Prove that the positive integers $n$ that cannot be written as a sum of $r$ consecutive positive integers, with $r &gt; 1$, are of the form n = $2^l$ for some $l ≥ 0$. I wrote just  k+(k+1)+.....(k+r-1) as $2^l$ and then by applying AP formula, it gets me to something like $r/2 ( 2k + (r-1))$ = $2^l$ which leads to nowhere.. .I did to get a contradiction. I have shown my approach and this is not a homework problem.I need some help to move from here or a new way to solve  the problem in a less cumbersome manner. Thank you.
122530
I have been wondering about this question for a while (I am almost sure that I read it in some contest mathematics book a while ago)     Determine which positive integers cannot be written as a sum of consecutive positive integers.   All numbers will be considered positive integers in the following unless stated otherwise.   I was thinking as follows: odd numbers will always work since if we take an odd $m$, we have  $$ m=2\cdot k+1=k+(k+1). $$   Then I realised that all numbers having at least one odd prime factor will also work. Take $n$ as a number having an odd prime factor and let this odd prime factor be $m$ and let $n=m\cdot k$. Then $$ n=\underbrace{k+k+\ldots+k}_{m \ \text{times}} $$ and since $m$ is odd, say $m=2\cdot l+1$ we have $$ n=\underbrace{k+k+\ldots+k}_{l \ \text{times}}+k+\underbrace{k+k+\ldots+k}_{l \ \text{times}} $$ and we can start moving $1$s one by one from the left side of the middle "$k$" to the right side of it $$ n=\underbrace{k+k+\ldots+(k-1)}_{l \ \text{times}}+k+\underbrace{(k+1)+k+\ldots+k}_{l \ \text{times}} $$ $$ n=\underbrace{k+k+\ldots(k-2)+(k-1)}_{l \ \text{times}}+k+\underbrace{(k+1)+(k+2)+\ldots+k}_{l \ \text{times}} $$ this way we end up in a sum of consecutive integers. It may happen that some integers on the left will be negative which is undesired but since these numbers are consecutive integers we always be able to cancel out the negative ones without disturbing the structure of the uncancelled ones. To see this in action let us take $22=11\cdot 2$. This would give $$ 2+2+2+2+2+2+2+2+2+2+2 $$ so $$ 2+2+2+2+1+2+3+2+2+2+2 $$ the next step gives $$ 2+2+2+0+1+2+3+4+2+2+2 $$ it seems that now we are out of numbers but we just continue moving an increasing number of $1$s to the right $$ 2+2+(-1)+0+1+2+3+4+5+2+2 $$ $$ 2+(-2)+(-1)+0+1+2+3+4+5+6+2 $$ finally $$ \underbrace{(-3)+(-2)+(-1)+0+1+2+3}_{=0}+4+5+6+7 $$ so we get $$ 22=4+5+6+7. $$  Now we have a method to generate the desired expression for all odd numbers and all the ones having at least one odd prime factor. Which numbers are left? The ones having ONLY even primes in their prime factorisation. These are of the form $2^n$. I was thinking what do these have in common and I thought that they are separated by a factor of $2$. The first of these are $1$ for $n=0$ and $1$ cannot be expressed in the desired for so if I could show that  $$ m \ \text{cannot be expressed}\Rightarrow 2m \ \text{cannot be expressed} $$ I am done. So I was trying to show the contrapositive $$ 2m \ \text{can be expressed}\Rightarrow m \ \text{can be expressed}. $$ So assume that $2m$ can be expressed. Now, eihter $m$ is odd in which case we are done since odds were easy to express or it is even, that is $m=2l$. From here the argument repeats. Eihter $l$ is odd in which case we are done or it is even, so $l=2k$ and so on. After a while we factored out all of the $2$'s and we end up with and odd number which we can express and we are done.  So integers of the form $2^n$ cannot be expressed in this form.  Is my reasoning correct?  If not please point out where I made a mistake. If it turns out to be correct then we are all happy I can sleep well tonight but it would be nice to see a more rigorous proof of this so if someone can provide one I am more than happy to read it. Since this is a contest question I expect it to have a really elegant solution and I do not consider mine an elegant one.
53246
I'm working through the problems in Niven's number theory book, and problem 46 in section 1.2 (page 19) has me stumped.     Prove that there are no positive integers $a, b, n &gt; 1$ such that $(a^n - b^n) | (a^n + b^n)$.   I've tried playing around a bit (e.g. noticing that $a^n - b^n$ must divide $2a^n$ and $2b^n$), but in general I've just been going around in circles.  Can anyone please provide a (small) hint in the right direction?
73198
Show that any non-negative rational integer may be written as the sum of two squares if it is of the form $$2^ip_1^{k_1}p_2^{k_2}\cdots p_m^{k_m}q_1^{2j_1}q_2^{2j_2}\cdots q_n^{2j_n}$$ where the $p_1,p_2,\dots,p_m$ are all primes congruent to $1\bmod 4$, the $q_1,q_2,\dots,q_n$ are all primes congruent to $3\bmod 4$ and $i,k_1,k_2,\dots,k_m,j_1,j_2,\dots,j_n$ are all non-negative integers.   I am completely blank on how to form a solution, so help would be much appreciated.
115678
I'm trying to apply an induction proof to show that $((n+1) 2^n - 1 $ is the sum of $(i 2^{i-1})$ from $0$ to $n$.   the base case: L.H.S = R.H.S we assume that $(k+1) 2^k - 1 $ is true. we need to prove that $(k+2) 2^{k+1} - 1$   My try to prove 3 is as follows:  $(k+2) 2^{k+1} - 1$  $(k+2) (2^k * 2) - 1$ , from 2: $2^k = 1/(k+1)$  $(k+2) (2 / (k+1)) - 1$  $(k+1) 2 - 1$  My question is, how could I get to $2^k$ from the last line to prove this formula is right?
40340
Prove that for any positive integers $m$ and $n$ , there exists a set of $n$ consecutive positive integers each of which is divisible by a number of the form $d^m$, where $d$ is some positive integer not equal to $1$.  I don't know how to approach this question.
19561
I am trying to prove that the product of three consecutive positive integers is never a perfect power. Can anyone point to gaps in my proof and/or post an alternate solution?  Let the three positive consecutive integers be $n - 1$, $n$ and $n + 1$ and let $(n - 1)n(n + 1) = h^k, k \ge 2$. Note that $\gcd(n - 1, n) = 1$ and $(n, n + 1) = 1$ implies that $n$ itself must be a perfect power and of the form $z^k$ (which is apparent once we look at the canonical representation of $h^k$). That means $n^2 - 1$ must be a perfect power itself and of the form $a^k$. If $n$ is odd, $n = 2m + 1$ for some $m \in \mathbb{N}$ i. e. $(n - 1)(n + 1) = 2m(2m + 2) = 2^2m(m + 1) = a^k$. Using the fact that $(m, m + 1) = 1$, $m$ and $m + 1$ must be perfect powers themselves and so $k\leq 2$. Coupled with fact that $k &gt; 1$, we infer $k = 2$. So, $n^2 - 1 = a^2$ which implies $n$ is not a natural number. We are left with the case when $n$ is even. Let $n = 2t + 1$. $(n + 1, n - 1) = 1$ as both are of (by the Euclidean algorithm). That means $n - 1$ and $n + 1$ are perfect powers. So, let $(n - 1) = b^k$ and $(n + 1) = l^k$. So, $l^k - b^k = 2$ for natural $l$ and $p$ and $k&gt;1$. I will prove that the diophantine equation has no solutions. Consider the function $f(k) = l^k - b^k - 2$. $f'(k) = \frac{1}{l}e^{k\log l} - \frac{1}{b}e^{k\log b}&gt;0$ if and only if $e^{k(log\frac{l}{b})}&gt;\frac{l}{b}$. Taking logarithm again, we get an equivalent condition $(k - 1)\log\frac{l}{b}&gt;0$ which is true as $k&gt;1$ and $l&gt;b$ as log is a monotonically increasing function.  Is my proof correct? Please feel free to chip in with your own solutions!
46126
QUESTION:   Prove that for all positive integers  $k\leq n$:  $\sum_{i=0}^{k}{n \choose i} (-1)^i = {n-1 \choose k}(-1)^k$  MY THOUGHTS:  INDUCTION : (assuming k even) - We assume:  ${n \choose 0} - {n \choose 1} + ... + {n \choose k-2} - {n \choose k-1} = {n-1 \choose k-1}$  We need to show:  ${n \choose 0} - {n \choose 1} + ...-{n \choose k-1} + {n \choose k} = {n-1 \choose k-1}$  My issue here is that we can say LHS equals ${n-1 \choose k-1} + {n \choose k}$, which is close, but not quite Pascal's Rule.   Another way: I thought looking to find some bijection that can be applied.   Let's assume n is even. Writing out the left and sum and moving all the odd (negative) terms to the RHS gives (note that the last term on the RHS is simply the original RHS:  ${n \choose 0} + {n \choose 2} + ... + {n \choose k} = {n \choose 1} + {n \choose 3} + ... + {n \choose k-1} + {n-1 \choose k}$    I am stuck here, but how I worked out in my mind was that:  $\bigg\{$set of even sized groups of  $\leq$ n people$\bigg\}$ -->  $\bigg\{$set of odd sized groups of  $\leq$ n-1 people$\bigg\}$ + ${n-1 \choose k}$  Then possibly, it would have something to do with adding or removing person n. Let me know your thoughts or suggestions.
106527
I believe I see that $a_n = 2^n(a_0+1) - 1$ but am somewhat uncertain where to proceed afterwards. I am a complete beginner at number theory and would appreciate it if someone could point me in the right direction--surely there is some obvious argument I am missing.
121709
Suppose $$ab \equiv 0 \pmod n$$ and that $a$ and $b$ are positive integers both less than $ n$. Does it follow that either $a | n$ or $b | n$?  If it does follow, give a proof.  If it doesn’t, then give an example.   For this question is this a suitable answer: No it doesn't. $3$ and $4$ are both less than $6$. $$3 \times 4 \equiv 0 \pmod {6}$$ but $4 \not \mid 6$.
49274
Let $n$, $p$ and  $r$ be three positive integers. Prove that for $n \geqslant 3, r&gt;1$, $$\sum_{k = 0}^{n} k! \neq p^\text{r}$$   SOURCE: BANGLADESH MATH OLYMPIAD (Preaparatory Question)  I am not so familiar with such a this kind of problem. Seeing that problem, I became little bit curious about what the text states and what will be its conception?   I couldn't solve the problem. Moreover, I don't know about the formula of $\sum k!$. Is there any? And I couldn't realize the essence of $p^\text{r}$ and what the reason is behind the fact the summation can't be equal to $p^r$ for some integer $p$.  Any kind of reference or conception will massively help me start with some approach to solve the above problem. Thanks for your support and effort in advance.
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.
1132
Show that $n$ does not divide $2^n - 1$ where $n$ is an integer greater than $1$?  Clearly $2^n - 1$ is an odd integer, therefore, let $n$ be an odd integer and it divides $2^n - 1$. We can write $2^n - 1 = (2-1)(2^{n-1} + \cdots \cdots + 1) = (2^{n-1} + \cdots \cdots + 1)$ From here on I don't know how to proceed?
54425
How to prove the following? Should I use induction or something else?     Let $n$ and $r$ be positive integers with $n \ge r$. Prove that   $$\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}.$$   Attempted start:  Basis step: ${\binom{1}{1}} = {\binom{2}{2}}$ true.  Where do I go from here?
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. :(
52105
For any positive integer $n$ (using integer division only), let $P(n)$ denote the number of ways in which the integer $(n^2+n)/4$ can be expressed as a sum of exactly $n /2$ distinct elements of the set $\{1,2,3,\dots, n\}$.    What is $P(n)$ in terms of n?  Specifically, how exponential is it? Is this less than $2^{n/2}$?
76288
Problem:Let $k$ be an odd positive integer. Prove that $(1+2+3...+n)|(1^k+2^k+3^k+...+n^k)$ for all positive integers $n$.  My Attempt: Proof (By Induction on $k$): Let $P(k)$ be the proposition that $(1+2+3...+n)|(1^k+2^k+3^k+...+n^k)$ for all positive integers $n$ when $k$ is odd. Then $P(1)$ and $P(3)$ are trivially true. So for $k\geq 5$, we assume that $P(k)$ holds. We would like to show that $P(k+2)$ also holds. In order to do that consider the following sums  $$1^{2}(1^k+2^k+3^k+...+n^k)=1^{k+2}+2^k+3^k+...+n^k$$ $$2^{2}(1^k+2^k+3^k+...+n^k)=2^2*1^k+2^{k+2}+2^2*3^k+...+2^2*n^k$$ $$..$$ $$n^2(1^k+2^k+3^k+...+n^k)=n^2*1^k+n^2*2^k+n^2*3^k+...+n^{k+2}$$ Note that $$1^2(1^k+2^k+3^k+...+n^k)+2^2(1^k+2^k+3^k+...+n^k)+...+n^2(1^k+2^k+3^k+...+n^k)\equiv 0\pmod {\frac{n(n+1)}{2}}$$  Therefore, if we denote $\sum_{i=1}^{n}i^2=s$, then    $(t)(1^k+2^k+3^k+...+n^k)-t+\sum_{i=1}^{n}i^{k+2}\equiv 0\pmod {\frac{n(n+1)}{2}}$ which implies that $P(k+2)$ holds. Hence by PMI we are done.  Now I am a bit hesitant in convincing myself whether this proof is correct or not. This is because I have never seen induction performed in this manner. Any reasons as to why this argument falls would be much appreciated.
6905
Let $(a_n)$ be a strictly increasing sequence of positive integers such   that: $a_2 = 2$ and $a_{mn} = a_m a_n$ for $m, n$ relatively prime. Show that $a_n = n$, for every positive integer $n$.   This is a result apparently due to Paul Erdős, and  has a proof by induction.  I tried like this, $a_{10}=a_2a_5$. After this what we can do?    [Editor's comment] Possibly due to the apparent simplicity of the conditions it may be difficult to appreciate the subtleties of this question. If we try and construct a counterexample like $a_3=4$, $a_4=5$, $a_5=6$, then the requirements dictate $a_6=8$, $a_{10}=12$, $a_{15}=24$. At this point we realize that we have been speeding. For monotonicity forces $a_9\le 11$, and therefore $a_{18}\le22&lt;a_{15}$, violating the requirements. It is not obvious why something similar ruins all the modifications to the sequence $a_n=n$. [/comment, JL]