<s>[INST] Square equal to sum of three squares For which integers $n$ there exists integers $0\le a,b,c < n$ such that $n^2=a^2+b^2+c^2$?
I made the following observations:

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*For $n=1$ and $n=0$ those integers doesn't exist.


*If $n$ is a power of 2 those integers doesn't exist. Let $n=2^m$ with $m>0$ the smallest power of 2 for which there exists $a,b,c$ such that $\left (2^m\right )^2=4^m=a^2+b^2+c^2$. Since $4^m$ is divisible by 4, $a^2+b^2+c^2$ has to be divisible by 4 too. This is only possible if $a^2\equiv b^2\equiv c^2\equiv 0\pmod 4$, so we can write $a=2a',b=2b',c=2c'$ with $a',b',c'\in \mathbb{N}$. But then we get $\left (2^{m-1}\right )^2=4^{m-1}=a'^2+b'^2+c'^2$, so $m=1$, otherwise $2^m$ wouldn't be the smallest power of two with this property. It is easy to check that $n=2$ doesn't work, so for $n=2^m$ the statement doesn't hold.


*I suspect (but can't prove) that for all other values the statement holds. It would be enough to prove that for all odd primes $p$ there exists $a,b,c$ such that $p^2=a^2+b^2+c^2$, since for all other values of $n$ there exist some $p,m$ such that $n=pm$. Then we get $n^2=(pm)^2=(ma)^2+(mb)^2+(mc)^2$.
 [/INST] If we want $a^2+b^2+c^2=n^2$, then $a^2+b^2=n^2-c^2=(n-c)(n+c)$.
So a way to generate solutions is to choose $a$ and $b$ and then try to find $n$ and $c$ that work.
Example: $a=10$, $b=11$. So $a^2+b^2=100+121=221$.
Now $221=13*17=(15-2)(15+2)$.  So a solution should be $10^2+11^2+2^2=15^2$.
You could also write $221=221=1*221=(111-110)(111+110)=111^2-110^2$.  So another solution would be $10^2+11^2+110^2=111^2$.
I will leave it to you to explore this idea further if you wish.
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