[ [ solving in integers ]

Solve the system in natural numbers

$a^{2}+b-c=100$,

$a+b^{2}-c=124$.

#
Subtracting and factoring, we get $(b-a)(a+b-1)=24$. The numbers $b-a$ and $a+b-1$ are of different parity, and $a+b-1>b-a$. Therefore,

$b-a=1, a+b-1=24$ or $b-a=3, a+b-1=8$.

In the first case, $a=12, b=13, c=57$, in the second case $a=3, b=6, c=-85$ - not a natural number.

## Answer

$(12,13,57)$.