Question: 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:
$$a=x^2$$
$$b=2xy$$
$$c=y^2$$

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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

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I would be extremely grateful if somebody can provide an as-easy-as-possible proof.

Answer: 
It looks like, the problem statement is not correct.

Take,

$$a=c=0, ~b=m^2, m≠0.$$
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](https://math.stackexchange.com/a/435697/165060), and can be generalized to a [much stronger result](https://math.stackexchange.com/questions/1157780/polynomials-such-that-pk-ql-for-all-integer-k).

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.)

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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.