Working through a question i got to this stage where i need to:

Show that if m > n then ((a^2)^n) + 1 divides ((a^2)^m)− 1

Should i be looking to expand ((a^2)^n) + 1?

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- Nov 5th 2009, 01:05 PMsirellwoodShow that if m > n then ((a^2)^n) + 1 divides ((a^2)^m)− 1
Working through a question i got to this stage where i need to:

Show that if m > n then ((a^2)^n) + 1 divides ((a^2)^m)− 1

Should i be looking to expand ((a^2)^n) + 1? - Nov 5th 2009, 01:13 PMtonio
- Nov 5th 2009, 01:16 PMsirellwood
argh sorry. my bad. edited :-)

- Nov 7th 2009, 03:25 AMflyingsquirrel
- Nov 7th 2009, 03:37 AMMedia_ManExponents not Associative
Sirellwood:

The operator "^" is NOT associative, that is, a^(2^n) does not equal (a^2)^n in general -- in Latex, $\displaystyle a^{2^n}\neq (a^2)^n$

Theorem: For $\displaystyle m>n, a^{2^n}+1|a^{2^m}-1$

Proof: $\displaystyle a^{2^m}-1=(a^{2^{m-1}}+1)(a^{2^{m-1}}-1)=(a^{2^{m-1}}+1)(a^{2^{m-2}}+1)(a^{2^{m-2}}-1)$...

Continue factoring the successive differences of squares k times until $\displaystyle m-k=n$. QED

Your theorem is quite true, if rendered to the page in proper notation. (Thinking) - Nov 7th 2009, 03:38 AMtonio
- Nov 7th 2009, 03:45 AMtonio

I supposed that if the OP had wanted $\displaystyle a^{2^n}$ he could have writte a^(2^n) and not (a^2)^n, which means $\displaystyle (a^2)^n=a^{2n}$. OTOH, he could have simply written a^(2n) and thus I think you're right and he meant $\displaystyle a^{2^n}$.

Even using simple ASCII one must be careful or even describe things by words when mathematical notation is cumbersome.

Tonio - Nov 7th 2009, 04:05 AMMedia_ManOn languageQuote:

one must be careful or even describe things by words when mathematical notation is cumbersome.