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 $1$.