I need help!  I need to prove that for any $2 \le n,k$ positive integers $(n - 1)^2 \mid n^k -1$ if and only if $(n - 1) \mid k$  Thanks!
14858
How do you prove this?  $$\left(n-1\right)^2\mid\left(n^k-1\right)\Longleftrightarrow\left(n-1\right)\mid k$$
15359
Show that if $a$ and $b$ are positive integers, then $$\left(a +\frac12\right)^n + \left(b+\frac{1}{2}\right)^n$$is an integer for only finitely many positive integers $n$.   I tried hard but nothing seems to work. :(
111685
I need to prove  $2n \leq 2^n$, for all integer $n≥1$ by mathematical induction?  This is how I prove this:  Prove:$2n ≤ 2^n$, for all integer $n≥1$  Proof: $2+4+6+...+2n=2^n$  $i.)$ Let $P(n)=1         P(1): 2(1)=2^1\implies 2=2$.        Hence, $P(1)$ is true.  $ii.)$  Assume that $P(n)$ is true for $n=k$, i.e, $2+4+6+...+2k=2^k$, and prove that $P(n)$ is also true for $n=k+1$, i.e, $2+4+6...+2(k+1)=2^{(k+1)}$  from the assumption add $2(k+1)$ on both sides so we have   $2+4+6...2k+2(k+1)=2^k+2(k+1)$  I'm confused with $2^k+2(k+1)$, I don't know how to make $2^k$ be equivalent to $2^{k+1}$.  I feel i'm doing something wrong.   Any help would be appreciated!
93701
Question: Prove $\forall n\geq 2,n\in\mathbb{Z}$, $(n+1)\mid(n^3+1)$    I know that it is possible to solve by factoring $n^3+1$ and showing that $n+1$ is a multiple, but I would like to show this via induction.  Attempt: Using induction:  Let $n=2:3\mid 9$, so true for $n=2$  (1) Assume true for $n=k:(k+1)\mid(k^3+1)$, so $(k+1)r=k^3+1$ for some $r\in\mathbb{Z}$  (2) Let $n=k+1:(k+2)\mid(k+1)^3+1$, so $(k+2)s=(k+1)^3+1$ for some $s\in\mathbb{Z}$  From (2) we have $(k+1)s+s=(k+1)^3+1$, subbing (1) into (2), I get $k^3+1+s=(k+1)^3+1$  Therefore $k^3+s=(k+1)^3$.  Setting $s=3k^2+3k+1$, this holds.  Where am I going wrong, is the proof valid, any help would be greatly appreciated.
92116
Prove that for fixed positive integers $k$ and $n$, the number of partitions of $n$ is equal to the number of partitions of $2n + k$ into $n + k$ parts.  I tried this with different values of n but can't seem to get a bijection between the two partitions.
29081
I want to prove by induction that $a^n \mid b^n$ implies that $a \mid b$ holds for all integers $n\geq 1$.  Clearly for $n=1$ this is true, since if $a \mid b$, then $a \mid b$.  Suppose this is true for some $n = k$. Then $a^k \mid b^k$, so $a|b$.  $a^k \mid b^k$ means there exists some integer $m$ such that $b^k = ma^k$, and $a \mid b$ means there exists an integer $r$ such that $b=ra$.  Then we can multiply both sides of the above equality by any constant, namely $b$.   Then $b \cdot b^k = bma^k$  so $b^{k+1} = (ra)ma^k = rma^{k+1}$  which means that $a^{k+1} \mid b^{k+1}$ completing the induction step, and thus proving that for all integers $n \geq1$ the original statement holds.  Are there any logical holes in the proof?
85926
My Number Theory book says that for $n, m$ be positive integers and $a&gt;1$, then     $(a^n -1)\mid(a^m -1)$ if and only if $n\mid m$.   I understand the proof for only if part, but in if part the autor says "it is clear". However a tried to prove that but a get stuck. Can you give a hint?
56685
Suppose $p$ is a prime and $a$ and $k$ are positive integers. Prove that, if $p \mid a^k$, then $p^k\mid a^k$ also.  I have already proven that if $a,b,n\in\mathbb{N}$ and if $a^n\mid b^n$, then $a\mid b$. I tried using induction:  $k=1 \Rightarrow p\mid a^1 \Rightarrow p^1\mid a^1$. Assume the statement holds for some $k$. Then if $p\mid a^{k+1}\Rightarrow p\mid a^ka$. Since $p$ is prime, $p\mid a$ or $p\mid a^k$. If $p\mid a$, we are done. If $p\mid a^k$ then by assumption, $p^k\mid a^k$ and so by the lemma, $p\mid a$ and so $p^{k+1}\mid a^{k+1}$.  Does this work?
98279
I have derived the inequality if $k&gt;1$,  ${n(n-1)⋯(n-k+1)\over k!} ({1\over n})^k&lt;{(n+1)n⋯(n-k+2)\over k!} ({1\over n+1})^k$  But, my problem is how to use this inequality to prove that if $n\geq1$, $(1+{1\over n})^n&lt;(1+{1\over n+1})^{n+1}$
55194
Suppose that when $n=k$  $(k\geq4)$, we have that $k!&gt;2^k$.  Now, we have to prove that $(k+1)!\geq2^{k+1}$ when $n=(k+1)$ $(k\geq4)$.  $$(k+1)! = (k+1)k! &gt; (k+1)2^k  \text{ (since }k!&gt;2^k)$$  That implies $(k+1)!&gt;2^k2$, since $(k+1)&gt;2$ because of $k$ is greater than or equal to 4.  Therefore $(k+1)!&gt;2^{k+1}$  Finally  we may conclude that $n!&gt;2^n$ for all integers $n\geq4$    Solved, thank you!
24893
Prove by induction that $\forall n \in\mathbb N$, $3\mid4^{n}-1$.  1) For $n = 1$ the statement is obviously true.  2) Now what about $n+1$? I was thinking of writing $4^{n}-1$ as $2^{2n}-1$ and then $4^{n+1}-1 = 2^{2n+2}-1$ but that got me nowhere.
3416
Can you help me to prove that $$(x+y)^n\leq 2^{n-1}(x^n+y^n)$$ for $n\ge1$ and $x,y\ge0$.  I tried by induction, but I didn't get a result.
94376
How can I show that   let $n$ and $k$ be integers.  $n(n+1)2^{n-2}=\sum_{k=1}^{n}k^2\dbinom{n}{k}$  It seems a bit confusing to me on the left hand side. You have a set of $n$ people on a team and you keep picking $k$ people who will play a match and you pick before that a captain $k^2$.
29500
Prove that for any positive integer $n$,       $$n^{1/n}  &lt; 1 + \sqrt{\frac{2}{n}}.$$   This due to Victor Linis, Eureka, Vol. 2, No. 2, February 1976, p. 29.   Hint:      Use the binomial theorem.
215
Use induction to prove the following: $1! + 2! + .... + n! &lt; (n + 1)!$  Base case:  $n = 1$  $1! &lt; 2!$ true   Inductive step: Assume that $1! + 2! + .... + k! \le  (k + 1)!$ is true   let $n = k + 1$  $1! + 2! + .... + (k + 1)! &lt; (k + 2)!$  $1! + 2! + .... + k! + (k + 1)! &lt; (k + 2)!$  $1! + 2! + .... + k! + (k + 1)! &lt; (k + 2)!$  $(k + 1)! + (k + 1)! &lt; (k + 2)!$  $2(k + 1)! &lt; (k + 2)!$  $2(k + 1)(k)(k - 1).. &lt; (k + 2)(k + 1)(k)(k - 1)....$  $2 &lt; (k + 2)$   the right is always bigger for k are non negative integer.
25053
$\forall m,n\in\mathbb Z$ , $m\ge1$ and $n\ge1$ how to prove that  $$\frac{(mn)!}{m!(n!)^m}$$ is an integer?  Thanks in advance.
64537
Prove that for every positive integer $n \ge 2$ $$\sum^n_{k=1}k \sqrt{\binom nk}\leq\sqrt{2^{n-1}n^3}$$ I tried it by induction but I didn't know how to end it.