I am trying to prepare for my final exam in NT. Any help would be appreciated. These questions are just for study purposes, I messed them up on my homework, etc.

Reduced Residue System Proof

Show that if a=b(mod n) then (a,n)= (b,n)

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- Jul 27th 2009, 11:06 AMdiddledabbleHelp Preparing for Final Exam
I am trying to prepare for my final exam in NT. Any help would be appreciated. These questions are just for study purposes, I messed them up on my homework, etc.

Reduced Residue System Proof

Show that if a=b(mod n) then (a,n)= (b,n) - Jul 27th 2009, 03:54 PMGamma
$\displaystyle a \equiv b $ (mod n) $\displaystyle \Leftrightarrow n|(b-a) \Rightarrow nk=b-a$ for some integer k

Now we show the set of divisors of a and n are the same as the set of divisors of b and n.

Let d divide a and n.

$\displaystyle nk=b-a \Rightarrow b=nk+a$ then d divides the RHS so it divides b as well.

Let d divide b and n.

$\displaystyle nk=b-a \Rightarrow b-nk=a$ then d divides the LHS so it divides a as well.

Thus the set of divisors is the same, so their greatest element is the same. so (a,n)=(b,n) - Aug 3rd 2009, 10:13 PMstreethot
First of all, assuming that $\displaystyle a\equiv b \ (mod \ m)$ we have that if $\displaystyle d|m$ and $\displaystyle d|a$ then $\displaystyle d|b$.

Proof:

$\displaystyle a-b=m.k$

$\displaystyle m=d.u$ and $\displaystyle a=d.s$

then

$\displaystyle b=d(s-u)$

Now, let $\displaystyle d=(a,m)$ and $\displaystyle e=(b,m)$. Then $\displaystyle d|m$, $\displaystyle d|a$ so $\displaystyle d|b$. Hence $\displaystyle d|e$.

Similarly, $\displaystyle e|m$, $\displaystyle e|b$ so $\displaystyle e|a$. Hence $\displaystyle e|d$. Therefore $\displaystyle d=e$ - Aug 4th 2009, 11:55 AMdiddledabbleAnother Proof
Show that if 2n+1 is prime, then n must be a power of 2.

- Aug 4th 2009, 12:01 PMGammaNot true
- Aug 4th 2009, 12:08 PMGamma
Maybe what you meant was $\displaystyle 2^n+1$ being prime implies $\displaystyle n=2^i$ for $\displaystyle i\in \{0,1,2,...\}$. If so, I would check out this:Fermat number - Wikipedia, the free encyclopedia

For completeness, it should state the n could also be 0. - Aug 4th 2009, 12:09 PMdiddledabble
- Aug 4th 2009, 12:12 PMGamma
You can't prove something is true if there is a counter-example... that is the beauty of math.

$\displaystyle 2|2^{n+1}$ so it can never be prime if n is a natural number (not including 0), so that would be a pretty dull question. Probably it is what I suggested above. - Aug 4th 2009, 12:13 PMstreethot
Gamma,

Show that**IF**2n+1 is prime, then n must be a power of 2. - Aug 4th 2009, 12:16 PMdiddledabble
Streethot, Everytime I read it I think of a different way to go with it. I think you have it though. 2^n+1 must be a prime to start with.

- Aug 4th 2009, 12:17 PMGamma
- Aug 4th 2009, 12:23 PMdiddledabble
But if it should have been $\displaystyle 2^n+1$ let n=2 that is not a square. I just don;t get this one and fermat seems way to tough to be on the exam.

- Aug 4th 2009, 12:30 PMstreethot
Sorry, i thought the question was about $\displaystyle 2^{m}+1$ with $\displaystyle m=2^n \ \ n\in\mathbb{N}$

- Aug 4th 2009, 12:33 PMGamma
The statement should be

IF $\displaystyle 2^n+1$ is prime, then n is a power of 2 or n=0.

the proof is given here Fermat number - Wikipedia, the free encyclopedia

There is nothing complicated about the proof supplied in the link. It is a proven fact, this has to be what the teacher was going for, the other possibilities are either not true or would never be asked on an exam by someone with a PhD in math.

I am not following what your problem is with this, it is a very straightforward if then statement. The proof goes by contradiction in supposing that n is not a power of two, then it has an odd prime factor. Then you show how you factor $\displaystyle 2^n+1$ which makes it NOT prime, a contradiction. - Aug 4th 2009, 12:35 PMGamma