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 $x=-1,0,1$.