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