Or look for decomposition of \(a^n\pm b^n\) for \(n\) odd or even.

For why \(x^{ab}=(x^a)^b\) think about definition of powers, as a repeated multiplication. To get \(x^n\) you multiply \(x\) by itself \(n\) times. Write down \(x\) \(n\) times, and group it in \(b\) groups of \(a\) items each. Now you have \(x^a\), written \(b\) times.

[QUOTE=bur;559892]<snip>
For that proof, I don't understand why [TEX]2^n = (2^a)^b[/TEX] if n = a*b with b being an odd integer. That it's equivalent to the product, I also don't get.[/QUOTE]You don't know the laws of exponents?

x[sup]a[/sup] * x[sup]b[/sup] = x[sup]a+b[/sup]

(x[sup]a[/sup])[sup]b[/sup] = x[sup]a*b[/sup]

The first law is initially proven for positive integers a and b; repeated application gives the second law.

Both laws can then be extended to all integer exponents by demanding that these laws continue to hold. In particular, x[sup]0[/sup] is 1.

The fact that x - y is an algebraic factor of x[sup]n[/sup] - y[sup]n[/sup] may be proved by mathematical induction, as indicated in [url=https://www.mersenneforum.org/showpost.php?p=515513&postcount=19]this post[/url], with a simplified argument in [url=https://www.mersenneforum.org/showpost.php?p=515627&postcount=25]this post[/url].

Substituting x[sup]a[/sup] for x and -y[sup]a[/sup] for y, and using the second law of exponents above, then tells you that

if b is [i]odd[/i], x[sup]a[/sup] + y[sup]a[/sup] is an algebraic factor of x[sup]ab[/sup] + y[sup]ab[/sup]