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How do we prove that this expression is always 'not prime'?

  • 4
    It's an even number which is not $2$. – 2017-01-24
  • 0
    Or a multiple of $8$ ... . – 2017-01-24
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    As a basic problem solving strategy, you should have verified this claim for a few small values of $X$. Then it's very likely you would have noticed the commonality that $X^2-1$ is always even. But there is no evidence that you put in this minimal basic effort. – 2017-01-24

3 Answers 3

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Suppse $x$ is odd. Then $x^2$ is odd and $x^2 - 1$ is even. The only time this can be prime is when $x^2 - 1$ is $2$, but this requires $x^2 = 3$ for which where are no solutions in the odd integers. Hence $x^2 - 1$ is not prime for all odd integers $x$.

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If $x$ is odd,, then $x = 2k+1$ for some $k\in\mathbb Z$. Then, we have that: $$x^2 -1= (2k+1)^2-1 = (4k^2+4k+1)-1 = 4k^2+4k = 4k(k+1)$$

As we have that $4\mid x^2-1$, it follows that this is even, not odd. Additionally, we know that it can't be prime, as it factors as $4\times k\times k+1$ (and may factor further depending on $k$). Now, we know that $k\in\mathbb Z$, so the expression $4k(k+1)$ is nearly always positive - it would be true if we ignored the cases when $k = \pm 1$. But they exist, so we can't ignore them.

As the answer isn't a - c, it must be d.

1

In general, you show that something isn't prime by finding a factor it has to have. Sometimes this means a specific number (for example, $1 + x^2 + x^4$ happens to always be divisible by $3$), and sometimes it's a polynomial.

In this case, remember that $x^2 - 1 = (x - 1)(x + 1)$ (this is a special case of what's usually called "difference of squares"). So $x^2 - 1$ is divisible by $x - 1$ and $x + 1$. The only way this can still be prime is if one of those factors are $1$ - so either $x - 1 = 1$ or $x + 1 = 1$. Then $x$ is either $2$ or $0$ - but neither of those are odd numbers, so neither are possible.