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

• Nov 5th 2009, 01:05 PM
sirellwood
Show 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 PM
tonio
Quote:

Originally Posted by sirellwood
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)^n)− 1

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

Tonio
• Nov 5th 2009, 01:16 PM
sirellwood
argh sorry. my bad. edited :-)
• Nov 7th 2009, 03:25 AM
flyingsquirrel
Quote:

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

If $\displaystyle a=2$, $\displaystyle n=2$ and $\displaystyle m=3$ then $\displaystyle a^{2n}+1 = 2^4+1= 17$, $\displaystyle a^{2m}-1=2^6-1=63 = 3^2\times 7$ but 17 does not divide 63.
• Nov 7th 2009, 03:37 AM
Media_Man
Exponents 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 AM
tonio
Quote:

Originally Posted by sirellwood
argh sorry. my bad. edited :-)

I'm almost sure that you must require quite a bit more, namely: not only m > n but in fact n | m...check this.

Tonio
• Nov 7th 2009, 03:45 AM
tonio
Quote:

Originally Posted by Media_Man
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)

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 AM
Media_Man
On language
Quote:

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