Arman wrote: Let $p$ and $n$ be a natural numbers such that $p$ is a prime and $1+np$ is a perfect square. Prove that the number $n+1$ is sum of $p$ perfect squares. $1+np=a^2$ implies $np=(a-1)(a+1)$ and so $a=kp\pm 1$ If $a=kp+1$, we get $1+np=(kp+1)^2$ and so $n+1=k^2p+2k+1$ $=(p-1)k^2+(k+1)^2$ If $a=kp-1$, we get $1+np=(kp-1)^2$ and so $n+1=k^2p-2k+1$ $=(p-1)k^2+(k-1)^2$ This proves your request, but only if you consider that $0$ is a perfect square (in order to manage the case where $k=1$) Else the request is wrong : just consider $(n,p)=(1,3)$ and $n+1=2$ is not the sum of three squares of natural numbers.

If $n>p$ it is true. Thanks!!!

http://artofproblemsolving.com/community/c6h369941p2038251 It is the first problem of IRAN MO 2001 .