## Task $2 / 82$

We are looking for all pairs of prime numbers $\left(p_{1} ; p_{2}\right)$ with $p_{1} \leq p_{2}$, for which $\left(4 p_{1}\right)^{2}+\left(4 p_{2}\right)^{2}=100000$.
The given equation is equivalent to the equation

$$
p_{1}^{2}+p_{2}^{2}=6250
$$

1) $p_{1} \neq 2$, because $p_{2}^{2}=6246$ does not yield a natural number $p_{2}$, let alone a prime number $p_{2}$.
2) $p_{1}=3$ yields $p_{2}^{2}=6241, p_{2}=79$. Since 79 is indeed a prime number, a first solution pair has been found: $\left(p_{11} ; p_{21}\right)=(3 ; 79)$.
3) If $p_{1}>3$, it is known that $p_{1} \equiv \pm 1(\bmod 6)$, thus $p_{1}^{2} \equiv 1(\bmod 6)$. Since $p_{2} \geq p_{1}$, the same applies to $p_{2}$. Therefore, it follows that

$$
p_{1}^{2}+p_{2}^{2} \equiv 1+1=2(\bmod 6)
$$

Since $6250 \equiv 4(\bmod 6)$, there can be no further solution pairs.