Prove that if $2|(x^2-1)$, then $8|(x^2-1)$.

Question

Prove that if $2\ |\ (x^2-1)$, then $8\ |\ (x^2-1)$.

Answer 1

well if $2$ divides $(x^2-1)=(x-1)(x+1)$ then being $2$ as prime it will divide either $x-1$ or $x+1$ atleast one. WLG, Let it divides $(x-1)$, then $(x-1)$ is even and thus $x-1+2=x+1$ is also even and thus $2$ also divides $x+1$ and thus $4$ divides $(x^2-1)$ So let $x^2-1=4k.$

$\textbf{Case 1-}$ $k$ is even, we are done.

$\textbf{Case 2-}$ $k$ is odd, let $k=2t+1$ then $x^2-1=4(2t+1)=8t+4$ $\implies x^2=8t+5$ . Now we show it is not possible for a perfect square to be congruent modulo $8$. Consider all possibilities that $x \equiv \hspace{.05cm} 0,1,2,3,4,5,6,7 \hspace{.07cm} \text{mod} \hspace{.05cm} 8$. Then $x^2 \equiv 0,1,4,1,0,1,4,1 \text{mod} 8$ resp. So no perfect square can give $5$ as a remainder when divided by $8$. Done.

I hope it is correct!

Answer 2

Hint:-

$x^{2}\equiv1 \pmod 2\\ \implies \gcd(x,2)=1\\\implies \gcd(x,4)=1\land \gcd(x,8)=1\\ \implies x^{\varphi(8)}\equiv1\pmod 8\land x^{\varphi(4)}\equiv1\pmod 4\\\implies x^{4}\equiv1\pmod 8\land x^{2}\equiv1\pmod 4$

By Euler's Theorem.

Answer 3

$2$ divides $x^2-1$ so $x$ must be odd. Writing $x=2k+1$, we have $$ x^2-1=(4k^2+4k+1)-1=4k(k+1). $$ Regardless of whether $k$ is odd or even, $k(k+1)$ is even, so the rightmost expression above is divisible by $8$.