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Old 2020-10-15, 10:01   #12
Romulan Interpreter
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Jun 2011

52·7·53 Posts

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.

Last fiddled with by LaurV on 2020-10-15 at 10:10
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Old 2020-10-15, 14:04   #13
Dr Sardonicus
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Feb 2017

24·271 Posts

Originally Posted by bur View Post
For that proof, I don't understand why 2^n = (2^a)^b if n = a*b with b being an odd integer. That it's equivalent to the product, I also don't get.
You don't know the laws of exponents?

xa * xb = xa+b

(xa)b = xa*b

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, x0 is 1.

The fact that x - y is an algebraic factor of xn - yn may be proved by mathematical induction, as indicated in this post, with a simplified argument in this post.

Substituting xa for x and -ya for y, and using the second law of exponents above, then tells you that

if b is odd, xa + ya is an algebraic factor of xab + yab
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