm 1, \pm p\}$ and $a+b+c=0$, but this is impossibile since $a,b,c$ are all odd. Now, let $p=2$. Then $a,b,c \in \{\pm 1, \pm 2\}$. It is clear that $a,b,c$ cannot all be equal (otherwise $a=b=c=0$) and cannot all be distinct (otherwise $a+b+c\neq 0)$. Suppose $a \geq b \geq c$. We have two cases.

(i) $a=b=1, c=-2$. This implies that $x,y,z$ are three consecutive integers, $x>y>z$. If $y=n>1$, where $n \in \mathbb{Z}$, the equation becomes
$$(n+1)^k-2n^k+(n-1)^k=2.$$ If $k=2$, we get an identity, so there are infinitely many positive integer solutions $(n+1,n,n-1)$. If $k>2$, then we have $$(n+1)^k+(n-1)^k > 2n^k+k(k-1)n^{k-2}>2n^k+2,$$ so the equation has no solution.

(ii) $a=2, b=c=-1$, so $x,y,z$ are three consecutive integers, $x>z>y$. If $z=n>1$, the equation becomes
$$-(n+1)^k-(n-1)^k+2n^k=2,$$ which gives $(n+1)^k+(n-1)^k=2(n^k-1)<2n^k$, contradiction.

In conclusion, the equation has never positive integer solutions $(x,y,z)$ if $p>3$ or $p=2, k>2$, and has infinitely many positive integer solution if $p=k=2$, namely $$(n+1,n,n-1), (n,n-1,n+1), (n-1,n+1,n), \qquad n>1.$$  

Mathematical Reflections 2013, Issue 4 - Problem J273

Problem:
Let $a,b,c$ be real numbers greater than or equal to $1$. Prove that
$$\dfrac{a^3+2}{b^2-b+1}+\dfrac{b^3+2}{c^2-c+1}+\dfrac{c^3+2}{a^2-a+1} \geq 9.$$

Proposed by Titu Andreescu.

Solution:

By the AM-GM Inequality, we have
$$\dfrac{a^3+2}{b^2-b+1}+\dfrac{b^3+2}{c^2-c+1}+\dfrac{c^3+2}{a^2-a+1} \geq 3\sqrt[3]{\left(\dfrac{a^3+2}{a^2-a+1}\right)\left( \dfrac{b^3+2}{b^2-b+1}\right)\left(\dfrac{c^3+2}{c^2-c+1}\right)}.$$
Since $x^3+2=(x-1)^3+3(x^2-x+1)$, we have $$\dfrac{x^3+2}{x^2-x+1}=\dfrac{(x-1)^3}{x^2-x+1}+3\geq 3$$ for all $x \geq 1$.
Therefore,
$$\dfrac{a^3+2}{b^2-b+1}+\dfrac{b^3+2}{c^2-c+1}+\dfrac{c^3+2}{a^2-a+1} \geq 9.$$

Mathematical Reflections 2013, Issue 3 - Problem S267

Problem:
Find all primes $p,q,r$ such that $7p^3-q^3=r^6$.

Proposed by Titu Andreescu.

Solution:
Suppose that $r=2$. Then, $7p^3=(q+4)(q^2-4q+16)$. Observe that both $p$ and $q$ are odd primes. Since $$q^2-4q+16=(q+4)(q-8)+48,$$ $\gcd(q+4,q^2-4q+16)|48$. Moreover, both factors are odd numbers, so $\gcd(q+4,q^2-4q+16) \in \{1,3\}$. If $\gcd(q+4,q^2-4q+16)=3$, then $p=3$ by Unique Factorization. Substituting these values into the original equation we get $q=5$. If $\gcd(q+4,q^2-4q+16)=1$, since both factors are greater than $1$, we get $p \neq 7$ and $$\begin{array}{rcl} q+4&=&7 \\ q^2-4q+16&=&p^3 \end{array} \qquad \begin{array}{rcl} q+4&=&p^3 \\ q^2-4q+16&=&7. \end{array}$$ It's easy to see that both systems of equations have no solution. Now, suppose that $r>2$. Then, exactly one between $p$ and $q$ is $2$ and the other is odd. Suppose that $p=2$. Then, $$56=(q+r^2)(q^2-qr^2+r^4).$$ Moreover both factors are greater than $1$, $q+r^2$ is even and $q^2-qr^2+r^4$ is odd, so the only possibility is $$\begin{array}{rcl} q+r^2&=&8 \\ q^2-qr^2+r^4&=&7 \end{array}$$ and clearly the first equation has no solution in odd primes. Now, suppose that $q=2$. Then, $$7p^3=(r^2+2)(r^4-2r^2+4).$$ Since
$$r^4-2r^2+4=(r^2+2)(r^2-4)+12,$$ then $\gcd(r^2+2,r^4-2r^2+4)|12$, but both factors are odd, so $\gcd(r^2+2,r^4-2r^2+4) \in \{1,3\}$. If $\gcd(r^2+2,r^4-2r^2+4)=3$, then $p=3$ by Unique Factorization, but there is no solution for $p=3,q=2$. If $\gcd(r^2+2,r^4-2r^2+4)=1$, since both factors are greater than $1$, we get $p \neq 7$ and $$\begin{array}{rcl} r^2+2&=&7 \\ r^4-2r^2+4&=&p^3 \end{array} \qquad \begin{array}{rcl} r^2+2&=&p^3 \\ r^4-2r^2+4&=&7. \end{array}$$ It's easy to see that both systems of equations have no solution. Therefore, the only primes which satisfy the given equation are $p=3,q=5,r=2$.

Mathematical Reflections 2013, Issue 3 - Problem S265

Problem:
Find all pairs $(m,n)$ of positive integers such that $m^2 + 5n$ and $n^2 + 5m$ are both perfect squares.

Proposed by Titu Andreescu.

Solution:
Suppose that $m=n$. Then $m^2+5m=m(m+5)$ must be a perfect square. Observe that $\gcd(m,m+5)=1$, otherwise $\gcd(m,m+5)=5$, which gives $m=5k$ for some $k \in \mathbb{N}^*$ and $m+5=5(k+1)$, but $m(m+5)=25k(k+1)$ cannot be a perfect square, since $k$ and $k+1$ are coprimes and must be perfect squares. Therefore, $m$ and $m+5$ are both perfect squares, and it's easy to see that this happens if and only if $m=4$.
Suppose without loss of generality that $m>n$. By an easy check, we can see that the only solution for $m \leq 4$ or $n \leq 4$ is $m=n=4$, so suppose that $m>n>4$. We have $m^2<m^2+5n<(m+3)^2$, so we obtain two cases.

(i) $m^2+5n=(m+1)^2$, which gives
$$\begin{equation} 5n=2m+1.                (1) \end{equation}$$
So, $6n>2m$, i.e. $3n>m$. Hence,
$$(n+3)^2<n^2+10n=n^2+4m+2<n^2+5m<n^2+15n<(n+8)^2.$$
We obtain four cases

(a) $n^2+5m=(n+4)^2$, i.e. $5m=8n+16$, and summing up this equation with (1), we get $3(m-n)=17$, contradiction. So, no solution in this case.
(b) $n^2+5m=(n+5)^2$, i.e. $5m=10n+25$, which is $m=2n+5$. From equation (1) we get $n=11$ and $m=27$.
(c) $n^2+5m=(n+6)^2$, i.e. $5m=12n+36$, which is $10m=24n+72$. Multiplying by $5$ equation (1), and summing up, we get $n=77$ and $m=192$.
(d) $n^2+5m=(n+7)^2$, i.e. $5m=14n+49$, and summing up this equation with equation (1), we get $3(m-3n)=50$, contradiction. So, no solution in this case.

(ii) $m^2+5n=(m+2)^2$, which gives
$$\begin{equation} 5n=4m+4.                (2) \end{equation}$$
So, $8n>4m$, i.e. $2n>m$. Hence,
$$(n+2)^2<n^2+5n<n^2+5m<n^2+10n<(n+5)^2.$$
We obtain two cases.

(a) $n^2+5m=(n+3)^2$, i.e. $5m=6n+9$, which is $20m=24n+36$. Multiplying by $5$ equation (2), and summing up, we get $n=56$ and $m=69$.
(b) $n^2+5m=(n+4)^2$, i.e. $5m=8n+16$, and summing up this equation with equation (2) we get $m-3n=20$, but $m-3n<0$, contradiction. So, no solution in this case.

In conclusion, all the pairs $(m,n)$ which satisfy the required conditions are
$$(4,4),(11,27),(27,11),(77,192),(192,77),(56,69),(69,56).$$

Mathematical Reflections 2013, Issue 3 - Problem J269

Problem:
Solve in positive integers the equation $$(x^2-y^2)^2-6\min(x,y)=2013.$$

Proposed by Titu Andreescu.

Solution:
Clearly, $x \neq y$. Suppose without loss of generality that $x<y$. Then, $$2013+6x=(x-y)^2(x+y)^2>(x+y)^2>4x^2,$$ which gives $4x^2-6x-2013<0$. Hence, $0<x<23$. Moreover, $$(x^2-y^2)^2=3(671+2x),$$ therefore $671+2x$ must be divisible by $3$ since the left member is a perfect square. This implies that $x=3k+2$ for some $k \in \mathbb{N}$, so $$(x^2-y^2)^2=9(225+2k),$$ and $225+2k$ must be a perfect square. If $k=0$ it's obvious. The least positive integer such that $225+2k$ is a square is $k=32$, but for this value we get $x=98>23$. Therefore $k=0$, $x=2$ and $(x^2-y^2)^2=2025=45^2$ which gives $y^2-x^2=45$, i.e. $y=7$. By symmetry, $x=7,y=2$ is another solution of the equation. So, all the positive integer solutions of the given equation are $(2,7),(7,2)$.

Mathematical Reflections 2013, Issue 3 - Problem J265

Problem:
Let $a,b,c$ be real numbers such that $$5(a+b+c)-2(ab+bc+ca)=9.$$
Prove that any two of the equalities $$|3a-4b|=|5c-6|, \qquad |3b-4c|=|5a-6|, \qquad |3c-4a|=|5b-6|$$
imply the third.

Proposed by Titu Andreescu.

Solution:
By symmetry, we can suppose that the first two equalities are given. Since $|x|=|y|$ if and only if $x^2=y^2$ for real numbers $x,y$, then
$$9a^2-24ab+16b^2=25c^2-60c+36, \qquad 9b^2-24bc+16c^2=25a^2-60a+36.$$ Summing up the two equalities and reordering, we have
$$25b^2-24(ab+bc)+36=16a^2-60(a+c)+9c^2+108.$$
Since $60(a+c)-24(ab+bc)=108-60b+24ca$, we get $$25b^2+(108-60b+24ca)+36=16a^2+9c^2+108,$$ and reordering we get $$(5b-6)^2=(3c-4a)^2,$$ i.e. $|5b-6|=|3c-4a|$, which is the third equality.

Mathematical Reflections 2013, Issue 2 - Problem U259

Problem:
Compute $$\lim_{n \to \infty} \dfrac{\left(1+\frac{1}{n(n+a)}\right)^{n^3}}{\left(1+\frac{1}{n+b}\right)^{n^2}}.$$

Proposed by Arkady Alt.

Solution:
We have $$\lim_{n \to \infty} \dfrac{\left(1+\frac{1}{n(n+a)}\right)^{n^3}}{\left(1+\frac{1}{n+b}\right)^{n^2}}= \dfrac{\lim_{n \to \infty} \left(1+\frac{1}{n(n+a)}\right)^{n^3}}{\lim_{n \to \infty} \left(1+\frac{1}{n+b}\right)^{n^2}}.$$
Since $\left(1+\frac{1}{n(n+a)}\right)^{n^3}=e^{n^3 \log \left(1+\frac{1}{n(n+a)}\right)} \sim e^n$ as $n \to \infty$ and $\left(1+\frac{1}{n+b}\right)^{n^2}=e^{n^2 \log \left(1+\frac{1}{n+b}\right)} \sim e^n$ as $n \to \infty$, we have
$$\lim_{n \to \infty} \dfrac{\left(1+\frac{1}{n(n+a)}\right)^{n^3}}{\left(1+\frac{1}{n+b}\right)^{n^2}} \sim \dfrac{e^n}{e^n}=1.$$





Note: the official problem was modified lately.

Mathematical Reflections 2013, Issue 2 - Problem S259

Problem:
Let $a,b,c,d,e$ be integers such that $$a(b+c)+b(c+d)+c(d+e)+d(e+a)+e(a+b)=0.$$ Prove that $a+b+c+d+e$ divides
$a^5 + b^5 + c^5 + d^5 + e^5 - 5abcde$.

Proposed by Titu Andreescu.

Solution:
Suppose that $a,b,c,d,e$ are the five roots $\alpha_1,\alpha_2,\alpha_3,\alpha_4,\alpha_5$ of a fifth degree polynomial $P(x)$.
Let $$\sigma_k=\sum_{i=1}^5 \alpha^k_i, \qquad s_k=\sum_{1 \leq j_1 < j_2 < \ldots < j_k \leq 5} \alpha_{j_1}\alpha_{j_2}\cdots\alpha_{j_k}.$$ With this notation, we know that $s_2=0$ and we want to prove that $\sigma_1$ divides $\sigma_5-5s_5$. We have $$P(x)=x^5-s_1x^4+s_2x^3-s_3x^2+s_4x-s_5,$$ so
$$\begin{array}{lcl} P(\alpha_1)&=&\alpha^5_1-s_1\alpha^4_1+s_2\alpha^3_1-s_3\alpha^2_1+s_4\alpha_1-s_5=0 \\ P(\alpha_2)&=&\alpha^5_2-s_1\alpha^4_2+s_2\alpha^3_2-s_3\alpha^2_2+s_4\alpha_2-s_5=0 \\ P(\alpha_3)&=&\alpha^5_3-s_1\alpha^4_3+s_2\alpha^3_3-s_3\alpha^2_3+s_4\alpha_3-s_5=0 \\ P(\alpha_4)&=&\alpha^5_4-s_1\alpha^4_4+s_2\alpha^3_4-s_3\alpha^2_4+s_4\alpha_4-s_5=0 \\ P(\alpha_5)&=&\alpha^5_5-s_1\alpha^4_5+s_2\alpha^3_5-s_3\alpha^2_5+s_4\alpha_5-s_5=0. \end{array}$$
Summing up the columns, we get $$\sigma_5-s_1\sigma_4+s_2\sigma_3-s_3\sigma_2+s_4\sigma_1-5s_5=0.$$ Since $s_1=\sigma_1, s_2=0$ and $\sigma_2=\sigma^2_1-2s_2=\sigma^2_1$ we obtatin $$\sigma_5-5s_5=\sigma_1(\sigma_4+s_3\sigma_1-s_4),$$ hence $\sigma_1|(\sigma_5-5s_5)$.

Mathematical Reflections 2013, Issue 2 - Problem J263

Problem:
The $n$-th pentagonal number is given by the formula $p_n = \dfrac{n(3n-1)}{2}$. Prove that there are infinitely many pentagonal numbers that can be written as a sum of two perfect squares of positive integers.

Proposed by Jose Hernandez Santiago.

Solution:
We have $$p_n=n^2+\dfrac{(n-1)n}{2}=n^2+T_{n-1},$$ where $T_{n-1}$ is the $(n-1)$-th triangular number. So, it is sufficient to prove that there are infinitely many triangular numbers which are perfect squares. Suppose that $$\dfrac{n(n+1)}{2}=m^2, \qquad m \in \mathbb{N}.$$ This equation is equivalent to $$(2n+1)^2-8m^2=1$$ and putting $x=2n+1, y=2m$ we have the Pell's equation $x^2-2y^2=1$, which has infinitely many solutions $x=P_{2k}+P_{2k-1}, y=P_{2k}$, where $$P_k=\dfrac{(1+\sqrt{2})^k-(1-\sqrt{2})^k}{2\sqrt{2}}$$ is the $k$-th Pell number. Therefore, there are infinitely many triangular numbers $m=P_{2k}/2$ which are perfect squares and we are done.

Mathematical Reflections 2013, Issue 2 - Problem J262

Problem:
Find all positive integers $m, n$ such that $${m+1 \choose n}={n \choose m+1}.$$

Proposed by Roberto Bosch Cabrera.

Solution:
If $m+1 \geq n$, we have ${m+1 \choose n}>0$. If $n<m+1$, then ${n \choose m+1}=0$, so $n \geq m+1$, i.e. $n=m+1$.
If $m+1 < n$, then ${m+1 \choose n}=0$, so it must be ${n \choose m+1}=0$, which gives $n < m+1$, a contradiction. Hence all positive integers which satisfies the given equation are the consecutive positive integers $m,m+1$.

Mathematical Reflections 2013, Issue 2 - Problem J260

Problem:
Solve in integers the equation $$x^4-y^3=111.$$

Proposed by Jos e Hern andez Santiago.

Solution:
We claim that the equation has no integer solution. As a matter of fact, suppose that the given equation has an integer solution. Let us consider the equation modulo $13$. Since $x^2 \equiv 0,\pm 1, \pm 3, \pm 4 \pmod{13}$, then $x^4 \equiv 0,1,3,-4 \pmod{13}$. Moreover $y^3 \equiv 0, \pm 1, \pm 5 \pmod{13}$. Hence, $$x^4-y^3 \equiv 0, \pm 1, \pm 2, \pm 3,\pm 4, \pm5, 6 \pmod{13},$$ but $111 \equiv -6 \pmod{13}$, a contradiction.

Mathematical Reflections 2013, Issue 2 - Problem J259

Problem:
Among all triples of real numbers $(x,y,z)$ which lie on a unit sphere $x^2+y^2+z^2=1$ find a triple which maximizes
$\min (|x-y|, |y-z|, |z-x|)$.

Proposed by Arkady Alt.

Solution:
Suppose without loss of generality that $\min (|x-y|, |y-z|, |z-x|)=|x-y|$. Let $f(x,y,z)=|x-y|$ and $g(x,y,z)=x^2+y^2+z^2-1$. Consider the Lagrangian function $$\begin{array}{lll} L(x,y,z,\lambda)&=&f(x,y,z)-\lambda g(x,y,z)\\&=&|x-y|- \lambda(x^2+y^2+z^2-1), \end{array}$$ with $\lambda \in \mathbb{R}$. By Lagrange Multipliers Theorem, a maximum or a minimum for $f(x,y,z)$ subject to the constraint $g(x,y,z)=0$ must be a stationary point of $L$. Therefore a maximum or a minimum satisfies
$$\begin{array}{rcl} \dfrac{\partial L}{\partial x} & = & 0 \\ \dfrac{\partial L}{\partial y} & = & 0 \\  \dfrac{\partial L}{\partial z} & = & 0 \\ \dfrac{\partial L}{\partial \lambda} & = & 0, \end{array}$$ i.e. $$\begin{array}{rcl} \pm 1 - 2\lambda x & = & 0 \\ \mp 1 - 2\lambda y & = & 0 \\ -2\lambda z & = & 0 \\ x^2+y^2+z^2-1 & = & 0. \end{array}$$ From the third equation we get $z=0$ since $\lambda=0$ would give a contradiction in the first two equations. From the first two equations we have $x=\pm 1/2\lambda, y=\mp 1/2\lambda$ and substituting these values into the fourth equation we get $\lambda=\pm \sqrt{2}/2$, so $x=\pm \sqrt{2}/2, y=\mp \sqrt{2}/2, z=0$ are two stationary points which satisfies the conditions. It's easy to see that these two triples maximize $f(x,y,z)$ since a minimum for $f$ subject to the constraint $g$ is $0$ (take $x=y=0, z=1)$. By symmetry we find that all triples which maximize $\min (|x-y|, |y-z|, |z-x|)$ are $$(\pm \sqrt{2}/2, \mp \sqrt{2}/2, 0), (\pm \sqrt{2}/2, 0, \mp \sqrt{2}/2), (0, \pm \sqrt{2}/2, \mp \sqrt{2}/2),$$ and $\max (\min (|x-y|, |y-z|, |z-x|))=\sqrt{2}$.

Mathematical Reflections 2013, Issue 1 - Problem U257

Problem:
a) Let $p$ and $q$ be distinct primes and let $G$ be a non-commutative group with $pq$ elements. Prove that the center of $G$ is trivial.

b) Let $p, q, r$ be pairwise distinct primes and let $G$ be a non-commutative group with
$pqr$ elements. Prove that the number of elements of the center of $G$ is either $1$ or a prime number.

Proposed by Mihai Piticari.

Solution:
We use the following

Lemma
If $G$ is a non-abelian group, then $G/Z(G)$ is not a cyclic group.

Proof
Homework!

a) Since $Z(G)$ is a subgroup of $G$, $|Z(G)|$ divides $|G|$, which means  $|Z(G)| \in \{1,p,q,pq\}$. Therefore, $|G/Z(G)| \in \{pq,q,p,1\}$ and since $G/Z(G)$ cannot be cyclic, it must be $|G/Z(G)|=pq$, i.e. $Z(G)$ is trivial.

b) As above, $|G/Z(G)|$ is a divisor of $G$, so, $|G/Z(G)| \in \{1,p,q,r,pq,qr,rp,pqr\}$. Since $G/Z(G)$ cannot be cyclic, the only possibilities are $|G/Z(G)| \in \{pq,qr,rp,pqr\}$, which means that $|Z(G)|$ is either $1$ or a prime number.

Mathematical Reflections 2013, Issue 1 - Problem U255

Problem:
Let $S_n$ be the group of permutations of $\{1,2,\ldots,n\}$. If $d > 1$ is an integer, let $H_d$ be the set of those $\sigma \in S_n$ for which there are $k \geq 1$ and $\sigma_1,\ldots,\sigma_k \in S_n$ with $\sigma = \sigma_1^d \cdots \sigma_k^d$. Find $H_2$ and $H_3$.

Proposed by Mihai Piticari and Sorin Radulescu.

Solution:
If $n=1,2$, clearly $H_2=H_3=S_n$. Let $n \geq 3$. We first prove that $H_d$ is a group for every integer $d > 1$. Indeed, if $\sigma, \tau \in H_d$, clearly $\sigma \tau \in H_d$. The associative property follows from the associativite property of the product of permutations. Moreover, $\textrm{id} \in S_n$ and $\textrm{id} = \textrm{id}^d$, so $\textrm{id} \in H_d$ for every $d > 1$. Finally if $\sigma_1,\ldots,\sigma_k \in H_d$ and $\sigma=\sigma_1^d \cdots \sigma_k^d$ for some $k \geq 1$, we have that $\sigma_1^{-1},\ldots,\sigma_k^{-1} \in S_n$ and $\tau=(\sigma_k^{-1})^d\cdots(\sigma_1^{-1})^d$ is the inverse of $\sigma$. Now, let $d=2$. It is clear that $H_2$ is a subgroup of $A_n$ since every permutation in $H_2$ is a product of even permutations and so is an even permutation. For every $a_1,a_2,a_3 \in \{1,\ldots,n\}$, we have also that $(a_1, a_2, a_3)=(a_1, a_3, a_2)^2$ so every $3$-cycle belongs to $H_2$. Since $A_n$ is generated by its $3$-cycles, it follows that $A_n$ is a subgroup of $H_2$, from which $H_2=A_n$. Now, let $d=3$. Obviously, $H_3$ is a subgroup of $S_n$. Moreover, for every $a_1,a_2 \in \{1,\ldots,n\}$ we have $(a_1,a_2)=(a_1,a_2)^3$, so every $2$-cycle belongs to $H_3$. Since $S_n$ is generated by its $2$-cycles, it follows that $S_n$ is a subgroup of $H_3$, which gives $H_3=S_n$.

Mathematical Reflections 2013, Issue 1 - Problem U253

Problem:
Evaluate $$\sum_{n>1} \dfrac{3n^2+1}{(n^3-n)^3}.$$

Proposed by Titu Andreescu.

Solution:
We observe that $$\dfrac{3n^2+1}{(n^3-n)^3}=\dfrac{1}{2}\left(\dfrac{1}{n^3(n-1)^3}-\dfrac{1}{(n+1)^3n^3}\right).$$
Hence, $$\begin{array}{lcl}\displaystyle \sum_{n=2}^\infty \dfrac{3n^2+1}{(n^3-n)^3}&=& \displaystyle \dfrac{1}{2} \sum_{n=2}^\infty \left(\dfrac{1}{n^3(n-1)^3}-\dfrac{1}{(n+1)^3n^3}\right)\\ &=& \displaystyle \dfrac{1}{2} \lim_{n \to \infty} \left(\dfrac{1}{8}-\dfrac{1}{(n+1)^3n^3}\right)\\&=& \dfrac{1}{16}. \end{array}$$

Mathematical Reflections 2013, Issue 1 - Problem S255

Problem:
Solve in real numbers the equation $$2^x+2^{-x}+3^x+3^{-x}+\left(\dfrac{2}{3}\right)^x+\left(\dfrac{2}{3}\right)^{-x}=9x^4-7x^2+6.$$

Proposed by Mihaly Bencze.

Solution:
We rewrite the equation in the form $$\left(2^{x/2}-2^{-x/2}\right)^2+\left(3^{x/2}-3^{-x/2}\right)^2+\left((2/3)^{x/2}-(2/3)^{-x/2}\right)^2=9x^4-7x^2.$$ Let $$f(x)=\left(2^{x/2}-2^{-x/2}\right)^2+\left(3^{x/2}-3^{-x/2}\right)^2+\left((2/3)^{x/2}-(2/3)^{-x/2}\right)^2$$ and $$g(x)=9x^4-7x^2.$$ Since $f(x)$ and $g(x)$ are even functions, it suffices to find the solutions when $x \geq 0$. It is easy to see that $x=0$ and $x=1$ are solutions of the given equation. Moreover, $f(x)$ is increasing for $x \geq 0$ since it is a sum of increasing functions, and $g(x)$ is increasing if $x \geq \sqrt{7/18}$ and decreasing if $0 \leq x \leq \sqrt{7/18}$, as can be seen from $g'(x)$. Since $f(x)$ and $g(x)$ are both injective functions if $x \geq 1$, then $h(x)=g(x)-f(x)$ is an injective function if $x \geq 1$, so $h(1)=0$ and $h(x) \neq 0$ for all $x>1$. Therefore the equation has no other solutions for $x \geq 0$, which means that the only solutions of the equation are $x=-1,0,1$.

Mathematical Reflections 2013, Issue 1 - Problem S253

Problem:
Solve in positive real numbers the system of equations:
$$\begin{array}{rcl}(2x)^{2013}+(2y)^{2013}+(2z)^{2013} & = & 3 \\ xy+yz+zx+2xyz & = & 1 \end{array}$$

Proposed by Roberto Bosch Cabrera.

Solution:
Let $2x=a, 2y=b, 2z=c$. Then, the given system of equations is equivalent to
$$\begin{array}{rcl}a^{2013}+b^{2013}+c^{2013} & = & 3 \\ ab+bc+ca+abc & = & 4. \end{array}$$ Using the AM-GM Inequality in the first equation, we get $abc \leq 1$ and from the second equation we get $ab+bc+ca \geq 3$. Suppose without loss of generality that $a \leq b \leq c$. By Chebyshev's Inequality, we have
$$\begin{equation} 1 \leq \dfrac{ab+bc+ca}{3} \leq \dfrac{(a+b+c)^2}{9},                           (1) \end{equation}$$
therefore $a+b+c \geq 3$. Using Chebyshev's Inequality once again, we have $$\dfrac{a^{n-1}+b^{n-1}+c^{n-1}}{3} \leq \left(\dfrac{a+b+c}{3}\right)\left(\dfrac{a^{n-1}+b^{n-1}+c^{n-1}}{3}\right) \leq \dfrac{a^n+b^n+c^n}{3}$$ for all $n \in \mathbb{N}^*$. Since $\dfrac{a^{2013}+b^{2013}+c^{2013}}{3}=1$, we get $$a^n+b^n+c^n \leq 3$$ for all positive integers $n < 2013$, and in particular $a+b+c \leq 3$. This gives $a+b+c=3$, and from (1) we get $ab+bc+ca=3$, which implies $abc=1$. Therefore, $a+b+c=3\sqrt[3]{abc}$, so $a=b=c=1$, i.e. $x=y=z=1/2$ is the only solution to the given system of equations.

Mathematical Reflections 2013, Issue 1 - Problem J256

Problem:
Evaluate $$1^2 2!+2^2 3! + \ldots + n^2(n+1)!.$$

Proposed by Titu Andreescu.

Solution:
We have $$\begin{array}{lcl} k^2(k+1)! & = & [(k^2+k-2)-(k-2)](k+1)!\\ & = & [(k-1)(k+2)-(k-2)](k+1)!\\ & = &(k-1)(k+2)!-(k-2)(k+1)! \end{array}$$ for all $k \in \mathbb{N}$. Therefore, $$1^2 2!+2^2 3! + \ldots + n^2(n+1)!=(n-1)(n+2)!+2.$$

Mathematical Reflections 2013, Issue 1 - Problem J254

Problem:
Solve the following equation $F_{a_1} + F_{a_2} + \ldots + F_{a_k} = F_{a_1+a_2+\ldots+a_k}$ , where $F_i$ is the $i$-th
Fibonacci number.

Proposed by Roberto Bosch Cabrera.

Solution:
If $k=1$ the equation has infinitely many solutions. Let $k \geq 2$ and suppose that $1 \leq a_1 \leq a_2 \leq \ldots \leq a_k$.
We have five cases.

(i) $k=2$. Suppose that $a_1+a_2 \leq 4$. It's easy to see that $(1,2)$ and $(1,3)$ are solutions. Now, suppose that $a_1+a_2 \geq 5$. If $a_1=1$, then $a_2 \geq 4$ and $F_{a_1+a_2-2}>F_{a_1}$. If $a_1>1$, then $a_2>2$, so $F_{a_1+a_2-1}>F_{a_2}$ and $$F_{a_1+a_2}=F_{a_1+a_2-1}+F_{a_1+a_2-2} > F_{a_2}+F_{a_1},$$ i.e. the equation has no solution if $a_1+a_2 \geq 5$.

(ii) $k=3$. There are no solutions when $a_3=1$. If $a_3=2$, we have the solution $(1,1,2)$. It's easy to see that there are no more solutions such that $a_2=1$ and $a_3>2$. Suppose that $a_3 \geq 2$ and $a_2 \geq 2$. If $a_3=2$, we get $F_{a_1+a_2+a_3-1}>F_{a_3}$ and if $a_3>2$, we get $F_{a_1+a_2+a_3-2}>F_{a_1+a_2}$, so $$F_{a_1+a_2+a_3}=F_{a_1+a_2+a_3-1}+F_{a_1+a_2+a_3-2}>F_{a_3}+F_{a_1+a_2} \geq F_{a_3}+F_{a_2}+F_{a_1},$$ i.e. no solution if $a_3>2$.

(iii) $k=4$. A simple check shows that there are no solutions if $a_4=1$. If $a_4 \geq 2$ and $a_3=1$ there are no solutions, and if $a_3 \geq 2$, we can argue similarly to the previous case and conclude that $$F_{a_1+a_2+a_3+a_4} > F_{a_4}+F_{a_1+a_2+a_3} \geq F_{a_4}+F_{a_3}+F_{a_2}+F_{a_1},$$ i.e. no solution in this case.

(iv) $k=5$. If $a_5=1$, we immediately see that $(1,1,1,1,1)$ is a solution. If $a_5 \geq 2$ and $a_4=1$, there are no solutions and if $a_4 \geq 2$ we get $$F_{a_1+a_2+a_3+a_4+a_5}>F_{a_5}+F_{a_1+a_2+a_3+a_4}\geq F_{a_5}+F_{a_4}+F_{a_3}+F_{a_2}+F_{a_1}.$$

(v) $k>5$. Clearly, $a_1+a_2+\ldots+a_k \geq k$. If $a_k=1$, there are no solutions since $a_1+a_2+\ldots+a_k=k$ and $F_k>k$. If $a_k \geq 2$ and $a_{k-1}=1$, then $a_1+a_2+\ldots+a_{k-1}=k-1$ and
$$\begin{array}{lcl} F_{a_1+a_2+\ldots+a_k}&=&F_{a_1+a_2+\ldots+a_k-1}+F_{a_1+a_2+\ldots+a_k-2}\\&>&F_{a_k}+F_{a_1+a_2+\ldots+a_{k-1}} =  F_{a_k}+F_{k-1} \\ & \geq & F_{a_k}+k-1 \\ & = & F_{a_k}+F_{a_{k-1}}+\ldots+F_{a_1}. \end{array}$$
If $a_k \geq 2$ and $a_{k-1} \geq 2$, we have
$$\begin{array}{lcl} F_{a_1+a_2+\ldots+a_k} & > & F_{a_k}+ F_{a_1+a_2+\ldots+a_{k-1}}\\ & > & F_{a_k}+F_{a_{k-1}} + F_{a_1+a_2+\ldots+a_{k-2}} \\ & \geq & F_{a_k}+F_{a_{k-1}}+F_{a_{k-2}}+F_{a_1+a_2+\ldots+a_{k-3}} \\ & \vdots & \vdots \\ & \geq & F_{a_k}+F_{a_{k-1}}+\ldots+F_{a_2}+F_{a_1}-1 , \end{array}$$ which gives $F_{a_1+a_2+\ldots+a_k} > F_{a_k}+F_{a_{k-1}}+\ldots+F_{a_2}+F_{a_1}$, so there are no solutions in this case.

Mathematical Reflections 2013, Issue 1 - Problem J253

Problem:
Prove that if $a,b,c>0$ satisfy $abc=1$, then $$\dfrac{1}{ab+a+2}+\dfrac{1}{bc+b+2}+\dfrac{1}{ca+c+2} \leq \dfrac{3}{4}.$$

Proposed by Marcel Chirita.

Solution:
By the AM-HM Inequality, we have $$\dfrac{1}{ab+1+a+1} \leq \dfrac{1}{4}\left(\dfrac{1}{ab+1}+\dfrac{1}{a+1}\right)=\dfrac{1}{4}\left(\dfrac{c}{c+1}+\dfrac{1}{a+1}\right)$$ $$\dfrac{1}{bc+1+b+1} \leq \dfrac{1}{4}\left(\dfrac{1}{bc+1}+\dfrac{1}{b+1}\right)=\dfrac{1}{4}\left(\dfrac{a}{a+1}+\dfrac{1}{b+1}\right)$$ $$\dfrac{1}{ca+1+c+1} \leq \dfrac{1}{4}\left(\dfrac{1}{ca+1}+\dfrac{1}{c+1}\right)=\dfrac{1}{4}\left(\dfrac{b}{b+1}+\dfrac{1}{c+1}\right).$$
Summing up the three inequalities, we obtain $$\dfrac{1}{ab+a+2}+\dfrac{1}{bc+b+2}+\dfrac{1}{ca+c+2} \leq \dfrac{1}{4} \left( \dfrac{a+1}{a+1}+\dfrac{b+1}{b+1}+\dfrac{c+1}{c+1}\right)=\dfrac{3}{4}.$$  

Mathematical Reflections 2012, Issue 6 - Problem O249

Problem:
Find all triples $(x,y,z)$ of positive integers such that $$\dfrac{x}{y}+\dfrac{y}{z+1}+\dfrac{z}{x}=\dfrac{5}{2}.$$

Proposed by Titu Andreescu.

Solution:
By the AM-GM Inequality it must be $$3\sqrt[3]{\dfrac{z}{z+1}} \leq \dfrac{x}{y}+\dfrac{y}{z+1}+\dfrac{z}{x}=\dfrac{5}{2},$$ i.e. $\dfrac{z}{z+1} \leq \dfrac{125}{216},$ which gives $z=1$.
Using the AM-GM Inequality once again we obtain
$$2\sqrt{\dfrac{x}{2}} \leq \dfrac{x}{y}+\dfrac{y}{2}+\dfrac{1}{x}=\dfrac{5}{2},$$
i.e. $\dfrac{x}{2} \leq \dfrac{25}{16}$, which gives $x=1,2,3$.

(i) If $x=1$, then $\dfrac{1}{y}+\dfrac{y}{2}+1=\dfrac{5}{2}$ and solving for $y$ we get $y=1,2$.
(ii) If $x=2$, then $\dfrac{2}{y}+\dfrac{y}{2}+\dfrac{1}{2}=\dfrac{5}{2}$ and solving for $y$ we get $y=2$.
(iii) If $x=3$, then $\dfrac{3}{y}+\dfrac{y}{2}+\dfrac{1}{3}=\dfrac{5}{2}$, which gives no solution.

Therefore, all the triples of positive integers which satisfy the given equation are $$(1,1,1),(1,2,1),(2,2,1).$$

Mathematical Reflections 2012, Issue 6 - Problem O247

Problem:
Solve in positive integers the equation $$xy+yz+zx-5\sqrt{x^2+y^2+z^2}=1.$$

Proposed by Titu Andreescu.

Solution:
Clearly, $xy+yz+zx \geq 3$. Rewriting the equation in the form $$(xy+yz+zx-1)^2=25(x^2+y^2+z^2),$$ we put $xy+yz+zx=5t+1, t > 0$, so that $x^2+y^2+z^2=t^2$. Then $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=t^2+2(5t+1)=(t+5)^2-23,$$ which gives $$(t+5-x-y-z)(t+5+x+y+z)=23.$$
Since $x,y,z,t$ are positive integers, it must be
$$\left\{\begin{array}{rcl} t+5-x-y-z & = & 1 \\ t+5+x+y+z & = & 23. \end{array} \right.$$
Summing up the two equations, we get $t=7$. So, $x+y+z=11$ and $xy+yz+zx=36$. Suppose without loss of generality that $x \leq y \leq z$. Clearly, $x\leq3$, so we have three cases.

(i) $x=1$. Therefore, $y+z=10, yz=26$, so there are no solution.
(ii) $x=2$. Therefore, $y+z=9, yz=18$, so $y=3, z=6$.
(iii) $x=3$. Therefore, $y+z=8, yz=12$, so there are no solution.

In conclusion, all the positive integer solutions are $$(2,3,6), (2,6,3), (3,2,6), (3,6,2), (6,2,3), (6,3,2).$$

Mathematical Reflections 2012, Issue 6 - Problem U249

Problem:
Let $(a_n)_{n \geq 1}$ be a decreasing sequence of positive numbers. Let $$s_n=a_1+a_2+\ldots+a_n,$$ and $$b_n=\dfrac{1}{a_{n+1}}-\dfrac{1}{a_n},$$ for all $n \geq 1$. Prove that if $(s_n)_{n \geq 1}$ is convergent, then $(b_n)_{n \geq 1}$ is unbounded.

Proposed by Bogdan Enescu.

Solution:
Since $(s_n)_{n \geq 1}$ is convergent, then $\lim_{n \to \infty} a_n=0$. Suppose that $(b_n)_{n \geq 1}$ is bounded. Then there exists $M \in \mathbb{R}^+$ such that $b_n \leq M$ for all $n \geq 1$. By the Problem 2.5.12 at page 97 of Radulescu, Andreescu - Problems In Real Analysis, the sequence $(a_n)_{n \geq 1}$ converges to zero if and only if the series $$\sum_{n=1}^\infty \left(1-a_{n+1}/a_n\right)=\sum_{n=1}^\infty a_{n+1}b_n$$ diverges. But
$$\sum_{n=1}^\infty a_{n+1}b_n \leq \sum_{n=1}^\infty a_nb_n \leq M\sum_{n=1}^\infty a_n < \infty,$$ contradiction.

Mathematical Reflections 2012, Issue 6 - Problem U247

Problem:
Let $a$ be a real number greater than $1$. Evaluate
$$\dfrac{1}{a^2-a+1}-\dfrac{2a}{a^4-a^2+1}+\dfrac{4a^3}{a^8-a^4+1}-\dfrac{8a^7}{a^{16}-a^{8}+1}+\ldots.$$

Proposed by Titu Andreescu.

Solution:
Let $$S_n=\sum_{k=1}^n (-1)^{k-1}\dfrac{2^{k-1}a^{2^{k-1}-1}}{a^{2^k}-a^{2^{k-1}}+1}.$$ Since
$$\begin{array}{rcr} \dfrac{1}{a^2-a+1}-\dfrac{1}{a^2+a+1}&=&\dfrac{2a}{a^4+a^2+1} \\  -\dfrac{2a}{a^4-a^2+1}+\dfrac{2a}{a^4+a^2+1}&=&-\dfrac{4a^3}{a^8+a^4+1} \\ \vdots & \vdots & \vdots \\ (-1)^{n-1} \dfrac{2^{n-1}a^{2^{n-1}-1}}{a^{2^n}-a^{2^{n-1}}+1}+(-1)^n \dfrac{2^{n-1}a^{2^{n-1}-1}}{a^{2^n}+a^{2^{n-1}}+1} & = & (-1)^{n+1} \dfrac{2^na^{2^n-1}}{a^{2^{n+1}}+a^{2^n}+1}. \end{array}$$
summing up the two columns, we get $$S_n-\dfrac{1}{a^2+a+1}=(-1)^{n+1} \dfrac{2^na^{2^n-1}}{a^{2^{n+1}}+a^{2^n}+1}.$$
Now, $$0 \leq \left|(-1)^{n+1}\dfrac{2^na^{2^n-1}}{a^{2^{n+1}}+a^{2^n}+1}\right| \leq \dfrac{2^na^{2^n}}{a^{2^{n+1}}}=\dfrac{2^n}{a^{2^n}}$$ and  $\dfrac{2^n}{a^{2^n}} \to 0$ as $n \to \infty$. So,
$$\sum_{k=1}^\infty (-1)^{k-1}\dfrac{2^{k-1}a^{2^{k-1}-1}}{a^{2^k}-a^{2^{k-1}}+1}=\lim_{n \to \infty} S_n = \dfrac{1}{a^2+a+1}.$$

Mathematical Reflections 2012, Issue 6 - Problem S251

Problem:
Find all triples $(x,y,z)$ of positive real numbers for which there is a positive real number $t$ such that the following inequalities hold simultaneously:
$$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+t\leq4, \qquad x^2+y^2+z^2+\dfrac{2}{t}\leq5.$$

Proposed by Titu Andreescu.

Solution:
From the two inequalities, it's easy to see that if there exists such a positive real number $t$, then $t$ satisfies
$$\begin{equation}\label{first} \dfrac{2}{5}<t<4.                                                              (1) \end{equation}$$
From the first inequality, using the AM-HM inequality we obtain $$\dfrac{9}{4-t} \leq \dfrac{9}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \leq x+y+z.$$
Now, suppose without loss of generality that $x \geq y \geq z$. Then, $x^2 \geq y^2 \geq z^2$ and $1/x \leq 1/y \leq 1/z$.
By Chebyshev's Inequality, we have $$x+y+z \leq \dfrac{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)(x^2+y^2+z^2)}{3} \leq \dfrac{(5-2/t)(4-t)}{3}.$$ Hence, $$\dfrac{9}{4-t} \leq \dfrac{(5-2/t)(4-t)}{3}$$ and after simple calculations we get
$$5t^3-42t^2+69t-32 \geq 0 \iff (t-1)^2(5t-32) \geq 0,$$ which gives
$$\begin{equation}\label{second} t=1, \qquad t \geq \dfrac{32}{5}.                                       (2) \end{equation}$$
From $(1)$ and $(2)$, we get $t=1$, from which $x+y+z=1/x+1/y+1/z=3$. But this implies that the arithmetic mean and harmonic mean are equal, so $x=y=z=1$ and the only triple is $(1,1,1)$.

Mathematical Reflections 2012, Issue 6 - Problem S249

Problem:
Find the minimum of $2^x-4^x+6^x-8^x-9^x+12^x$ where $x$ is a positive real number.

Proposed by Titu Andreescu.

Solution:
We have $$\begin{array}{rcl}2^x-4^x+6^x-8^x-9^x+12^x&=&(4^x-3^x)(3^x-2^x)-[(4^x-3^x)+(3^x-2^x)]\\&=&(4^x-3^x-1)(3^x-2^x-1)-1. \end{array}$$
Let $f_n(x)=(n+1)^x-n^x-1$, where $n$ is a positive integer. $f_n(x)$ is increasing since
$$f'_n(x)=(n+1)^x\log(n+1)-n^x\log n > 0 \iff \left(1+\dfrac{1}{n}\right)^x>\dfrac{\log n}{\log (n+1)}$$ for all $x \in \mathbb{R}^+$. Moreover $f_n(1)=0$, so $f_n(x) \geq 0$ if $x \in [1,\infty)$ and $f_n(x) \leq 0$ if $x \in (0,1]$, for all $n \in \mathbb{N}^*$. Therefore, $$f_3(x)f_2(x)-1 \geq -1 \qquad \forall x \in \mathbb{R}^+$$ and the equality occurs if and only if $x=1$.

Mathematical Reflections 2012, Issue 6 - Problem S248

Problem:
Let $\mathcal{C}(O,R)$ be a circle and let $P$ be a point in its plane. Consider a pair of diametrically
opposite points $A$ and $B$ lying on $\mathcal{C}$. Prove that while points $A$ and $B$ vary on the
circumference of $\mathcal{C}$, the circumcircles of triangles $ABP$ pass through another fixed point.

Proposed by Ivan Borsenco.

Solution:
Let us consider another pair of diametrically opposite points $A'$ and $B'$ on $\mathcal{C}$. Then, $O$ is the midpoint of $AB$ and $A'B'$, so in the triangles $ABP$ and $A'B'P$ the points $O$ and $P$ are fixed, and this implies that the line $PO$ is fixed. Therefore, the line $PO$ intersects the circumcircle of the triangle $A'B'P$ in another point $Q$, which is fixed. By the arbitrarity of the pair of diametrically opposite points $A',B'$, we obtain that all the circumcircles of triangles $ABP$ pass through the point $Q$.

Mathematical Reflections 2012, Issue 6 - Problem S247

Problem:
Prove that for any positive integers $m$ and $n$, the number $8m^6 + 27m^3n^3 + 27n^6$ is
composite.

Proposed by Titu Andreescu.

Solution:
For the homogeneity of the expression, we set $t=m/n$. Then,
$$\begin{array}{rcl} 8t^6+27t^3+27 &= & 8t^6-27t^3+27+54t^3\\ &=& (2t^2)^3+(-3t)^3+3^3-(2t^2)(-3t)(3)\\&=&(2t^2-3t+3)(4t^4+6t^3+3t^2+9t+9). \end{array}$$ The second factor is clearly greater than $1$ and $2t^2-3t+3=2(t-1)^2+t+1>1$, so multiplying both sides of the equation by $n^6$ we obtain the conclusion.

Mathematical Reflections 2012, Issue 6 - Problem J251

Problem:
Let $a,b,c$ be positive real numbers such that $a \geq b \geq c$ and $b^2>ac$. Prove that
$$\dfrac{1}{a^2-bc}+\dfrac{1}{b^2-ca}+\dfrac{1}{c^2-ab}>0.$$

Proposed by Titu Andreescu.

Solution:
By AM-GM Inequality,
$$\dfrac{1}{a^2-bc}+\dfrac{1}{b^2-ca}+\dfrac{1}{c^2-ab} \geq \dfrac{3}{a^2+b^2+c^2-ab-bc-ca} > 0,$$
where the last inequality follows from the fact that $a,b,c$ cannot all be equal.

Mathematical Reflections 2012, Issue 6 - Problem J249

Problem:
Find the least prime $p>3$ that divides $3^q-4^q+1$ for all primes $q>3$.

Proposed by Titu Andreescu.

Solution:
Since $3^5-4^5+1=-780=-(2^2\cdot3\cdot5\cdot13)$, and $3^7-4^7+1$ is not divisible by $5$, we only have to show that $p=13$ divides $3^q-4^q+1$ for all primes $q>3$. Since $q$ is a prime, $q=6k \pm 1$ for some $k \in \mathbb{N}^*$. Moreover, $3^{6k} \equiv 1 \pmod{13}$ and $4^{6k} \equiv 1 \pmod{13}$, then
$$\begin{array}{rrrrrr} 3^{6k+1}-4^{6k+1}+1 & \equiv & 3-4+1 \equiv & 0 & \pmod{13} \\ 3^{6k-1}-4^{6k-1}+1 & \equiv & 9-10+1 \equiv & 0 & \pmod{13}, \end{array}$$
which gives the conclusion.

Mathematical Reflections 2012, Issue 6 - Problem J248

Problem:
Let $f:[1,\infty) \longrightarrow \mathbb{R}$ be defined by $f(x)=\dfrac{\{x\}^2}{\lfloor x \rfloor}$. Prove that $f(x+y) \leq f(x)+f(y)$, for any real numbers $x$ and $y$.

Proposed by Sorin Radulescu.

Solution:
Since $\{x+y\} \leq \{x\}+\{y\}$ and $\lfloor x \rfloor + \lfloor y \rfloor \leq \lfloor x+y \rfloor$ for all $x,y \in \mathbb{R}$, using the well known inequality $$\dfrac{a^2}{x}+\dfrac{b^2}{y} \geq \dfrac{(a+b)^2}{x+y} \qquad \forall a,b,x,y \in \mathbb{R}, x,y>0$$ we have $$f(x+y)=\dfrac{\{x+y\}^2}{\lfloor x+y \rfloor} \leq \dfrac{(\{x\}+\{y\})^2}{\lfloor x \rfloor + \lfloor y \rfloor} \leq \dfrac{\{x\}^2}{\lfloor x \rfloor} + \dfrac{\{y\}^2}{\lfloor y \rfloor}=f(x)+f(y)$$ for all $x,y \in [1,\infty)$.

Mathematical Reflections 2012, Issue 6 - Problem J247

Problem:
Let $a$ and $b$ be distinct zeros of the polynomial $x^3-2x+c$. Prove that $a^2(2a^2+4ab+3b^2)=3$ if and only if $b^2(3a^2+4ab+2b^2)=5$.

Proposed by Titu Andreescu.

Solution:

Let $\alpha$ be the other root of the polynomial $x^3-2x+c$. Then $\alpha^3-2\alpha+c=0$ and $$a+b=-\alpha, \qquad ab=\alpha^2-2.$$ Therefore, $$\begin{array}{rcl} a^2(2a^2+4ab+3b^2)+b^2(3a^2+4ab+2b^2) & = & 2(a+b)^2(a^2+b^2)+2a^2b^2 \\ & = & 2\alpha^2 [\alpha^2-2(\alpha^2-2)]+2(\alpha^2-2)^2 \\ &=& 2[\alpha^2-(\alpha^2-2)]^2 \\ &=& 8, \end{array}$$ and the statement follows.