The divisibility of the values of quadratic polynomials in $x$, for integer $x$

Question

I would like to know method of finding validity of the statement by proofs.

1) $8$ does not divides $x^2 - 7$ for any integral value of $x$? 2) For any odd integer $x;$ the term $(x-1)^2$ is always divisible by $4.$

May I know proof please.

Answer 1

1) If $x$ is even, then $x^2-7$ is odd, so it is not divisible by $8$.

If $x=2k-1$ for $k\in\mathbb Z$, then $$x^2-7=(2k-1)^2-7=4k^2-4k-6=2(2(k^2-k-1)-1)$$ is not divisible by $8$ because $2(k^2-k-1)-1$ is odd.

2) If $x=2k-1$ for $k\in\mathbb Z$, then $(x-1)^2=(2k-2)^2=4(k-1)^2$ is divisible by $4$.

Answer 2

Hints:

1) Consider squares modulo 8. $1^2 \equiv 1 \bmod 8, 2^2 \equiv 4 \bmod 8, 3^2 \equiv 1 \bmod 8, 4^2 \equiv 0 \bmod 8$ and the pattern repeats. (because $(4k+i)^2 \equiv i^2 \bmod 8$ for $i = 1, 2, 3, 4$.
Now, subtract 7 from each case, modulo $8$. What do you get?

2)$x$ is an odd integer $\implies (x-1) $ is an even integer $\implies (x-1) = 2k$ for some $k$. Now, what can you conclude about $(x-1)^2$?

Answer 3

consider $x^2\equiv 0,1,4 \mod 8$ and for b) set $x=2m+1$ where $m$ is an integer. Sonnhard.

Answer 4

$$x^2-7\equiv0\pmod8\iff x^2\equiv7\equiv-1\pmod8$$

But $x\equiv0,\pm1,\pm2,\pm3,4\pmod8\implies x^2\equiv0,1,4$