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Let $a = x^2$ since $a$ is a perfect square. Then $a - 1 = x^2 - 1$.

Claim: $1 < x + 1 < x^2 - 1$.

Since $x^2 \ge 9$, $x \ge 3$.

Since $x \ge 3$, $1 < x - 1$.

Multiply both sides of $1 < x - 1$ by $x + 1$:

$x + 1 < x^2 - 1$.

Since $1 < x + 1 < x^2 - 1$ and $x + 1|x^2 - 1, x^2 - 1$ is composite.

Please, see if this proof is correct.

VladimirPutin
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2 Answers2

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$x^2\ge9\implies $either $x\ge3$ or $x\le-3$

If $x\ge3, x+1>x-1\ge2$

If $x\le-3, x-1<x+1\le-2$

So in either case, $x^2-1=(x+1)(x-1)$ has two non-trivial factors $(x+1),(x-1)$ which are both $\ne1$

lab bhattacharjee
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  • Given the nature of the problem, I'd have thought that $x$ was implicitly assumed to be a natural number. – WillO Jul 09 '14 at 16:15
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    @WillO, Nothing wrong in a little generalization to integers, right? – lab bhattacharjee Jul 09 '14 at 16:16
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    No, nothing wrong with it --- I just wouldn't want the OP (who seems to be new at this) to think his proof was inadequate, whereas it was probably just fine (i.e. it was just fine if the default assumption in the course he's taking is that $x$ is a natural number, which is probably the case). – WillO Jul 09 '14 at 16:20
  • For the second case, do we say "since $x^2 - 1 > x^2 - 9 \ge 0 > -2 \ge x + 1, x^2 - 1 > x + 1$? – VladimirPutin Jul 09 '14 at 16:24
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It's correct, but not structured in the best way. You never explicitly mention the key idea of the proof, which is to use the factorization:

$$x^2-1=(x+1)(x-1)$$

You should introduce that factorization first, to then motivate checking that each factor is greater than $1$. You did it the other way around, which seems counter-intuitive.

Jack M
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