Prove that for every prime number $p>3$, $4p^2+1$ can be written as the sum of three square numbers Given that $p>3$ prove that $4p^2+1$ can be written as the sum of three distinct positive square numbers. 
Plugging in $5$ I get $101=49+36+16=7^2+6^2+4^2$
I also know that all primes greater than $3$ can be written in the form $3k+1$ and $3k+2$ but plugging those values in I get:
$36k^2+24k+5$,  $36k^2+48k+17$ and the solution probably lies in arranging these numbers in such a way that we get the desired squares, but I can't come up with a combination ,or is my "idea" not even in the right direction? 

If $p=6k\pm 1$ then $4p^2+1 = 144k^2\pm 48k +5$.
So you might want to write $$4p^2+1 = (ak+1)^2+(bk+2)^2+(ck)^2=(a^2+b^2+c^2)k^2+(2a+4b)k+5$$
So you need to find $a,b,c$ so that $a^2+b^2+c^2=144, a+2b=\pm 24$, and $c\neq 0$.
I'll leave it to you to solve for $a,b,c$. 
But you might get a quess of what they are by looking at the case $p=7$ along with your solution for $p=5:$
$$4\cdot 7^2+1 = 9^2+10^2+4^4$$
