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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?

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marked as duplicate by Steven Stadnicki, vrugtehagel, Namaste discrete-mathematics Nov 21 '18 at 19:41

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  • $\begingroup$ Does $0$ count as a square here? I mean $4\cdot 7^2+1 = 0^2+1^2+14^2$, and $0,1,14$ are distinct. If yes, this is quite trivial: $4p^2+1=0^2+1^2+(2p)^2$ for any integer $p\neq 0$. $\endgroup$ – user614671 Nov 21 '18 at 16:42
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    $\begingroup$ It is possible to ignore $0$, as I did in my answer @Snookie . $\endgroup$ – Mohammad Zuhair Khan Nov 21 '18 at 16:46
  • $\begingroup$ no you can not include 0 $\endgroup$ – Marko Škorić Nov 21 '18 at 16:48
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Notice that $5=1^2+2^2,$ so:
$4(6k+1)^2+1=144k^2+48k+5=(ak)^2+(bk+1)^2+(ck+2)^2=(a^2+b^2+c^2)k^2+(2b+4c)k+5$
$\therefore 144=a^2+b^2+c^2 \qquad$ and $\qquad 2b+4c=48$

By trial and error, I found that $a=4, b=8, c=8$.
So $4(6k+1)^2+1=(4k)^2+(8k+1)^2+(8k+2)^2$

Now for $p=6k-1,$
$4(6k-1)^2+1=144k^2-48k+5=(ak)^2+(bk-1)^2+(ck-2)^2=(a^2+b^2+c^2)k^2-(2b+4c)k+5$
$\therefore 144=a^2+b^2+c^2 \qquad$ and $\qquad -48=-(2b+4c)$

So, $4(6k-1)^2+1=(4k)^2+(8k-1)^2+(8k-2)^2$

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