Does $30$ always divide  $n^5-n$ where $n \in \mathbb{Z^+}$ ?
10357
How can I prove that prove $n^5 - n$ is divisible by $30$? I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$ Now, $n(n-1)(n+1)$ is divisible by $6$. Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$. My guess is using Fermat's little theorem but I don't know how.
70311
I need to show that $n$ divides $\phi(a^n -1)$ where $a, n$ are positive integer without using concepts of abstract algebra  I know that $$a^n\equiv 1\pmod {a^n-1}$$  How do I proceed from there?
109607
Consider the set of positive integers $n \in \mathbb {Z}&gt;0$ such that $\dfrac{n}{2}$ is a perfect square, $\dfrac{n}{3}$ is a perfect cube, and $\dfrac{n}{5}$ is a perfect fifth power; that is to say, such that there exist integers $a, b, c \in \mathbb{Z}$ such that $\dfrac{n}{2} = a^2$, $\dfrac{n}{3} = b^3$, $\dfrac{n}{5} = c^5$.  Let $n_0$ denote the smallest element in this set (it exists by the well-ordering principle). Determine the numerical value of $n_0$.
53577
We now that Fermat's last theorem is true so there are not positive integer solutions to $$x^n+y^n=z^n$$ for $n\in\mathbb{N}$ and $n&gt;2$.  But what about if $n\in\mathbb{R}$ or $n\in\mathbb{R}^+$?
109360
I need to show that it doesn't exist any $k \in \mathbb{Z}$ that satisfies $41 \mid (k^5-2)$.  Can I apply Fermat's Little Theorem to prove that in anyway? How? The only thing that came close when I was thinking was that $$41 \mid (k^5-1)(k^5+1)(k^{10}+1)(k^{20}+1)$$ (because $41\mid (k^{40}-1)$ by Fermat's Little Theorem), but from that I couldn't conclude anything. Thanks.
39269
Let $z$ be a complex number, and let $n$ be a positive integer such that $z^n = (z + 1)^n = 1$. Prove that $n$ is divisible by 6.  For this problem I am stumped...how should I begin? Also there's a hint for it:  From $z^n = 1$, prove that $|z| = 1$. What does the equation $(z + 1)^n = 1$ tell you? What do the resulting equations tell you about $z$?  Could someone give me a hint on where to begin? thanks in advance
76065
Find all $n\in\mathbb N^+$ such that the sum of the digits of $5^n$ is equal to $2^n$.   I've solved this, but I think the proof is a bit weird and I wonder if there's a better one.    My proof:  We can see that $n\neq 1$, $n\neq 2$, $n=3$, $n\neq 4$ and $n\neq 5$.  Let $n&gt;5$. Notice that $5^6$ has $5$ digits, while $\frac{2^6}{9}=7\frac{1}{9}$, so $5^6$ has to have at least $8$ digits in order for the sum of the digits to be able to at least reach $2^6$.  Everytime $1$ is added to the exponent of $5$, the amount of digits $5^n$ gets is at most $1$, while $\frac{2^{n+1}}{9}-\frac{2^n}{9}=\frac{2^n}{9}&gt;3$, so the lowest amount of digits it has to have to reach $2^n$ becomes higher by at least $4$.  When $n=6$, the amount of digits $5^n$ has is already lower than the amount it has to have in order to reach $2^n$, and I've just proved that the amount it has to have in order to reach $2^n$ grows faster than the amount of digits $5^n$ has, therefore $5^n$ will never be able to reach $2^n$ when $n&gt;5$.  Hence the only solution is $n=3$.
62722
Use induction to prove that, for all $n \in \mathbb{Z}^+$, $2\mid (n^2 − n)$.   That is, I am supposed to use induction to prove that $(n^2 − n)$ can be divided by $2$ when $n$ is a positive integer.  I've tried the following: $$\begin{split} (n+1)^2 − (n+1) &amp;= (n+1)(n+1) - (n+1)\\ &amp;=(n+1)(n-1)+2 - (n+1)\\ &amp;=n^2 +n -n -1 +2 -n -1\\ &amp;=n^2 -n, \end{split}$$ but I'm not particularly good at maths so I have no idea if this is even correct, and if it is, what it even means.
34404
I tried to prove this question by first considering the possible last digit of $p$ when $p=n^2+5$, but that reasoning got me nowhere. Then I tried to prove it by contrapositive, and however I just couldn't really find where to start. Hence I'm here asking for some hints (only hints, no solution please).  Many thanks, D.
65408
Let $a,b,m,n$ be positive integers such that g.c.d.$(a,b)=1$ and $a^m+b^m|a^n+b^n$ , then is it true that $m$ divides $n$ ?
9981
$n\in \Bbb N$ Prove that if $n^2$ is divided by 3, then also n can also be divided by 3.  I started solving this by induction, but I'm not sure that I'm going in the right direction, any suggestions?
10019
I did some mathematical induction problems on divisibility   $9^n$ $-$ $2^n$ is divisible by 7. $4^n$ $-$ $1$ is divisible by 3. $9^n$ $-$ $4^n$ is divisible by 5.   Can these be generalized as $a^n$ $-$ $b^n$$ = (a-b)N$, where N is an integer? But why is $a^n$ $-$ $b^n$$ = (a-b)N$ ?  I also see that $6^n$ $- 5n + 4$ is divisible by $5$ which is $6-5+4$ and $7^n$$+3n + 8$ is divisible by $9$ which is $7+3+8=18=9\cdot2$.  Are they just a coincidence or is there a theory behind?  Is it about modular arithmetic?
98002
Find the set of all positive integers $n$ with the property that the set $M=\{n, n + 1,n + 2,n + 3,n + 4,n + 5\}$ can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.   If $n=1$ them $M=\{1,2,3,4,5,6\}$ and there is no such partition, so $n \ge 2$.  If the prime $p|n$ then either $p|2$ or $p|3$ or $p|5$ which means either $p=2$ or $p=3$ or $p=5$.  Suppose $n=2k$. Then $M=\{2k, 2k + 1,2k + 2,2k + 3,2k + 4,2k + 5\}$. I have no idea how to proceed.
18071
How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divisible by $4$ so $8|(n^3 - 1)(n^3 + 1)$. I'm stuck at proving divisibility by $9$ and $7$
44709
I'm trying to prove using induction that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have:   Base case: $P(1) = (1) + (8) + (27) = 36, 36$ can be divided by $9$ so the base case is valid Inductive step: let $P(n)$ be the statement $9$ divides $n^3 + (n+1)^3 + (n+2)^3$. Assume $P(k)$ is true, so $9$ divides $k^3 + (k+1)^3 + (k+2)^3$.   And this is where I'm stuck. I'm not sure how to demonstrate that $9$ divides $P(n)$ when $n = k+1$. If someone could step me in the right direction that would be awesome. Thanks!
25419
Prove by induction that $5^n - 1$ is divisible by $4$.   How should I use induction in this problem. Do you have any hints for solving this problem?  Thank you so much.
54489
In $\mathbb{Z}_n$ the elements are fully partitioned between the units and the zero-divisors. I believe this is the case, am I correct?  Now, I take it this does not hold true in general, there may be rings with elements that are neither units nor zero divisors?