I stumbled upon this problem when reading the resource provided by AoPS, on    Here is the problem:  Show that if $a$ and $b$ are positive integers, then  $$\left(a + \frac12\right)^n + \left(b + \frac12\right)^n$$  is an integer for only ﬁnitely many positive integers $n$.  I'm unsure on how to begin - any help would be much appreciated!   Thanks
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. :(
108730
I have found two formulas. I want to connect them!  The number of ways in which a given positive integer $n≥3$ can be expressed as a sum of three positive integers $x,y,z$ (i.e. $x+y+z=n$) , subject to $x≤y≤z$ is the integer closest to $\frac{n^2}{12}$. If $x+y+z=n$ where $x,y,z$ are postive integers then the number of ways is ${n-1 \choose 2}$ .  The difference between these two formula is the condition $x≤y≤z$. Can we connect these two ?? Please help me! I mainly wants to know how one implies the other.
123218
I am having a problem understanding this proof.   Suppose that $\sqrt{2}$ was rational.   Then there would exist positive integers $a,b$ such that $\sqrt{2}=\frac{a}{b}$. Consequently, the set $S=\{k\sqrt{2} : k \text{ and } k\sqrt{2}$ are positive integers $\}$ is nonempty set of positive integers. Therefore, by the well-ordering property, $S$ has a smallest element, say, $s = t \sqrt{2}$.   We have  $s \sqrt{2}-s = s\sqrt{2}-t\sqrt{2}=(s-t)\sqrt{2}$.  Because $s\sqrt{2} = 2t$ and $s$ are both integers,  $s\sqrt{2} - s = s\sqrt{2}-t\sqrt{2} = (s-t)\sqrt{2}$  must also be an integer.   Furthermore, it is positive, because $s\sqrt{2}-s = s(\sqrt{2}-1)$ and $\sqrt{2} &gt; 1$.   It is less than $s$, because $\sqrt{2} &lt; 2$ so that $\sqrt{2}-1 &lt; 1$. This contradicts the choice of $s$ as the smallest positive integer in $S$. It follows that $\sqrt{2}$ is irrational.  My main problem here is how they got the set $S$. My second confusion is the algebra part $s \sqrt{2}-s$. How did they come up with this?   If someone can explain this proof to me it would be great! thank you
63872
I am currently working on this problem from Hardy's Course of Pure Mathematics and have gotten stuck near the end. I was wondering if someone could help me determine what to go next.  Question  If $a_1, a_2, ...,a_n$ are all positive and $S_n=a_1+a_2+...+a_n$ then:  $(1+a_1)(1+a_2)...(1+a_n) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}$  My Attempt  Proof by induction:  I had first shown that it was true for $n=1$ and $n=2$.  Now suppose that it is true for n. Then it must also be true for $n+1$  $(1+a_1)(1+a_2)...(1+a_n)(1+a_{n+1}) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!} + \dfrac{(S_{n})^{n+1}}{(n+1)!}$  Define: $(1+a_1)(1+a_2)...(1+a_n)=x$ and $1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}=y$  Then: $x+xa_{n+1} \le y+\dfrac{(S_{n})^{n+1}}{(n+1)!}$  By the induction assumption, we know that $x \le y$.  What I can't figure out  From this point, what I think the next natural step would be is to show that  $xa_{n+1} \le \dfrac{(S_{n})^{n+1}}{(n+1)!}$  I know this rearranges to:  $xa_{n+1} \le \dfrac{(S_{n})^{n}}{n!} \times \dfrac{S_n}{(n+1)} $  And feel there may be something I can do here, but haven't been able to figure anything out.
25198
For each $n$ consider the expression $$\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]$$  I am trying to prove by induction that this is an integer for all $n$.  In the base case $n=1$, it ends up being $1$.   I am trying to prove the induction step:   if $\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]$ is an integer, then so is $\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$.   I have tried expanding it, but didn't get anywhere.
89238
Given any $30$ positive integers we can find some integers so their sum is divided by $30$.  I don't have any idea where to start. First I would like just some hints, then If possible I would like do discuss different ways to solve the problem.
28336
Find all positive integers $n$ such that $a \equiv a^{-1} \pmod{n}$ for all invertible $a$ modulo $n$.   I found that $n = 1,2,4,6,8,12,24$ satisfy this. How can we prove that these are all of them?
115559
I have been trying to show these two inequalities hold for all positive integers n, but I don't know how to proceed at all... I have tried playing around with them but I didn't find anything helpful. I have also tried induction but I couldn't make it work.
9583
So lets say I have $$ n^{2} \le n! $$  For what positive integers is this not true? $n=2$ and $ 3$  Base case? $$n=4 \implies  16 \le 24 $$  What is the inductive hypothesis and how do I show the inductive proof? Thanks
104504
How to find all positive integers $m,n$ such that $n(n+1)(n+2)=6m^3$ ?    I can see that $m=n=1$ is a solution, but is it the only solution ?
40575
Show that if $a$ and $b$ are positive integers with $(a,b)=1$ then $(a^n, b^n) = 1$ for all positive integers n  Hi everyone, for the proof to the above question,  Can I assume that since $(a, b) = 1$, then in the prime-power factorization of a and b, they have no prime factor in common, when they are taken to the $nth$ power, they will still have no prime factors in common, and so $(a^n, b^n) = 1$ for all positive integers n.  I think I'm jumping to conclusions here again, if so, leave some tips on how to do the proof properly, thanks :)  And also, I do not know how to approach the reverse problem where if $(a^n, b^n) = 1 then (a, b) = 1$, any guidance will be much appreciated!
93512
Show that if $a,b$ are rational numbers and $x$ is a positive real number then $x^a$$x^b$ $=$ $x^{(a+b)}$  I honestly have no idea how to even do this. Anyone have any hints or a good explanation?  Thanks in advance!
16666
Prove that for for all $p,q\in \mathbb{Z}$, $q&gt;0$ we have: $$ \left| \sqrt{2}-\frac{p}{q} \right| &gt; \frac{1}{3q^2}. $$  To be honest, I do not know where to start - any help would be appreciated.
10213
Let $m$ be a positive integer and $n$ a nonnegative integer. Prove that $$(n!)\cdot(m!)^n|(mn)!$$  I can prove it using , but I have to use the lemma that $$ \dfrac{\displaystyle\left(\sum_{i=1}^na_i\right)!}{\displaystyle\prod_{i=1}^na_i!} \in \mathbb{N} $$ I believe that it can be proved using the lemma, since  he has mentioned so at the end of his answer. Any help will be appreciated. Thanks.
114720
I got stuck on this inequality for a day. If $p$ is positive integer, then the problem becomes too easy, but I can't find how we deal with the general case when p can be any positive number. Can someone give me a hint? I really appreciate     Prove that $|x^p - y^p| \le p|x-y|(x^{p-1} + y^{p-1})$ provided that $1 \le p \lt \infty$ and $x, y \ge 0$
42158
What is the smallest positive integer $n$, such that there exist positive integers $a$ and $b$, with $b$ obtained from $a$ by a rearrangement of its digits, so that   $a – b = 11\dots 1$ (The number of '$1$'s equal to $n$)?   Is there any example that can satisfy the equation?
102108
Could anybody help me some ideas on the below problem:  Let $n$ be a positive integer. Prove that there are infinite pairs of positive integer $(x\, y)$ such that $$x^2+x+1=(y^2+y+1)(n^2+n+1).$$ Thanks in advance.