$n^4+2n^3+2n^2+2n+1=m^2$
Matthew Barrera 
Find all integer solutions $(n,m)$ to $$n^4+2n^3+2n^2+2n+1=m^2$$
My attempt,
$$n^4+2n^3+2n^2+2n+1=(n^2+n+1)^2-n^2$$
$$=(n^2+n+1-n)(n^2+n+1+n)$$
$$=(n^2+1)(n^2+2n+1)$$
$$=(n^2+1)(n+1)^2$$
So I ended up, $(n^2+1)(n+1)^2=m^2$
I'm stuck here. Can anyone continue for me or is there another way to solve this? Thanks in advance.
$\endgroup$3 Answers
$\begingroup$If $n=-1$ then $m=0$. On the other hand, if $n\not=-1$ then $$\sqrt{n^2+1}=\frac{|m|}{|n+1|}\in\mathbb{Q}.$$ So $n^2+1$ should be a perfect square which implies that $n=0$ and $m=\pm 1$.
$\endgroup$ $\begingroup$Since the other two terms are perfect squares, you now have to ask yourself, when is $n^2+1$ a perfect square?
$\endgroup$ $\begingroup$For $(n,m) = (0,1), (0,-1), (-1,0)$ are the solutions. Other than that, $\left(\dfrac{n^2}{n+1}\right)^2 = n^2+1\implies \left(n-1 + \dfrac{1}{n+1}\right)^2 = n^2+1$. The left side is not an integer while the right side is. This is enough to say the equation has no integer solutions.
$\endgroup$ 3
