Why is it true that if 7 divides 91 then $(2^7-1) $ divides $(2^{91}-1)$?  1) $2^{91}-1$  $7|91  \implies (2^7-1)|(2^{91}-1)$  $\implies 2^7-1$ is factor   2) $2^{1001}-1$  $7|1001  \implies (2^7-1)|(2^{1001}-1)$  $\implies 2^7-1$ is factor
17137
In respect to a larger proof I need to prove that $(n-m) \mid (n^r - m^r) $ (where $\mid$ means divides, i.e., $a \mid b$ means that $b$ modulus $a$ = $0$). I have played around with this for a while now but to no avail and would be grateful if you could help me.
517
Let $a,b&gt;2$ be positive integers. We need to show that $2^a +1$ is not divisible by $2^b-1$.  Could any one give me hint?
19148
Why is it that $\forall n \in N$, $2^n$ is not divisible by $3$?  I can prove it easily by induction, but I don't understand the intuition of why this is true. Could anyone please supply the intuition? [This has been answered.]  Moreover, I understand that there's also a proof of this fact that uses modular arithmetic. How does this proof go? [Also answered - it follows immediately from the definition of the modulus operation.]
75422
I need help with this question.     Let $a$ and $b$ be positive integers such that $a^{n}+n$ divides $b^{n}+n$ for every natural number $n$. Show that $a=b$.   Any help would be appreciated! Thanks!
41621
Is it possible for some integer $n&gt;1$ that $2^n-1\mid 3^n-1$ ?   I have tried many things, but nothing worked.
69922
Let $a &gt; 2$  be an odd number and let $ n $ be a positive integer. Prove that $ a $ divides $1^{a^{n}} + 2^{a^{n}} + ... + (a-1)^{a^{n}}$ ? (ref. Titu Andreescu, Number theory, page no. 5).  I am not getting any clue as to how to approach the solution.
66390
I am stuck on this problem, and was wondering if anyone could help me out with this. The question is as follows:  Let $n$ be an integer such that $n ≥ 1$. Prove that $6$ divides $n(n + 1)(n + 2)$. Note: An integer $a$ divides an integer $b$, written $a|b$, if there exists $q ∈ Z$ such that $b = qa$. Alternatively, $a|b$ if dividing $b$ by $a$, $b ÷ a$, results in an integer.  Should I do a proof by induction? All help/input is appreciated!
66574
Show that $13$ divides $2^{70} + 3^{70}$.   My main problem here is trying to figure out how to separate the two elements in the sum, and then use Fermat's little theorem. So how can I separate the two?  Thanks!
1391
A little bird told me that if $2^n+1$ is prime, then $n$ is a power of $2$. I tend not to trust talking birds, so I'm trying to verify that statement independently.  Suppose $n$ is not a power of $2$. Then $n = a \cdot 2^m$ for some $a$ not a power of $2$ and some integer $m$. This gives $2^n+1 = 2^{a \cdot 2^m}+1$. Now I suspect there's a way to factor that, but I don't see how. Can someone give me a hint?
67668
I've been struggling with this all day today. I imagine it's not very hard, but my algebra skills are terrible. So, how can I show that if $m&gt;n$ and $a$ is a positive integer, then $$a^{2^n}+1 \mid a^{2^m}-1.$$ I just can't get a coherent picture of what I'm supposed to do. If this too (see yesterday's question) boils down to the Unique Factorization Theorem, I would be very grateful for any intuition on how to think about this!
957
Show that    $$1^{p-1} + 2^{p-1} +\ldots + (p-1)^{p-1} \equiv -1 \mod p$$   So, I use Fermat's little theorem, that is if $p$ does not divide $a$, then $a^{p-1}$ is congruent to $1$ (mod $p$). But I'm still confused. If $1^{p-1}$ is congruent to $1$ (mod $p$), $2^{p-1}$ is congruent to $1$ (mod $p$), and $p$ does not divide $p-1$, so $(p-1)^{p-1}$ is congruent to $1$ (mod $p$). How does that equal $-1$ (mod $p$)?
123037
It is well known that $(n!)^2$ divides $(2n)!$.  Does it follow that $(n!)^3$ divides $(3n)!$ and so on up to $(n!)^n$ dividing $(n^2)!$?  If yes or no, could you provide the details behind the argument?  As a quick test, I tried $n=3$ and $n=4$ which worked.  Thanks,  -Larry
69085
Prove by mathematical induction that $2^{3^n}+1$ is divisible by $3^{n+1}.$   I'm currently trying to finish a task that requires me to use mathematical induction to prove that $2^{3^n}+1$ is divisible by $3^{n+1}$. This is the first time that the divider is not a simple integer and I'm having trouble trying to prove it.   This is what I managed to do:     First step:      For $n=1$:      $2^3+1=9$      $3^2=9$      $L=P$      Second step: I assume that for some numbers $n&gt;=1$  $2^{3^n}+1$ is divisible by $3^{n+1}$      Third step: (induction hypothesis):   $2^{3^{n+1}}+1$ is divisible by $3^{n+1}$   I'm getting stuck here, trying to do some algebra and simplify the dividend:     $2^{3^{n+1}}+1 = 2^{3^n*3}+1$
117753
Prove that $$gcd(2^{2^m}+1, 2^{2^n}+1)=1$$ if $m, n$ are positive integers such that $m \neq n$.  A Hints to solve the problem is also given in the book as follows:  Let $m&gt;n$. Then $2^m=2^{n}2^{m-n}=2^n.2k$ for some integer $k$, (since $m&gt;n$). Let $2^{2^n}=x$. Then $2^{2^m}-1=x^{2k}-1$ and it is divisible by $x+1$.  I am still unable to solve. Please help.
36993
Prove that if $p$ and $q$ are odd primes and $q$ can divide $a^p-1$, then either $q$ divides $a-1$ or else $q=2kp+1$ for some integer $k$.  I thought the supposition is $a\equiv 1 (mod\ q)$. So $$(a-1)(a^{p-1}+a^{p-2}+...+1) \equiv 0 (mod\ q)$$ Then I proved $q$ may divide $a-1$. But I don't know how to check that $q=2kp+1$. thks
9261
Hi I'm practicing some number theory and I've come across the following problem,  Suppose $n \geq 1$ is an odd integer that divides $a^2 - 1$ for some integer $a &gt; 1$. Prove that,  $$n = \gcd(a-1, n)\gcd(a+1, n)$$  Now I know that the $\gcd(a-1, a+1) = 1$ if $a$ is even and $2$ if $a$ is odd and I believe this is relevant but I can't solve the problem. I've not really any idea where to start so even just help with that would be great.  Thanks
88266
This is a part from a homework. I solved all examples apart from this one. So the task is: We know that $3$ divides $a^2 + b^2$. Prove that $3$ divides $a$ and $3$ divides $b$. I cannot think of anything useful. I know that $a^2 + b^2 = (a + b)^2 - 2ab$, but I don't see how it can help me :(  Best regards, Petar