Problem 2. Determine the non-zero natural numbers $a, b, c$ for which

$$
\frac{a+b}{2}+\frac{a^{2}+b^{2}}{2}=\frac{7 c+1}{c+1}
$$
Solution. Let $M=\frac{a+b}{2}+\frac{a^{2}+b^{2}}{2}=\frac{a+b+a^{2}+b^{2}}{2}=\frac{a(1+a)+b(1+b)}{2} \in \mathbb{N}$, since $a(1+a)$ and $b(1+b)$ are products of consecutive natural numbers, hence even.

On the other hand, $M=\frac{7 c+1}{c+1}=7-\frac{6}{c+1} \in \mathbb{N}$, so $\frac{6}{c+1} \in\{1,2,3,6\}$ $1 p$ $2 p$

If $\frac{6}{c+1}=1$, then $M=6$, so $a(1+a)+b(1+b)=12$ with the solution $a=b=2$ and $c=5$.

If $\frac{6}{c+1}=2$, then $M=5$, so $a(1+a)+b(1+b)=10$ without a solution.

If $\frac{6}{c+1}=3$, then $M=4$, so $a(1+a)+b(1+b)=8$ with the solutions $a=1, b=2, c=1$ and $a=2, b=1, c=1$.

If $\frac{6}{c+1}=6$, then $c=0$ does not work.