Find a positive integer $n$ such that $n/2$ is a square, $n/3$ is a cube and $n/5$ is a ﬁfth power.  Above problem was given in the text “An Introduction to the Theory of Numbers” by “Ivan Niven and Zuckerman".  I tried to generalize the problem statement as:  Find a positive integer $n$ such that $n/2$ is a square, $n/3$ is a cube, $n/5$ is a ﬁfth power, $n/7$ is a seventh power ... $n/p_{k}$ is a $p_{k}$-th power.   I have found the solution for this and trying to generalize it more. Now, How can I know that this work is authentic i.e. it has not been done by anyone earlier? Any sort of help is welcomed.
122922
Show that there are infinitely many $N$ such that $\frac{N}{2}$ is a perfect square, $\frac{N}{3}$ is a perfect cube, and $\frac{N}{5}$ is a perfect fifth power.    A hint is given with this question, which is: $7^{30}$ is a perfect square, a perfect cube and a perfect fifth power.  Can anyone solve this? Thanks!
114441
Let $a \geq 2$ be a positive integer such that $\log_{10}a$ is irrational. Then for any positive integer $N$ there exist infinitely many powers $a^n$ of $a$ whose decimal representation begins with $N$.   I am not sure how to use the fact that $a^n$ begins with $N$. How should I prove this?
65526
Are there any positive $n$ for which $ n^4+n^3+n^2+n+1$ is a perfect square?  I tried to simplify   \begin{align*} n^4+n^3+n^2+n+1 &amp;= n^2(n^2+1)+n(n^2+1)+1\\  &amp;= (n^2+n)(n^2+1)+1 \\   &amp;= n(n+1)(n^2+1)+1 \end{align*} Then I assumed that the above expression is a square; then  $$ n(n+1)(n^2+1)+1 = k^2$$  $$  \begin{align*} n(n+1)(n^2+1) &amp;= (k^2-1) \\    &amp;= (k+1)(k-1) \end{align*} $$   Then trying to reason with prime factors, but cannot find a concrete proof yet.
110783
If an irrational number of the form $a+\sqrt{b}$ is raised to the power $n$, with $a, b \in \mathbb{Z}$, and $n \in \mathbb{N}$, what can be said about the integral part of the resulting number? For instance, is it an even number? If yes what are the conditions on $n$? If odd, what are the conditions?  I think I should start from considering $(a+\sqrt{b})^n= p+q$ where p is an integer and q is a proper fraction. I don't know how to proceed further. Please help. And also suggest a website that can give more of such properties.
21066
We have the sum of squares of $n$ consecutive positive integers: $$S=(a+1)^2+(a+2)^2+ ... +(a+n)^2$$ Problem was to find the smallest $n$ such, that $S=b^2$ will be square of some positive integer. I found an example for $n=11$. Now, I'm trying to prove, that if $2&lt;n&lt;11$ there is't any solution. So I need some help with this: If $n,a,b \in \mathbb{N}$ and $n&lt;11$ prove that equation  $$ n\cdot a^2 + n(n+1)\cdot a + \frac{1}{6}n(n+1)(2n+1)=b^2 $$  can't be solved. Or if I'm wrong, find counterexample.  The only idea I have is: to consider the remains $\operatorname{Mod}[b^2,n]$ and $\operatorname{Mod}[\frac{1}{6}n(n+1)(2n+1),n]$. For each $2&lt;n&lt;11$, but it is very long.
5493
I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam.     If $n = 1 + m$, where $m$ is the product of four consecutive positive   integers, prove that $n$ is a perfect square.   Now since $m = p(p+1)(p+2)(p+3)$;     $p = 0, n = 1$ - Perfect Square      $p = 1, n = 25$ - Perfect Square      $p = 2, n = 121$ - Perfect Square   Is there any way to prove the above without induction? My approach was to expand $m = p(p+1)(p+2)(p+3)$ into a 4th degree equation, and then try proving that $n = m + 1$ is a perfect square, but I wasn't able to do it. Any idea if it is possible?
7034
Let $n &gt; 1$ be an integer. Then $2^n - 1\nmid 3^n - 1$. I don't know how to prove it. Can anybody help me, please?  In general, for a fixed positive integer $a &gt; 1$, has $a^n - 1|(a +1)^n - 1$ any integer solutions?
61352
If $n$ be a positive integer greater than $1$, then prove that $$2^n&gt;1+n\sqrt{2^{n-1}}$$    I found this problem under $AM \ge GM$ chapter. Help me to solve this problem using  $AM \ge GM$. Thanks in advance.
71373
If $n\geq2$, then $n!$ is not a perfect square. The proof of this follows easily from Chebyshev's theorem, which states that for any positive integer $n$ there exists a prime strictly between $n$ and $2n-2$. A proof can be found .  Two weeks and four days ago, one of my classmates told me that it's possible to prove that $n!$ is never a perfect square for $n\geq2$ without using Chebyshev's theorem. I've been trying since that day to prove it in that manner, but the closest I've gotten is, through the use of the prime number theorem, showing that there exists a natural number $N$ such that if $n\geq N$, $n!$ is not a perfect square. This isn't very close at all. I've tried numerous strategies over the past weeks and am now trying to use the Sylow theorems on $S_n$ to somehow show that $|S_n|$ can't be square (I haven't made any progress).  Was my classmate messing with me, or is there really a way to prove this result without Chebyshev's Theorem? If it is possible, can someone point me in the right direction for a proof?  Thanks!
79685
I have to prove that for any positive integer $n$ at least one of $n+3$ and $n^2+3n+3$ is not a perfect cube. Is there a methodical way to solve this problem?  I managed to solve it by contradiction, trying to force it into some sort of algebraic identity.  The solution I found is as follows:     If $n+3$ and $(n^2+3n+3)$ are both cubes then the product is a cube. The product is $n^3+6n^2+12n+9=(n+2)^3+1$. There are no two positive cubes with distance one between them, this is the contradiction.   For some reason this seems rather artificial, it makes me think that if one wished to make a very hard problem like this then he could look for much more complicated identities, and then actually finding the contradiction would be a lot harder. Is there a "safer" way to tackle these problems, without having to rely on "luck". (Of course it is not 100% luck because practice makes you a lot faster at finding the way to put together the expression)
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!
85930
An integer $n$ is a perfect square if and only if it can be expressed as the square of some other integer i.e. there exists some integer $a$ such that $n=a^2$.    Prove that if $n$ is a perfect square, then either $n \equiv 1 ~(\mod 4)$, or $n \equiv 0 ~(\mod 4)$.  How can I prove this?
47264
Let $a$ and $b$ be positive integers such that $a | b^2, b^2 | a^3, a^3 | b^4, b^4 | a^5, \cdots $ Prove that $a = b$.  My way is as follows: Let $A=v_p(a), B=v_p(b)$ be the exact power of a prime $p$ dividing $a$ and $b$ respectively. Then by assumption, $(2n-1)A \leq 2nB, n=1,2,3,\cdots.$ Hence $A \leq (2n/(2n-1))B$. Taking limit as $n$ tends to infinite yields $A \leq B.$ Likewise, $2nB \leq (2n+1)A, n=1,2,3,\cdots,$ by a similar argument, we have $B \leq A.$ So that $A=B$ and we are done. But I am rather unsatisfied with the proof because it involves limit. I wanna ask if there is any other alternative proof, thank in advances.
57512
This is a 2015 IMO problem.  It seems difficult to solve.     Find all triples of positive integers $(a, b, c)$ such that each of the numbers $ab-c$, $bc-a$, and $ca-b$ is a power of $2$.   Four such triples are $(a,b,c)=(2,2,2),(2,3,2),(3,5,7),(2,6,11)$.
118824
Prove that for any positive integer $n$, the number of odd integers among the binomial coefficients $\binom{n}{k}$ $(0 \leq k \leq n)$ is a power of $2$.   I am thinking of trying to prove that the number of odd binomial coefficients is equal to 2 to the power of the number of $1$s in the binary form of $n$. How should I go about proving this?
67668
I've been struggling with this all day today. I imagine it's not very hard, but my algebra skills are terrible. So, how can I show that if $m&gt;n$ and $a$ is a positive integer, then $$a^{2^n}+1 \mid a^{2^m}-1.$$ I just can't get a coherent picture of what I'm supposed to do. If this too (see yesterday's question) boils down to the Unique Factorization Theorem, I would be very grateful for any intuition on how to think about this!
92212
While playing around with programming, a problem suddenly came across my mind.      Given $n$ let's say $2^{64}$ (64-bit unsigned integer). Find a positive integer $x$, $1 \leq x \leq n$, such that $x$ has maximum divisors.   My attempt was: Let $n = p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}.$ The number of divisors is given by: $$(a_1 + 1) \times (a_2 + 1) \times ... \times (a_k + 1)$$  We need to find the maximum of these products. Since $(a_1 + 1)_{max} \implies {a_1}_{max}$, the product becomes: $a_1 \times a_2 \times .... \times a_k$ Let $P = a_1 \times a_2 \times .... \times a_k$. The maximum of $P$ means $a_i = a_j$ where $1 \leq i, j \leq k$.   So I guess my question is how can we find a number that is less than $n$ which has this property?     Thanks,