1. Find all prime numbers $p, q$ and $r$, for which $p+q^{2}=r^{4}$.
I/1. We rewrite the equation as $p=r^{4}-q^{2}=\left(r^{2}-q\right)\left(r^{2}+q\right)$. Since $p$ is a prime number, it must be that $r^{2}-q=1$ and $r^{2}+q=p$. We rewrite the first equation as $q=r^{2}-1=(r-1)(r+1)$. Since $q$ is a prime number, it must be that $r-1=1$. Therefore, $r=2$ and $q=3$. From the equation $r^{2}+q=p$, we get $p=7$.
Expression $p=\left(r^{2}-q\right)\left(r^{2}+q\right)$ ..... 1 point
Conclusion $r^{2}-q=1$ ..... 3 points
Factorization $q=(r-1)(r+1)$ ..... 1 point
Conclusion $r=2$ ..... 1 point
Conclusion $q=3, p=7$ ..... 1 point

2. We observe that one of the primes must be 2. We eliminate the possibilities that $p=2$ or $q=2$ (using congruences or by rearranging the equation similarly to the first solution). We get that $r=2$ and $p+q^{2}=16$. We check the possibilities and get $q=3$ and $p=7$.

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Justification $p \neq 2$..........................................................................................................................................

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