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.