$n(n+1)(n+2)$ is never a square I want to show that for $n,m$ 2 non zeros integers that the equation $n(n+1)(n+2)=m^2$ has no solution. 
I tried to solve but could only find that if it was the case $m$ should be divisible by 18 . Any help
 A: Lemma: if integers $a,b > 0$ and $ab = k^2$ and $\gcd(a,b) = 1,$ then both $a,b$ are squares. This is by unique factorization.
No need to consider negative $n,$ as $n \leq -3$ gives a negative product. With $n \geq 1:$ As $n^2 + 2n = (n+1)^2 - 1,$ we find
$$  \gcd(n+1, n^2 + 2n) = 1. $$
If the product were a square, we would require $n^2 + 2n$ to be a square. However, $n^2 + 2n+1$ really is a square, the only consecutive squares are $0,1,$ which forces $n=0.$ Done.
A: A positive integer $n$ is a square if and only if all of its prime factors occur with even multiplicity.
So this would be true of the integer $d:=n(n+1)(n+2)$ if it is a square. Since consecutive integers are coprime, and "twins" (i.e. $n$ and $n+2$) have only $2$ as a common factor, you can analyze the (odd) primes $p$ occurring in $d$: if $p|d$, then $p$ divides one of $n$, $n+1$, or $n+2$, but notice it has to occur in exactly one of these unless $p=2$. If $p=2$ then either it occurs in both $n$ and $n+2$, or only occurs in $n+1$. So you have a few cases.
Aince these three consecutive integers can't share (odd) prime factors, this shows that all of $n$, $n+1$, and $n+2$ would have to be squares. But this is impossible.
A: By reducing mod $2$ and $3$ you can show that $2$ and $3$ divide $m$, hence $m^2$ has actually to be divisible by $36$. But this is not really helpful towards your result.
Let $p>2$ be a prime that divides $n$. Then $p$ does not divide $(n+1)$ and $(n+2)$. If $n=p^k \cdot q$ with $(q,p)=1$, then $k$ has to be even (since every prime appears with even power in the decomposition of $m^2$). We didn't make any assumption on $p$ except $p>2$, so every prime except possibly $2$ appears with even multiplicity! So $n=s^2$ or $n=2\cdot s^2$ for some $s \in \mathbb{N}$. 
We can make the same argument for $n+1$ and $n+2$, so there are $t$ and $u$ such that $n+1=t^2$ or $2t^2$ and $n+2=u^2$ or $2t^2$. Now, note that the difference between any two squares of elements of $\mathbb{N}$ is always bigger than $3$.
If $n$ is even, then $n+1=t^2$. But then $n$ cannot be a square, so $n=2s^2$, and $n+2=2u^2$. But then $n+2-n = 2u^2 - 2s^2$ so $1 = u^2 - s^2$, another contradiction. 
If $n$ is odd then $n=s^2$ and $n+2=u^2$, a contradiction.
