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...

  • $\begingroup$ Is it not allowed to use the fact that any non-negative integer is a sum of 4 squares? $\endgroup$ – Hw Chu Feb 28 '18 at 18:50
  • $\begingroup$ Can you solve the case $p=2$? $\endgroup$ – Hagen von Eitzen Feb 28 '18 at 18:53

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 $$n+1 = \underbrace{l^2 + l^2 + \cdots + l^2}_{p-1 \text{times}.} + (l \pm 1)^2.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.