mersenneforum.org > Math Question about number of the form 2^n + 1
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 2020-10-15, 10:01 #12 LaurV Romulan Interpreter     Jun 2011 Thailand 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
2020-10-15, 14:04   #13
Dr Sardonicus

Feb 2017
Nowhere

24·271 Posts

Quote:
 Originally Posted by bur 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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