Integer $x$ for which $x^4+x^3+x^2+x+1$ is perfect square. Integer values of $x$ for which $\bf{x^4+x^3+x^2+x+1}$ is a Perfect Square.
$\underline{\bf{My\; Try}}$:: Let $\bf{x^4+x^3+x^2+x+1 = k^2}$, where $k\in \mathbb{Z}$
$4x^4+4x^3+4x^2+4x+1 = 4k^2 = (2k)^2$
Now How can I proceed after that
Help Required,
Thanks
 A: Hint: For all but finitely many integer values of $k$, we have
$$ (2k^2 + k)^2 < 4k^4 + 4k^3 + 4k^2 + 4k + 4 < (2k^2 + k + 2 ) ^ 2 $$
A: We can just change this into a quadratic equation (or cubic equation, but that becomes very highly complicated).
$$\because x^4 + x^3 + x^2 + x + 1 = k^2$$
then this means that $x^2 + x + (1 - k^2 - x^3 - x^4) = 0$. This means that:
$$x = \frac{-1 \pm \sqrt {1 - 4(1 - k^2 - x^3 - x^4)}}{2}$$
And this means that:
$2x + 1 = \pm \sqrt {1 - 4(1 - k^2 + x^3 + x^4)}$ meaning that this root is odd if $x$ is an integer. But if that is so, then, the expression:
$$-1 \pm \sqrt{1 - 4(1 - k^2 - x^3 - x^4)}$$
is even $(2x)$. But is $x$ itself an even integer? We can test to see if $x$ is odd or even, but I will only be testing if it is even. So let's call this even number $2r$ and assume that:
$$(2r)^4 + (2r)^3 + (2r)^2 + 2r + 1 = k^2$$
$$16r^4 + 8r^3 + 4r^2 + 2r + 1 = k^2$$
Since all other values on the left hand side of the equation are even (except $1$), then this means that $k^2$ is odd because the sum of all the even numbers is another even number $2n$. And we know that $2n + 1$ is odd. And since $k^2$ is odd, then $k$ is odd. Where:
$$2n = r(16r^3 + 8r^2 + 4r + 2)$$
$$\implies 2n = r(r(16r^2 + 8r + 4) + 2)$$
$$\implies 2n = r(r(r(16r + 8) + 4) + 2)$$
$$\therefore \frac{2n}{r^3} = 16r + 8 + 4 + 2 = 16r + 14$$
Now we can form a real quadratic equation. If $2n/r^3 = 16r + 14$ then:
$$4r^2 + 14^{1/3}r - 2n = 0$$
$$r = \frac{-14^{1/3} \pm \sqrt{14^{2/3} - 32n}}{8}$$
$$r = - \frac{14^{1/3}}{2} \pm \sqrt {2n + \frac{14^{2/3}}{4}}$$
The equation just above this sentence is given by "completing the square". By simplifying this, we will be able to simplify the quadratic expression much easier. If so, we get that:
$$r = \frac{-14^{1/3} \pm \sqrt {8n + 14^{2/3}}}{2}$$
$\therefore$ $x$ cannot be an even integer since:
$$x = 2r = -14^{1/3} \pm \sqrt {8n + 14^{1/3}} = \frac{-1 \pm \sqrt {1 - 4(1 - k^2 - x^3 - x^4)}}{2}$$
$$\implies 2x = 4r = 2(-14^{1/3} \pm \sqrt {8n + 14^{1/3}}) = -1 \pm \sqrt {1 - 4(1 - (x^2 + x + 1))}$$
$$\therefore 2x = 4r = -112^{1/3} \pm \sqrt {32n + 896^{1/3}} = -1 \pm \sqrt {4(x^2 + x + 1) - 3}$$
To conclude, the equation $x^4 + x^3 + x^2 + x + 1 = k^2$ has no solutions in integers if $x$ is even and $k$ is odd, so we know that there is no value for $n$ to satisfy the equation. And just from the looks of it, I don't think there would be solutions if $x$ was odd and $k$ is even, but that is just out of pure assumption. For equations like yours, you can use quadratic formulas and "complete the squares" to solve for $x$ and use the example provided. Hope I helped! :)
