Prove that , if $an^4+bn^2+c $ is a perfect square $\forall n \in\mathbb Z$, then $a=x^2, b=2xy, c=y^2$ Hello mathexchange users!
Let be $a,b,c$ integer numbers , $a\ne0$ such that:
$$an^4+bn^2+c$$ is a perfect square for every $n$ positive integer.
Prove that there is $x,y$ integers such that:
$$a=x^2$$
$$b=2xy$$
$$c=y^2$$

Now i will include my work on the problem:
I considered the polynomial
$$P(n)=an^4+bn^2+c$$
and tried to compute:
$\sqrt{P(n+1)}-\sqrt{P(n)}$ but it goes to infty so got nothing
XX
I would be extremely grateful if somebody can provide an as-easy-as-possible proof.
 A: It looks like, the problem statement is not correct.
Take,
$$a=c=0, ~b=m^2, m≠0.$$
A: There was an early answer that proved the statement under the assumption that an integer coefficent polynomial $P(x)$ must be a perfect square (i.e. $P(x) = Q(x)^2$ for some integer coefficient polynomial) if $P(n)$ is a perfect square for all $n\in \mathbb{N}$. That answer was later deleted, presumably because the answerer couldn't convince themselves that this assumption was correct. It turns out that it is in fact true, and can be generalized to a much stronger result.
Using this lemma, we get that $P(x) = a x^4 + b x^2 + c$ is the square of some quadratic $Q(x) = (m x^2 + l x + n)^2$. Squaring it and matching the coefficients, we find that $m^2 = a$, $2lm = 2ln = 0$, $2mn + l^2 = b$, and $n^2 = c$. Assuming that $a$ and $b$ are non-zero, we must have $l=0$. This completes the proof. (Note that the case where $m=n=0$ and $l\neq 0$ accounts for the counterexample @lonestudent found.)

Unfortunately, combined with the cited proof of the lemma, this might not be an "as-easy-as-possible" proof. However, you can adapt the proof of the lemma to the special case of $P(x) = a x^4 + b x^2 + c$ and not have to use some of the harder results related to resultants in full generality.
