prove that 5n+3, where n is a prime, can never be a perfect square
Hello,
Hmmm I don't see how to apply Fermat's little theorem here o.O
So you want to prove that there is no x such that $\displaystyle x^2=5n+3$, that is $\displaystyle x^2\equiv 3 (\bmod 5)$
If x=0(mod5), then x²=0(mod5)
If x=1(mod5), then x²=1(mod5)
If x=2(mod5), then x²=4(mod5)
If x=3(mod5), then x²=4(mod5)
If x=4(mod5), then x²=1(mod5)
thus it x doesn't exist.
Hmm if you want to apply FLT, it's actually possible.
Assume there is x such that $\displaystyle x^2\equiv 3(\bmod 5)$
Square it :
$\displaystyle x^4\equiv 9(\bmod 5) \equiv 4(\bmod 5)$
But we know, by FLT, that for any x coprime with 5, $\displaystyle x^4\equiv 1(\bmod 5)$
And if it is not coprime with 5, it means that it's a multiple of 5. In which case, $\displaystyle x^4\equiv 0(\bmod 5)$
So we have a contradiction and there is no x such that $\displaystyle x^2\equiv 3(\bmod 5)$
Is it more similar to what you're doing in class ?