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 $\max (\min (|x-y|, |y-z|, |z-x|))=\sqrt{2}$.