Questions about Fermat numbers?

**JRichardson1729** · Oct 24th 2010, 03:05 PM

I have recently happened upon an interesting problem that I cannot solve. Here it is:
a. Factorise $x^3+1$ .
b. Factorise $x^n+1$ when ${n}\geq{3}$ is odd. Use this to prove the following:
i. If $2^n+1$ is prime, then $n$ must be a power of $2$ .
ii. Show that $2^3^2+1$ is not prime.

I can't seem to make any headway on the problem, in particular the very last part. All help is appreciated on this problem.

**Dinkydoe** · Oct 24th 2010, 04:10 PM

For (a),(b) show that $x^n+1 = (x+1)(x^{n-1}-x^{n-2}+\cdots +x^2-x+1)$ , as n is odd.

i) Use (b) to show that if $n=2^kq$ , where q is odd, then

the factorisation of $x^n+1$ is obtained by substituting $x=x^{2^k}$ into $(x^q+1)$

So this means if n is not a power of 2....?

ii) You can show that $2^{32}+1$ is divisable by 5.

Can you do modulo-calculations?

**JRichardson1729** · Oct 25th 2010, 08:04 AM

Originally Posted by Dinkydoe

For (a),(b) show that $x^n+1 = (x+1)(x^{n-1}-x^{n-2}+\cdots +x^2-x+1)$ , as n is odd.

i) Use (b) to show that if $n=2^kq$ , where q is odd, then

the factorisation of $x^n+1$ is obtained by substituting $x=x^{2^k}$ into $(x^q+1)$

So this means if n is not a power of 2....?

ii) You can show that $2^{32}+1$ is divisable by 5.

Can you do modulo-calculations?

I cannot do modulo-calculations, however I don't think that $2^3^2+1$ is divisible by $5$ .

**undefined** · Oct 25th 2010, 09:47 AM

Originally Posted by JRichardson1729

I cannot do modulo-calculations, however I don't think that $2^3^2+1$ is divisible by $5$ .

$2^3^2+1$ was proven composite by Euler. Explanation in this PDF

http://www.maa.org/editorial/euler/H...oring%20F5.pdf

**JRichardson1729** · Oct 25th 2010, 10:40 AM

Originally Posted by undefined

$2^3^2+1$ was proven composite by Euler. Explanation in this PDF

http://www.maa.org/editorial/euler/H...oring%20F5.pdf

I knew of this, however I don't see how it uses the factorisation of $x^n+1$ when ${n}\geq{3}$ and is odd as the question asks for.

**undefined** · Oct 25th 2010, 11:13 AM

Originally Posted by JRichardson1729

I knew of this, however I don't see how it uses the factorisation of $x^n+1$ when ${n}\geq{3}$ and is odd as the question asks for.

Since 32 is not odd, the property does not directly apply. Maybe it is an error in the problem statement.

**Dinkydoe** · Oct 25th 2010, 01:49 PM

Sorry, made a lame error in (ii)....but indeed. I agree with undefined. I don't see any relation between (ii) and the previous.

**JRichardson1729** · Oct 25th 2010, 02:24 PM

I found this video, YouTube - the 5th Fermat number is not a prime (german), that shows this, but, unfortunately, I don't think that it uses the factorization of $x^n+1$ when ${n}\geq{3}$ and is odd and I don't know how to adapt it so that it does (if that is possible at all).

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