Let

$$3 x+4 y=m^{2}, \quad 4 x+3 y=n^{2}$$

Then

$$7(x+y)=m^{2}+n^{2} \Rightarrow 7 \mid m^{2}+n^{2}$$

Considering $m=7 k+r, \quad r \in\{0,1,2,3,4,5,6\}$ we find that $m^{2} \equiv u(\bmod 7), \quad u \in$ $\{0,1,2,4\}$ and similarly $n^{2} \equiv v(\bmod 7), \quad v \in\{0,1,2,4\}$. Therefore we have either $m^{2}+n^{2} \equiv 0 \quad(\bmod 7)$, when $u=v=0$, or $m^{2}+n^{2} \equiv w(\bmod 7), w \in\{1,2,3,4,5,6\}$. However, from (2) we have that $m^{2}+n^{2} \equiv 0(\bmod 7)$ and hence $u=v=0$ and

$$m^{2}+n^{2} \equiv 0 \quad\left(\bmod 7^{2}\right) \Rightarrow 7(x+y) \equiv 0 \quad\left(\bmod 7^{2}\right)$$

and consequently

$$x+y \equiv 0 \quad(\bmod 7)$$

Moreover, from (1) we have $x-y=n^{2}-m^{2}$ and $n^{2}-m^{2} \equiv 0\left(\bmod 7^{2}\right)$ (since $u=v=0$), so

$$x-y \equiv 0 \quad(\bmod 7) .$$

From (3) and (4) we have that $x+y=7 k, x-y=7 l$, where $k$ and $l$ are positive integers. Hence

$$2 x=7(k+l), 2 y=7(k-l)$$

where $k+l$ and $k-l$ are positive integers. It follows that $7 \mid 2 x$ and $7 \mid 2 y$, and finally $7 \mid x$ and $7 \mid y$.

## ALGEBRA