Solve the equation

$$
\sqrt{x+y \sqrt{2}}+\sqrt{u+v \sqrt{2}}=1
$$

in integers.
In the task, we will use the fact that if $a, b, c, d$ are rational numbers, and $a+b \sqrt{2}=c+d \sqrt{2}$, then $a=c$ and $b=d$. If $b=d$, this is obviously true; otherwise, $\sqrt{2}=\frac{a-c}{d-b}$, which cannot be, because $\sqrt{2}$ is irrational, while $\frac{a-c}{d-b}$ is rational.

For the equation $\sqrt{u+v \sqrt{2}}=1-\sqrt{x+y \sqrt{2}}$, squaring and rearranging gives:

$$
\sqrt{x+y \sqrt{2}}=\frac{1+x-u+(y-v) \sqrt{2}}{2}
$$

Thus, $\sqrt{x+y \sqrt{2}}$ can certainly be written in the form $\frac{2 m}{2}+\frac{2 n}{2} \sqrt{2}$, where $2 m$ and $2 n$ are integers.

Then: $2 m=1+x-u$ and $2 n=y-v$.

$$
\sqrt{x+y \sqrt{2}}=m+n \sqrt{2}, x+y \sqrt{2}=m^{2}+2 n^{2}+2 m n \sqrt{2}
$$

From which $x=m^{2}+2 n^{2}, y=2 m n$.

$y=\frac{1}{2}(2 m)(2 n)$ is an integer, so $2 m$ or $2 n$ is even, i.e., $m$ or $n$ is an integer.

If $m$ is an integer, $x-m^{2}=\frac{1}{2}(2 n)^{2}$ is also an integer, so $2 n$ is even, hence $n$ is an integer.

If $n$ is an integer, $x-2 n^{2}=\frac{1}{4}(2 m)^{2}$ is also an integer, so $2 m$ is even, hence $m$ is an integer.

In summary: $\sqrt{x+y \sqrt{2}}=m+n \sqrt{2}$ where $m$ and $n$ are integers.

$$
0 \leq \sqrt{x+y \sqrt{2}}=m+n \sqrt{2} \leq \sqrt{x+y \sqrt{2}}+\sqrt{u+v \sqrt{2}}=1
$$

i.e.

Case I: $n=0$

Then $0 \leq m \leq 1$, so $m=0$ or $m=1$.

If $m=0, x=m^{2}+2 n^{2}=0, y=2 m n=0, 2 m=1+x-u$ i.e., $u=1, 2 n=y-v$, i.e., $v=0$.

If $m=1, x=m^{2}+2 n^{2}=1, y=2 m n=0, 2 m=1+x-u$, i.e., $u=0, 2 n=y-v$, i.e., $v=0$.

We have found two solutions: $u=1, x=y=v=0$ and $x=1, y=u=v=0$.

Case II: $n \neq 0$

Then $n$ is an integer, so $\sqrt{2} n$ is irrational, i.e., not an integer. From the inequality $0 \leq m+n \sqrt{2} \leq 1$, it follows that $m=-[\sqrt{2} n], 2 \quad x=m^{2}+2 n^{2}=2 n^{2}+[\sqrt{2} n]^{2}, \quad y=2 m n=-2 n[\sqrt{2} n], \quad u=1+x-2 m=2 n^{2}+(1+[\sqrt{2} n])^{2}$, $v=y-2 n=-2 n(1+[\sqrt{2} n])$.

Since our transformations are equivalent, these indeed satisfy the conditions of the problem.

In summary: $x=2 n^{2}+[\sqrt{2} n]^{2}, \quad y=-2 n[\sqrt{2} n], \quad u=2 n^{2}+(1+[\sqrt{2} n])^{2}$, $v=-2 n(1+[\sqrt{2} n])$ where $n$ is any non-zero integer. In addition to these, the solutions found in Case I are also valid: $u=1$, $x=y=v=0$ and $x=1, y=v=u=0$.

Péter Várkonyi (Fazekas M. Fóv. Gyak. Gimn., IV. o.t.)