Question

If $2n+1$ is a prime number, show that $1^2,2^2,3^2,....n^2$ when divided by $2n+1$ leave different remainder.

Answer 1

$⟹$ Let $r,n$ be two of the numbers $1,2,3,.....n;$ $r^2-n^2$ is divisible by $2n+1.$ Now $2n+1$ is a prime number; Hence either $r+s$ or $r-s$ must be divisible by $2n+1;$ But $r$ and $s$ are each less than $n,$ So that $r+s$ or $r-s$ is each less than $2n+1;$ Hence $r^2-s^2$ cannot be divisible by $2n+1$ that is $r^2$ and $s^2$ cannot leave the same remainder when divided by $2n+1$