Show that if $x,y,z$ are positive integers, then $(xy + 1)(yz + 1)(zx + 1)$ is a perfect square if and only if $xy +1, yz +1, zx+1$ are all perfect squares. Find infinitely many triples (a,b,c) of positive integers such that a, b, c are in arithmetic progression and such that $ab + 1, bc +1, ca+1$ are perfect squares.   First of all I tried to prove that $xy +1, yz +1, zx+1$ were relatively prime. I didn't succeed.  $$\gcd(xy+1, yz+1) = \gcd(y(x-z)+1,yz+1)$$  For the second problem, I have discovered the solution $(2,4,12)$ which isn't in arithmetic progression.  I have tried googling it but didn't find an easy and satisfactory solution.  Any help would be appreciated.
123678
Show that if $x, y, z$ are positive integers, then $(xy + 1)(yz + 1)(zx + 1)$ is a perfect square if and only if $xy + 1, yz + 1, zx+1$ are perfect squares.
60238
For any integer $a,b$ let $N_{a,b}$ denote the number of positive integer $x&lt;1000$ satisfying $x= a \mod 27$ and $x=b \mod 37$. Then which of them is correct         there exist $a,b$ such that $N_{a,b} =0$    for all $a,b$, $N_{a,b}=1$   for all $a,b$, $N_{a,b}&gt;1$   there exist $a,b$ such that $N_{a,b}$ and  there exist $a,b$ such that $N_{a,b}=2$.      Since we know that by Chinese Remainder Theorem if $m_1, m_2, .., m_k$ are pairwise relatively prime positive integers, and if $a_1, a_2, .., a_k$ are any integers, then the  simultaneous congruences $x ≡ a_1 (mod\;m_1), x ≡ a_2 (mod\;m_2), ..., x ≡ a_k (mod\;m_k)$ have a solution, and the solution is unique modulo $m$, where  $m = m_1m_2⋅⋅⋅m_k$. Then how we will approach by acoording to this theorem. So please someone help me and explain it.
30520
I haven't found any useful method to solve the following problem: Prove that if $x,y,z\in\mathbb{Z}$ and $x^3+y^3=3z^3$ then $xyz=0$.  Source:
89037
Let $n$ be an integer such that for any integer $d$, if $d | n$ then $d + 1 | n + 1$.   Show that $n$ is a prime number or equal to $1$  My workout...  Suppose $n = xy$ for some postive integers $x$ and $y. x|n$, and hence we assume without loss of generality $x + 1|n + 1.$   Then $x|xy + 1$. However, $xy + x$ is divisible by $x$.  This implies $x − 1$ is divisible by $x$, which is an impossibility if $x$ is not equal to $1$.  Therefore whenever $n$ is decomposed into two factors, one must turn out to be $1$. Hence $n$ must be a prime.  I am looking for any other method which solves this quickly and more quickly...
41594
I want to prove $gcd(a, a+2)$ = 1 or 2 for every integer $a$.   I know that (a, a+2) is equal to the least positive value of $ax + (a+2)y$, where $x, y \in \mathbb{Z}$. However, I'm a little stuck at this point. I'm thinking I should split apart the problem into two cases, one where $a$ is odd and one where $a$ is even? Any suggestions? Thanks.
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!
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.
85116
how to prove $\forall n$  : $p_1p_2...p_n +1$ isn't the perfect square. $p_i$ is $i$th prime number.
15235
I am asked to prove that:       For integers $n, x,y &gt; 0$, where $x,y$ are relatively prime,        every $n \ge (x-1) (y-1)$ can be expressed as $xa + yb$, with nonnegative integers $a,b \ge0$.   How should I approach this? I have very limited knowledge in number theory.
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.
49227
Prove that for each $n$, there exist $n$ consecutive integers, each of which is divisible by a perfect square larger than $1$. Can any one help with this? I was stuck because I know how to solve for a particular $n$ value, but I was helpless in the generalized form.
61717
Prove that $x^x+y^y=z^z$ doesn't have integer solutions  To be honest, I don't see any way to start this problem, I tried for hours but it's not as easy as I thought.  Any hints?  As you can see in the comments, there is a solution for natural numbers, the problem is when the set is extended to integers solutions.  *This is a problem for a national olympiad in my country, this olympiad usually puts all the solutions but i didn't found a solution for this problem in the page, i don't know why.  Possible answer:  Property:   Be $A,B,C$ $\in \Bbb R$ $\ne 0$ such that $A+B=C$. Then exist $U,V \in {A,B,C}$ such that:  $$2|U| \gt 2|V| \ge |U|$$  Proof: Write $C'=-C$, then $A+B+C'=0$. The numbers $A,B,C \ne 0$ by hypothesis, and they can't be the three of the same sign because then they can't add $0$, so there are two of the numbers $A, B, C'$ that have the same sign and the other has a different sign.  Amplifying the equality $A+B+C'= 0$ $\,$ by $-1$ if its necessary, we can assume that there are two of them (which we will call $X,Y$) that are positive and the third negative (we will call it $-Z$, with $Z\gt 0$). Then we have $Z=X+Y$ with $X,Y,Z \gt 0$ and that $X,Y,Z$ are equal to $|A|, |B| , |C|$ in some order.  WLOG we can assume that $X\le Y$ , then:  $$Z=X+Y\le 2Y \lt 2Y+2X = 2Z$$  Obtaining $2Z \gt 2Y \ge Z$ as we wanted.  Solution: Assume that  $$(1): \; x^x+y^y=z^z$$  has integer solutions $\ne 0$, let $(x,y,z) = (a,b,c)$. According to the property, there exists integers $\ne 0$ $(u,v) \in (a,b,c)$ such that $2|u^u| \gt 2|v^v| \ge |u^u|$. Dividing by 2 and taking $\log$ we see that there are non-zero integers $(u,v)$ such that: $$u\log |u| \gt v\log |v| \ge u\log |u| - \log 2$$ That can be written as: $$(2): \;0 \lt u\log |u| - v\log |v| \le \log 2$$  The inequality (2) says that $u\log|u| \gt 0$ or $v\log|v| \lt 0$, but if $v \log |v| \lt 0$ then: $$u \log |u| - v \log |v| = (-v)\log|-v|\;-\;(-u)\log |-u|$$ with $(-v)\log|-v| \gt 0$, so unless we make a substitution, we can assume that $u \log |u| \gt 0$. In this case, necessarily $u \ge 2$, so   the second inequality of (2) says that: $$v \log |v| \ge u \log |u| - \log 2 \ge 2 \log 2 - \log 2 \gt 0$$  Matching this with the first equality of (2) we get: $$ log 2 \ge u \log |u| - v \log |v| \gt (v+1)\log |v| - v\log|v| = \log|v| \ge \log2$$ a contradiction. Thus $x^x+y^y=z^z$ doesn't have any integers solution.
45365
Let $p$ be a prime number. Find all $p$'s such that $p^4 + p^3 + p^2 + p +1$ is a perfect square.   I tried rewriting the expression as  $p^4 + p^3 + p^2 + p +1 = x^2 \iff (p^2 + p)(p^2 + 1)=(x - 1)(x + 1)$. Then I think I need to bound this using $x$ but am not sure how to.
88941
Let $ S = 1! ~2!~\dotsm ~100! $. Prove that there exists a unique positive integer $k$ such that $S/k!$ is a perfect square.   I thought this was a cute, fun problem and I did solve it, but any alternative methods that you guys would use?
61973
Find all integers $x$ such that $x^2+3x+24$ is a perfect square.    My attempt:    $x^2+3x+24=k^2$ $3(x+8)=(k+x)(k-x)$    Now, do I find solution treating cases? But that doesn't seem very easy. Please help.
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)
69195
Prove that for any set of $2n+3$ integers from the interval $[-2n-1,2n+1]$ there is a triple $(x,y,z)$ such that $x+y+z=0$.    Example  : Choose $5$ number from $\{-3,-2,-1,0,1,2,3\}$ there are $x,y,z$ with $x+y+z=0$.