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

sirellwood · #1 $Old$ November 5th, 2009, 03:05 PM

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?

tonio · #2 $Old$ November 5th, 2009, 03:13 PM

Quote:

Originally Posted by sirellwood $View Post$

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?

You realize there is no m in your mathematical expressions, right?

Tonio

sirellwood · #3 $Old$ November 5th, 2009, 03:16 PM

argh sorry. my bad. edited :-)

flyingsquirrel · #4 $Old$ November 7th, 2009, 05:25 AM

Quote:

Originally Posted by sirellwood $View Post$

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

If $a=2$ , $n=2$ and $m=3$ then $a^{2n}+1 = 2^4+1= 17$ , $a^{2m}-1=2^6-1=63 = 3^2\times 7$ but 17 does not divide 63.

Media_Man · #5 $Old$ November 7th, 2009, 05:37 AM

Sirellwood:

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

Theorem: For $m>n, a^{2^n}+1|a^{2^m}-1$
Proof: $a^{2^m}-1=(a^{2^{m-1}}+1)(a^{2^{m-1}}-1)=(a^{2^{m-1}}+1)(a^{2^{m-2}}+1)(a^{2^{m-2}}-1)$ ...
Continue factoring the successive differences of squares k times until $m-k=n$ . QED

Your theorem is quite true, if rendered to the page in proper notation.

tonio · #6 $Old$ November 7th, 2009, 05:38 AM

Quote:

Originally Posted by sirellwood $View Post$

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

tonio · #7 $Old$ November 7th, 2009, 05:45 AM

Quote:

Originally Posted by Media_Man $View Post$

Sirellwood:

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

Theorem: For $m>n, a^{2^n}+1|a^{2^m}-1$
Proof: $a^{2^m}-1=(a^{2^{m-1}}+1)(a^{2^{m-1}}-1)=(a^{2^{m-1}}+1)(a^{2^{m-2}}+1)(a^{2^{m-2}}-1)$ ...
Continue factoring the successive differences of squares k times until $m-k=n$ . QED

Your theorem is quite true, if rendered to the page in proper notation.

I supposed that if the OP had wanted $a^{2^n}$ he could have writte a^(2^n) and not (a^2)^n, which means $(a^2)^n=a^{2n}$ . OTOH, he could have simply written a^(2n) and thus I think you're right and he meant $a^{2^n}$ .
Even using simple ASCII one must be careful or even describe things by words when mathematical notation is cumbersome.

Tonio

Media_Man · #8 $Old$ November 7th, 2009, 06:05 AM

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.