A perfect problem

Suppose y^2= x^4+x^3+x^2+x+1 ( y being an integer). Right hand side of the expression can be written as (x^2+x/2+3/8)^2+(5x/8+55/64). Multiplying both sides by 64, expression becomes: (8y)^2=(8x^2+4x+3)^2+(40x+55) [ 8y=n & (8x^2+4x+3)= m say] Now (40x+55)=0 not possible( as it does not give integer x). Therefore, abs(40x+55)>= 2(8x^2+4x+3)-1 [because n^2 and m^2 will differ by AT LEAST 2m-1] Which gives ((4-√66)/4)<=x<=((4+√66)/4). Check the integer values for x in this region. x= -1, 0, 3 So the sum is -1+0+3=2.

For clarity let $y^2=x^4+x^3+x^2+x+1$ .

Note

$(x^2+\dfrac{1}{2}x)^2 =x^4+x^3+\dfrac{1}{4}x^2 < x^4+x^3 + \dfrac{1}{4} x^2 + \dfrac{3}{4} (x + \dfrac{2}{3})^2 + \dfrac{2}{3} = y^2$

And: $y^2 \le x^4+x^3+\dfrac{9}{4} x^2 + x +1 = (x^2 + \dfrac{1} {2} x +1)^2$

So we have $(x^2+\dfrac{1}{2}x)^2 < y^2 \le (x^2+\dfrac{1}{2}x+1)^2$ with equality when x=0.

We see $x=0$ is indeed a solution. For $x \neq 0$ , the inequality shows that x cannot be even. If x is odd, y must be in between the two "consecutive" numbers.

So we have $y=x^2+\dfrac{1}{2}x+\dfrac{1}{2}$

Plugging in to the original equation, we get:

$(x^2+\dfrac{1}{2}x+\dfrac{1}{2})^2 = x^4+x^3+x^2+x+1$

$\implies \dfrac{1}{4} x^2 - \dfrac{1}{2} x - \dfrac{3}{4} =0$

$x^2 - 2x -3 =0$

$(x-3)(x+1)=0$

So $x=0, -1, 3$

The above expression will be a perfect square only for x=0,-1,3.

$x^4+x^3+x^2+x+1=n^2$ $x^3(x+1)+x(x+1)=n^2 -1$ $(x^3+x)(x+1)=(n-1)(n+1)$ $x(x^2+1)(x+1)=(n-1)(n+1)$ consider all possibilities $x(x^2+1)=n-1,x+1=n+1. solving \quad for\quad x =-1$ $x(x+1)=n-1,x^2+1=n+1. solving \quad for\quad x =-1$ $x(x+1)(x^2+1)=n+1.n-1=1 solving \quad for\quad x =\quad not\quad integral$ $x(x^2+1)=n+1,x+1=n-1. solving \quad for\quad x =\quad not\quad integral$ $x(x+1)=n+1,x^2+1=n-1. solving \quad for\quad x =3$ $(x+1)(x^2+1)=n+1.,n-1=x solving \quad for\quad x =0$ all other either is not integral or is repeated from $-1,0,3$ so, $-1+0+3=\boxed{2}$

For relatively large values we realize this value is between two consecutive squares.