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?
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?
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.
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
Funny, if you read pre-twentieth century math books and proofs, they rarely rely on notation and speak almost solely in plain english words. "One more than an integer raised to a power of two must evenly divide one less than that same number raised any higher power of two..." Mathematical notation is a language unto itself that has extremely strict rules of grammar.one must be careful or even describe things by words when mathematical notation is cumbersome.