4A. Find all pairs of coprime natural numbers $a$ and $b$ such that $a+b$ is a divisor of $a^{4}+b^{4}$.

Solution We use $a^{4}-b^{4}=(a-b)(a+b)\left(a^{2}+b^{2}\right)$, from which it follows that $a+b$ is a divisor of $a^{4}-b^{4}$. We get that $a+b$ is a divisor of $\left(a^{4}+b^{4}\right)+\left(a^{4}-b^{4}\right)=2 a^{4}$ and $a+b$ is a divisor of $\left(a^{4}+b^{4}\right)-\left(a^{4}-b^{4}\right)=2 b^{4}$, or $(a+b) \mid$ GCD $\left(2 a^{4}, 2 b^{4}\right)$. From GCD $(a, b)=1$ it follows that GCD $\left(a^{4}, b^{4}\right)=1$, hence $(a+b) \mid$ GCD $\left(2 a^{4}, 2 b^{4}\right)=2$. Finally, $a=1, b=1$.

## II year

1 AB. Find the integer values of the parameter $a$ for which the solutions of the system:

$$
\left\{\begin{array}{l}
a(x+y-1)=x-2 y-1 \\
a(x-y-3)=x+2 y-3
\end{array}\right.
$$

are integers.
Solution. The given system is equivalent to:

$$
\left\{\begin{array}{l}
(a-1) x+(a+2) y=a-1 \\
(a-1) x-(a+2) y=3 a-3
\end{array}\right.
$$

We have

$$
\left\{\begin{array}{l}
2(a-1) x=4 a-4 \\
2(a+2) y=-2 a+2
\end{array}\right.
$$

If $a \neq 1$, then $x=2$, and if $a=1$, then $x \in \mathbf{R}$. If $a=-2$, there is no solution for $y$ and thus no solution for the system. If $a \neq -2$, we have $y=\frac{-a+1}{a+2}=-1+\frac{3}{a+2}$. Since the solutions must be integers, it follows that $a+2$ must be a divisor of 3, and thus $a+2 = \pm 1, \pm 3$. Therefore, $a=-5, -3, -1, 1$. We get: $a=-5: x=2, y=-2 ; a=-3: x=2, y=-4 ; a=-1: x=2, y=2 ; a=1: x \in \mathbf{R}, y=0$.

2 AB. Let $a, b, c$ be real numbers such that: $a b-a=b+119, b c-b=c+59$, $c a-c=a+71$. Calculate all possible values of $a+b+c$.