I have no idea where to start. Any hint(s) or suggestions? Prove if $2\mid(x^2-1) $, then $4\mid(x^2-1)$
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5Hint: $x^2-1=(x-1)(x+1)$. – Thomas Andrews Sep 15 '14 at 01:23
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3One can actually show that $8|(x^2-1)$. – Quang Hoang Sep 15 '14 at 01:34
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You can simply consider two cases. What happens if $x$ is even (i.e., $x=2n$ for some $n$)? What happens for $x=2n+1$ (=odd)? Something similar as in this answer: http://math.stackexchange.com/questions/99716/the-square-of-an-integer-is-congruent-to-0-or-1-mod-4/99741#99741 – Martin Sleziak Sep 15 '14 at 06:49
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One way is that the assumption implies that $x$ must be odd, so $x=2n+1$ and thus $$x^2=4n^2+4n+1$$
Rene Schipperus
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Suppose $2\mid (x^2-1)$. Then $2\mid (x-1)(x+1)$.
By Euclid's lemma, either $2\mid (x+1)$ or $2\mid (x-1)$.
But $x+1$ is even if and only if $x-1$ is even. Thus, both must be even.
The product of two even numbers must be divisible by four, so $$4\mid (x^2-1)$$
Zubin Mukerjee
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wait. I'm confused here. for the euclid's lemma, is that supposed to be and? – Aaron Sep 15 '14 at 01:34
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Euclid's lemma only implies that at least one of the two is true, so I used the inclusive or. – Zubin Mukerjee Sep 15 '14 at 01:39
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It happens to be the case that both of the statements are true, in your problem. But Euclid's lemma doesn't imply that in general, because it's not true in general. For example, $6$ is even, but that doesn't mean both $2 \mid 2$ and $2 \mid 3$ (the latter is clearly false). – Zubin Mukerjee Sep 15 '14 at 01:43
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$x-1$ and $x+1$ are separated by $2$ so they are either both odd or both even. The first case means that the product will not be divisible by two, so the second case must hold true.
Hence $2|(x-1)(x+1) \implies 2|(x+1)$ and $2|(x-1)$ so $4|(x-1)(x+1)$.
Deepak
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$x^2\equiv1\pmod2\iff x\equiv1\pmod2,x=2a+1$ where $a$ is any integer
$$\implies x^2=(2a+1)^2=8\frac{a(a+1)}2+1\equiv1\pmod8$$
lab bhattacharjee
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