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Let $p>3$ is prime number. Prove that number $4p^2+1$ can show as sum of squares of three different numbers.
Only what I know that every prime number $p>3$ can show as $p=6k+1$ or $p=6k-1$, such that $k \in \mathbb Z$.
If I put $p=6k+1$, then $4(6k+1)^2+1=(12k)^2+(24k+3)^2-(24k+2)^2$, here I did not show what they want in task.
For $p=6k-1$ things do not change, do you have some idea?