How to prove the following?:  Let $m,n\in \mathbb{N}$ ; $ \;m^{2}\mid n^{2}\Longrightarrow \;\;m\mid n$  Just a hint please. I tried two ways but did not work.
124253
Let $a$ and $b$ be positive integers. Prove that: If $a^2$ divides $b^2$, then $a$ divides $b$.   Context: the lecturer wrote this up in my notes without proving it, but I can't seem to figure out why it's true. Would appreciate a solution.
7687
I came across the following in the book Handbook of Mathematical Induction: $$ 19\mid (2^{2^n}+3^{2^n}+5^{2^n}),\quad n\in\mathbb{Z^+}\tag{1} $$ Apparently, this problem is not so bad if you think about it in a somewhat "outside of the box" sort of way, but it is supposed to be quite challenging otherwise.   Question: How might one prove $(1)$ using induction or otherwise? The "outside of the box" answer has been provided but with details oranged out so as not to spoil it for anyone who may come up with that solution individually.
25374
I'm trying to prove by induction that for all $a,b \in \mathbb{Z}$ and $n \in \mathbb{N}$, that $(a-b) \mid (a^n-b^n)$. The base case was trivial, so I started by assuming that $(a-b) \mid (a^n-b^n)$. But I found that: \begin{align*} (a-b)(a^{n-1}+a^{n-2}b + a^{n-3}b^2+...+b^{n-1}) &amp;= a^n-b^n. \end{align*}  Doesn't this imply that $(a-b) \mid (a^n-b^n)$ as $a^{n-1}+a^{n-2}b+\dots+b^{n-1}$ is clearly an integer? This obviously isn't a proof by induction, but is there anything wrong with taking this approach to prove this result, other than the fact that it isn't what is being asked?   Aside, I'm having trouble getting started with the proof by induction. I've tried making the above equivalence useful during the induction step, but it doesn't seem to be helpful, so any hints would be greatly appreciated!
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Prove that for for all $p,q\in \mathbb{Z}$, $q&gt;0$ we have: $$ \left| \sqrt{2}-\frac{p}{q} \right| &gt; \frac{1}{3q^2}. $$  To be honest, I do not know where to start - any help would be appreciated.
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I am getting crazy with this one.     Suppose $a_n=(1^2+2^2+3^2+\ldots+n^2)^n$ and $b_n=n^n(n!)^2$. Show that $a_n&gt;b_n$ for all $n$.    They suggest to use the AM-GM inequality.
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How do I prove if $\sup\{ (n!)^{1/n} | n \in\Bbb N\} = +\infty$?  I don't know how to properly explain it
110541
I tried proving inductively but I didn't really go anywhere. So I tried:  Let $3|n(n+1)(n+2)$. Then $3|n^3 + 3n^2 + 2n \Longrightarrow  3|(n(n(n+3)) + 2)$  But then?
111645
Prove by induction that $3\mid n^3 - n$.   I'm having an argument with my professor whether my exam was right or not. Before I sign a formal complain to get a review on my exam, I'd like to be sure it's correct. My answer:  Proof by induction:  Proposition: $\forall n \in\mathbb{ N}$ (belongs to natural numbers), $3 | (n^3 - n)$;  Basic Step: $P(0) = 0^3 - 0$ is divisible by 3.  Inductive Step:     $p(n+1) = (n+1)^3 - (n+1)$      $= n^3 + 3n^2 + 3n + 1 - n - 1$      $= n^3 + 3n^2 + 2n$      $= n^3 + 3n^2 + 2n - (n^3 - n)$      $= 3n^2 + 3n$   By induction, $p(n+1)$ is true if:      $3\mid n$ and $3\mid m$ and $3 | n - m$     to prove that, assume that $n$ is $3p$ and $m$ is $3q$, therefore $3 | 3p - 3q$.  My professor wanted:     $= (n^3 -n) + 3n^2 + (2n + n)$
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$A$$\rightarrow$$(B$ $\vee$ $C$ ) , $B$ $\rightarrow$ $C$ $\vDash$ $A$ $\rightarrow$ $D$  I think it's wrong but I have no idea how to prove.
84845
I'm having some difficulty with this proof. Could you help me?     Let $a,b&gt;0$ and $c_n = (a^n + b^n)^{1/n}$. Show that $$\lim_{n\to \infty}c_n = \max \{a,b\}.$$   Thank you.
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$n\in \Bbb N$ 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?
74920
Suppose that $0\neq x\in\mathbb{R}$ and $x + \frac1x\in\mathbb{Z}$. Prove that, for all  $n\ge1$, $x^n + \frac1{x^n}\in\mathbb{Z}$.  I can't figure out and understand the question. Can you give me some hints ?
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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.
99342
$a \mid b$ and $b \mid a$ if and only if $a= \pm b$.  How do I go about this proof? Step by step explanation please!
111036
Prove that for any $m,n\in\mathbb{Z}^+$, the number $2^n-1$ is divisible by $(2^m-1)^2$ iff the number $n$ is divisible by $m(2^m-1)$.  My work so far:  It's an if and only if statement, so the common approach is to split the proof into two "if then" proofs ($A\rightarrow B \text{ and } B\rightarrow A$).  Starting with $A\rightarrow B$, for a direct proof, I would assume $A$ and prove $B$. I can use the definition of divisibility:  Assume $n=m(2^m-1)k$ for some integer $k$  Then I need to prove that $2^n-1=(2^m-1)^2h$ for some integer $h$  So I'm thinking I need to start with the expression $2^n-1$ and transform it to $(2^m-1)^2h$, or vice versa, and I need to use the assumption $n=m(2^m-1)k$ somehow.  I notice that the LS of the equality contains the variable $n$ while the RS contains $m$, and the assumption relates $n$ to $m$, so that could be a way forward. For example, simple substitution gives $2^{m(2^m-1)k}-1$, but I'm not sure what I could do from here. Perhaps there's a divisibility theorem that could help?  Other things I notice: $2^n-1$ and $(2^m-1)^2$ are always odd. Also, the divisors in both cases are multiples of $2^m-1$.  Any help/hints appreciated.
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I am looking to prove the following statement false:  Let $k$ be a positive integer, then $k\mid n$ if and only if $k\mid n^2$.  So I am trying to find a $k$ where this does not hold but after aimlessly trying numbers I was curious if there is a methodical way to find $k$.
59317
I was asked to show that $\forall n\in \mathbb N$ there exist a $p\in \mathbb N^\ast$ such that $$(1+\sqrt{2})^n = \sqrt{p} + \sqrt{p-1}$$   I used induction but it wasn't fruitful, so I tried to use the binomial expansion of $(1+\sqrt{2})^n$ but it seems I lack some insight to go further.   Any hint is welcomed.