This question is pulled from a number theory practice set that did not provide an answer key. Link:  My reasoning is to prove that $n^7$ - n is divisible by 2,3, and 7 because they are the prime factors of 42. I've proven divisibility by 2, but I cannot prove divisibility by 3 or 7. Please let me know if this is the right path, and any ideas are welcomed proving divisibility by 3 or 7.
71362
I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
58168
I have just proved the divisibility rule for 3 and 9.  Let $n\in\mathbb{N}$. Let $m$ be a number that appears when you shuffle the digits in $n$. Show that $n-m$ is a multiple of 9.  Can anyone offer some help?
21480
I'm trying to prove the following, but can't seem to understand it. Can somebody help?     Prove $6^n - 1$ is always divisible by $5$ for $n \geq 1$.   What I've done:  Base Case: $n = 1$: $6^1 - 1 = 5$, which is divisible by $5$ so TRUE.  Assume true for $n = k$, where $k \geq 1$: $6^k - 1 = 5P$.  Should be true for $n = k + 1$  $6^{k + 1} - 1 = 5Q$  $= 6 \cdot 6^k - 1$  However, I am unsure on where to go from here.
26540
I have seen other criteria for divisibility by 7. Criterion described below present in the book Handbook of Mathematics for IN Bronshtein (p. 323) is interesting, but could not prove it. Let $n = (a_ka_{k-1}\ldots a_2a_1a_0)_{10} = \displaystyle{\sum_{j=0}^{k}}a_{k-j}10^{k-j}$. The expression $$ Q_{3}^{\prime}(n) = (a_2a_1a_0)_{10} - (a_5a_4a_3)_{10} + (a_8a_7a_6)_{10} -\ldots  $$ are called alternating sum of the digits of third order of $n$. For example,  $$ Q_{3}^{\prime}(123456789) = 789-456+123=456 $$ Proposition: $7 | n \ \Leftrightarrow \ 7 | Q_{3}^{\prime}(n)$.  proof. ??  Thanks for any help.
5854
Why is it that a number such as $108$, that is divisible by $2$ and $3$, is also divisible by $6$?   Is this true that all numbers divisible by two integers are divisible by their product?
32197
For what values of $m,n$ natural, do $2^n - 1$ is divisible by $2^{m+n} - 3^m$?  Thank you very much.
58624
I am trying to prove old test exercises and this one is resisting me.  I know you can believe that I meant $a^ 2 + b ^ 2$ is divisible by 4, but no, in the test it says that 4 is divisible by $a ^ 2+ b^2$.  Suppose that $ a,b \in \mathbb{Z}$  If 4 is divisible by $ a^2 + b^2$, then $a$ and $b$ are not odd.  I started like this:  Suppose that a and b are even, then $a = 2m$ and $b = 2n$ for some $m,n \in \mathbb{Z} $.   Thus  $$ a^2 = (2m)^2 \ \mbox{and}\ \  b^2= (2n)^2$$ $$ a ^2 + b^2= (2m)^2 + (2n)^2$$ $$ a^2 + b^2 = (4 m^2 + 4n^2) = 2(2m^2+2n^2) $$  Now $$2m ^ 2 + 2n^2\in \mathbb{Z} = k$$ Then  $$ a^2+b^2= 2k$$  Is this a correct proof?/ Any better ways to prove this.  Sorry for the lousy English. Thanks.
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:     $2^{3^{n+1}}+1 = 2^{3^n*3}+1$
32987
Prove that:  If $n$ is greater than 2, then $(3n)!$ is divisible by $n! \times  (n + 1)! \times  (n + 2)!$  From Barnard &amp; Child's "Higher Algebra".  I know that the highest power of a prime $p$ contained in $N$! is  $$ \left\lfloor{\frac{N}{p}}\right\rfloor + \left\lfloor{\frac{N}{p^2}}\right\rfloor + \left\lfloor{\frac{N}{p^3}}\right\rfloor ... $$  I'm unable to show that the formula above gives a higher value for $N = 3n$ than for the sum of its values when $N$ = $n$ , $n + 1$ and $n + 2$, considering the condition that $n &gt; 2$.
115930
Question :   Determine all natural numbers n such that 3   divides $n\cdot2^{n}+1$  Actually I don't have any ideas to approach but  my efforts :   I see $n=1,2,7,8,13,14$ so I think :   $n=6k+1$ and $n=6k+2$ $k\in \Bbb{N} $  If I'm not wrong but I don't know how ? I prove it ?
117078
Show using induction that $n^3-n$ is divisible by 6 $\forall n\ge1, \quad n \in \mathbb{N}$   First off i show that the basis step: $1^3-1=0, \quad \frac{0}{6}=0$  Now I factorised it and set it equal to a multiple of 6: $\mathbf{n(n+1)}(n-1)=6A$  Assuming the result is true for k terms, and trying for $k+1$ terms:   $\mathbf{k(k+1)}(k+2)=6B$  I'm stuck here, I realise that the bold terms are the same, but $k+2$ and $n-1$ are not. Could someone show me what do to next to solve this.  Also is it possible to prove this using modular arithmetic?  Thanks,
8423
If $n\in\mathbb N$ and $4^n+2^n+1$ is prime, prove that there exists an $m\in\mathbb N\cup\{0\}$ such that $n=3^m$.  I.e. if $4^n+2^n+1$ is prime, prove that $n=3^m$, where $m\in\mathbb N\cup\{0\}$.  I don't know how I could solve this, I don't have any ideas. I know that $4^n+2^n+1$ is obviously odd. Assume $m=0$, i.e. $n=1$. Then $3$ is prime. So that doesn't disprove what we're trying to show.  So assume $m&gt;0$. Then $4^n+2^n+1&gt;3$, so $4^n+2^n+1$ is either $3k+1$ or $3k+2$, where $k\in\mathbb N$. So either $4^n+2^n$ or $4^n+2^n-1$ is divisible by 3. I think it's not something that could help out at solving this. Just a little observation. I don't have any ideas.
93780
$7^{2n} -48n - 1$ is divisible by 2304 for all $n \in N$  so I did, P(n) : $7^{2n}-48n-1=2304k$ (k meaning there is an integer which will depend on n)  Prove base case $P(1): 7^2 - 48(1)-1 = 0$, proving that $k=0$ in the meaning of 'divides'   $n=k+1$  $7^{2(k+1)}-48(k+1)-1 = 2304l$ (l being for some integer)  I don't know how to proceed any further. could someone show me whats next?  thanks alot
4839
We can prove that there is no integer $n&gt;1$ such that $2^n-1\mid 3^n-1$. This leads to the following question:     Is it true that for every pair of primes $p&lt;q$ there are only finitely many integers $n$ such that $p^n-1\mid q^n-1$? Are there two primes $p&lt;q$ and an integer $n&gt;p+q$ such that $p^n-1\mid q^n-1$?Is it true that if $n&gt;6$ then $2^n-1\nmid 5^n-1$?   Edit: Here are some examples: \begin{array}{ll} 2^{36}-1\mid 41^{36}-1, &amp;3^{12}-1\mid 97^{12}-1,\\ 5^{6}-1\mid 37^{6}-1, &amp;7^{4}-1\mid 151^{4}-1. \end{array}  Now I prove there is no integer $n&gt;1$ such that $2^n-1\mid 3^n-1.$  Proof: Denote $A=2^n-1$ and $B=3^n-1$. If $n$ is even then $3$ divides $A$ but not $B$, a contradiction.  If $n$ is odd then $A\equiv -5\pmod {12}$. Since every prime greater than $3$ is $\equiv \pm1,\pm5 \pmod {12}$, some prime factor $p$ of $A$ must be congruent to $\pm5 \pmod {12}$. As $p \mid B$, we have $3^{n+1}\equiv 3 \pmod p$. Since $n+1$ is even, we get $(\frac{3}{p})=1$, a contradiction again.
40705
I have to prove that any number of the form $3n+2$ has a prime factor of the form $3m+2$. Ive started the proof  I tried saying by the division algorithm the prime factor is either the form 3m,3m+1,3m+2.  Case 1: $3m$      suppose $3m$ does divide $3n+2$. That means $3mr=3n+2$ where $r$ is some integer. But we get $mr-n= 2/3$ and since $mr-n$ is an integer that's a contradiction. Thus $3m$ does not work.  Case 2: $3m+1$ Same argument here  Case 3: $3m+2$ By the fundamental theorem of arithmetic every integer is divisible by a prime. But not sure what to do from here.
11718
Show that $n^2-1$ is divisible by $8$, if $n$ is an odd positive integer. Please help me to prove whether this statement is true or false.
120494
For which natural n is $3^n+5^n$ divisible by $3^{n-1}+5^{n-1}$? I only got that $3^n+5^n=8k$ when $n$ is odd. How to solve this one?