, 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.