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,
$$\dfrac{a^3+2}{b^2-b+1}+\dfrac{b^3+2}{c^2-c+1}+\dfrac{c^3+2}{a^2-a+1} \geq 9.$$