Determinate all the positive integers $x$ such that $x^4+6x^3+11x^2+3x+11$ is a perfect square.
My try
With pure intuition I can say that there is no $ x $ that meets this condition, i tried by brute force some numbers and none worked, but I would like to know if what I'm guessing is right or not, and what would be a way to prove it, because i don't see the way to prove this.
Suppose that exist $n\in\Bbb N$ such that $$x^4+6x^3+11x^2+3x+11=n^2$$
See $\text{mod } 3$ the equation and you can prove that $n^2\not\equiv 2 \text{ mod } 3$ for all $n\in\Bbb N$. In fact, if $n=3k+a$ with $a=0,1,2$ then $n^2=9k^2+6ka+a^2\equiv a^2 \text{ mod } 3$ and this values are $0,1,1$ respectively.
Now,
If $x\equiv 0 \text{ mod } 3$ then $x^4+6x^3+11x^2+3x+11\equiv 0^4+6.0^3+11.0^2+3.0+11\equiv 11\equiv 2 \text{ mod } 3$
If $x\equiv 1 \text{ mod } 3$ then $x^4+6x^3+11x^2+3x+11\equiv 1^4+6.1^3+11.1^2+3.1+11\equiv 1+6+11+3+11\equiv 2 \text{ mod } 3$
If $x\equiv 2 \text{ mod } 3$ then $x^4+6x^3+11x^2+3x+11\equiv 2^4+6.2^3+11.2^2+3.2+11\equiv 16+0+44+0+11\equiv 2 \text{ mod } 3$.
This means that no have solutions
Hint : $x^{3} + 11x$ is always divisible by 3.
