$n,p \in \mathbb{N}$ , $p$ prime and $1+np$ perfect square $\implies n+1$ is the sum of $p$ perfect squares.

Question

I was looking to the following problem of Spain's mathematics olympiad (for high-school students):

Let $n,p$ be positive integers with $p$ prime such that $1+np$ is a perfect square. Then $n+1$ is the sum of $p$ perfect squares.

Can you give me any hints to solve it. In principle, I've thought of double induction, formula for the sum of squares...

Answer 1

If $1 + np = m^2$, then $m \equiv \pm1 \pmod p$, hence $m = lp\pm 1$ for some $l$. Then $np = lp(lp\pm2)$, so $n = l^2p\pm2l$, so $$n+1 = \underbrace{l^2 + l^2 + \cdots + l^2}_{p-1 \text{times}.} + (l \pm 1)^2.$$

Answer 2

Assuming that the smallest natural number is one, counterexamples can be found: Suppose n=11 and p=13. Then np+1 = 144 is a perfect square as required, but n+1 = 12 cannot be written as the sum of 13 perfect squares, the smallest of which is 1.

If zero is permitted as a natural number (not in my tradition) strings of zero squared can be used to pad out anywhere required, and the statement is less interesting.

The problem in the proof as given might lie with the fact that if l=1 [and m= -1 (mod p)], then l-1 gives zero.