Problem 1123

Solution:

Let
$$3x + 4y = m^2, \quad 4x + 3y = n^2$$
Then
$$7(x + y) = m^2 + n^2 \Rightarrow 7 \mid m^2 + n^2$$
Considering $m = 7k + r$, $r \in \{0, 1, 2, 3, 4, 5, 6\}$, we find that $m^2 \equiv u \pmod{7}$, $u \in \{0, 1, 2, 4\}$ and similarly $n^2 \equiv v \pmod{7}$, $v \in \{0, 1, 2, 4\}$. Therefore, we have either $m^2 + n^2 \equiv 0 \pmod{7}$, when $u = v = 0$, or $m^2 + n^2 \equiv w \pmod{7}$, $w \in \{1, 2, 3, 4, 5, 6\}$.

However, from (2) we have that $m^2 + n^2 \equiv 0 \pmod{7}$ and hence $u = v = 0$ and
$$m^2 + n^2 \equiv 0 \pmod{7^2} \Rightarrow 7(x + y) \equiv 0 \pmod{7^2}$$
and consequently
$$x + y \equiv 0 \pmod{7}$$
Moreover, from (1) we have $x - y = n^2 - m^2$ and $n^2 - m^2 \equiv 0 \pmod{7}$ (since $u = v = 0$), so
$$x - y \equiv 0 \pmod{7}$$
From (3) and (4) we have that $x + y = 7k$, $x - y = 7l$, where $k$ and $l$ are positive integers. Hence
$$2x = 7(k + l), \quad 2y = 7(k - l)$$
where $k + l$ and $k - l$ are positive integers. It follows that $7 \mid 2x$ and $7 \mid 2y$, and finally $7 \mid x$ and $7 \mid y$.