1. Find all natural numbers $m$ and $n$ for which $m^{2}+n^{5}=252$.

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II/1. 1. way. Since $m^{2}=252-n^{5}$ is a nonnegative number, it follows that $n^{5} \leq 252$, so $n<4$. If $n=1$, we get $m^{2}=251$, for $n=2$ we get $m^{2}=220$, and for $n=3$ we get $m^{2}=9$. The only possibility is therefore $m=n=3$.
Writing $m^{2}=252-n^{5}$ or $n^{5}=252-m^{2}$ ..... 1 point
Conclusion $n^{5} \leq 252$ ..... 1 point
Conclusion $n<4$ ..... 1 point
Handling the case $n=1$ ..... 1 point
Handling the case $n=2$ ..... 1 point
Handling the case $n=3$ ..... 1 point
Determination that $m=n=3$ is the only solution 1 point

2. way. Since $n$ is a natural number, $m^{2}=252-n^{5} \leq 252$. Therefore, $m<16$. If we substitute values from 1 to 15 for $m$ in sequence, we get the values for $n^{5}$ as $251, 248, 243, 236, 227, 216, 203, 188, 171, 152, 131, 108, 83, 56, 27$. Since $1^{5}=1, 2^{5}=32, 3^{5}=243$, and $4^{5}=1024$, we get a solution only when $m=3$, which gives $n=3$.
Writing $m^{2}=252-n^{5}$ or $n^{5}=252-m^{2}$ ..... 1 point
Conclusion $m^{2} \leq 252$ ..... 1 point
Conclusion $m<16$ ..... 1 point
Calculation of $n^{5}$ for $m=1,2,3,4,5$ ..... 1 point
Calculation of $n^{5}$ for $m=6,7,8,9,10$ ..... 1 point
Calculation of $n^{5}$ for $m=11,12,13,14,15$ ..... 1 point
Determination that $m=n=3$ is the only solution ..... 1 point