Find the number of positive integers $x$ for which $x^4+x^3+x^2+x+1$ is a perfect square.
My attempts:
Let $x^4+x^3+x^2+x+1=k^2\implies (x+1)^2(x^2-x+1)=(k-x)(k+x)$
I analysed this a bit found $x=0$ as one which satisfy all condition, how do I find other, please help, try to continue this further, if any other elegant method then add that too.
Note that the problem is equivalent to finding integer solutions to $$4y^2=4x^4+4x^3+4x^2+4x+4$$ Now proceed to note that if $x>3$, we can find $$(2x^2+x)^2=4x^4+4x^3+x^2 < 4x^4+4x^3+4x^2+4x+4=4y^2$$ And $$4y^2 < 4x^4+4x^3+5x^2+2x+1=(2x^2+x+1)^2 $$ Since $4x^4+4x^3+4x^2+4x+4$ is stuck between squares of two consecutive numbers, it cannot be a square itself, which is a contradiction.
Thus we have that if $x$ is a positive integer it must be less than, or equal to, $3$. Trial and error gives us $x=3, y=11 $ is a valid solution.