Only $0$, because $1=0^{2}+1^{2}=1^{2}+0^{2}$.

But I see that $2$ also is a solution of problem. Note: 1)Parts of sum must are positive integers. 2)$1^{2}+2^{2}$ and $2^{2}+1^{2}$ are same. Now try again!

Two well known and already posted theorems: a) every integer can be written as sum of four squares. b) every integer $n>N$ (with e.g. $N=169$) can be written as sum of five nonzero squares. Thus every big integer can be written in two ways, so we are left to check some small numbers, done.

ZetaX wrote: b) every integer $n>N$ (with e.g. $N=169$) can be written as sum of five nonzero squares. Wow, please post that link here, thanks

http://www.mathlinks.ro/Forum/viewtopic.php?t=121595

Hello, ZetaX wrote: Two well known and already posted theorems: a) every integer can be written as sum of four squares. b) every integer $n>N$ (with e.g. $N=169$) can be written as sum of five nonzero squares. Thus every big integer can be written in two ways, so we are left to check some small numbers, done. I think that if it is possible to use four squares theorem, it's useless to use b) and check up to 169. There is something simpler : For $n\geq 16$, if n can be written as $a^{2}+b^{2}+c^{2}+d^{2}$ with either 1, either 2, either 3, either 4 not in the set $\{a,b,c,d\}$, then n can be written as $u^{2}+v^{2}+w^{2}+t^{2}+x^{2}$ where $x$ is the missing number (1,2,3 or 4) and $(u^{2},v^{2},w^{2},t^{2})$ the 4 or less squares of expression of $n-x^{2}$. Because $x$ is not in in the set $(a,b,c,d)$, the two expressions are different. So we have just to check numbers 1 to 15 plus the number $30=1^{2}+2^{2}+3^{2}+4^{2}$. $1=1$ is OK $2=1+1$ is OK $3=1+1+1$ is OK $4=1+1+1+1=4$ is NOK $5=4+1=1+1+1+1+1$ is NOK $6=4+1+1$ is OK $7=4+1+1+1$ is OK $8=4+4=4+1+1+1+1$ is NOK $9=9=4+4+1$ is NOK $10=9+1=4+4+1+1$ is NOK $11=9+1+1=4+4+1+1+1$ is NOK $12=9+1+1+1=4+4+4$ is NOK $13=9+1+1+1+1=4+4+4+1$ is NOK $14=9+4+1=4+4+4+1+1$ is NOK $15=9+4+1+1$ is OK $30=16+9+4+1=25+4+1$ is NOK And the requested set is $\{1,2,3,6,7,15\}$ (+ $0$ is $0$ is to be considered) A question now is : can this problem be solved without using four squares theorem ? -- Patrick