Question

If $p$ is a prime number, and a prime to $p$, and if a square number $c^2$ can be found such that $c^2-a$ is divisible by $p$, show that $a^{\frac{1}{2}(p-1)}-1$ is divisible by $p$.

Answer 1

$\Rightarrow p.,$

$c^2-a=kp$(given)

$\therefore a=c^2-kp$

$\therefore a^{\frac{1}{2}(p-1)}-1={(c^2-kp)}^{\frac{p-1}{2}}-1$

$=c^{p-1}-M\left(p\right)-1$

Since $p-1$is an even integer

Also a is prime to p, so that c must be prime to p;hence $c^{p-1}-1=M\left(p\right)$ by Fermat's theorem, and the result at once follows.