Slovenija 2008

Let us assume there exist numbers $m$ and $n$, one even and one odd, such that $A=(m+3n)(5m+7n)(7m+5n)(3m+n)$ is a perfect square. Let $d$ be the greatest common divisor of $m$ and $n$, and write $m=d{m}_1$, $n=d{n}_1$. Then $m_1$ and $n_1$ are coprime and $A$ can be written as $$A=d^4(m_1+3n_1)(5m_1+7n_1)(7m_1+5n_1)(3m_1+n_1).$$ Let $B=(m_1+3n_1)(5m_1+7n_1)(7m_1+5n_1)(3m_1+n_1)$. Since $A$ is a perfect square, so is $B$.

Let $p$ be a positive integer that divides both $m_1+3n_1$ and $5m_1+7n_1$. Since $m_1$ and $n_1$ have different parity, the numbers $m_1+3n_1$ and $5m_1+7n_1$ are odd and $p$ must be odd, too. On the other hand we have $$p\mid 5\cdot (m_1+3n_1)-(5m_1+7n_1)=8n_1,$$ $$p\mid 3\cdot (5m_1+7n_1)-7(m_1+3n_1)=8m_1,$$ so $p\mid m_1$ and $p\mid n_1$, which implies $p=1$. We have shown that $m_1+3n_1$ and $5m_1+7n_1$ are coprime.

Let $q$ be a positive integer such that $q\mid m_1+3n_1$ and $q\mid 7m_1+5n_1$. Again, $q$ must be odd and $$q\mid 7(m_1+3n_1)-(7m_1+5n_1)=16n_1,$$ $$q\mid 3(7m_1+5n_1)-5(m_1+3n_1)=16m_1.$$ So, $q\mid n_1$ and $q\mid m_1$, which implies $q=1$. We conclude that $m_1+3n_1$ and $(5m_1+7n_1)(7m_1+5n_1)$ are coprime.

Similarly, one can show that $3m_1+n_1$ and $(5m_1+7n_1)(7m_1+5n_1)$ are coprime. Now, assume $r\mid m_1+3n_1$ and $r\mid 3m_1+n_1$. Then $r\mid 3(m_1+3n_1)-(3m_1+n_1)=8n_1$ and $r\mid 3(3m_1+n_1)-(m_1+3n_1)=8m_1$, so $r\mid n_1$, $r\mid m_1$ and $r=1$.

We have shown that the two factors $3m_1+n_1$ and $m_1+3n_1$ are coprime to any other factor of $B=(m_1+3n_1)(3m_1+n_1)(5m_1+7n_1)(7m_1+5n_1)$, and since $B$ is a perfect square, we see that $m_1+3n_1$ and $3m_1+n_1$ must be perfect squares as well. Denote them by $a^2$ and $b^2$, respectively. Then $$(a-b)(a+b)=a^2-b^2=2(n_1-m_1).$$ The right-hand side is divisible by $2$ but not by $4$ because $n_1$ and $m_1$ have different parity. On the other hand, $a-b$ and $a+b$ have the same parity and since their product is divisible by $2$, it must also be divisible by $4$. This contradiction shows that $$(m+3n)(5m+7n)(7m+5n)(3m+n)$$ cannot be a perfect square.