Kee Wei's ordered pairs

Notice that $(x+1)^3-x(x+1)=x^3+2x^2+2x+1$ . This motivates the substitution (just to make expressions neater) $z=x+1$ . Using this, our given expression becomes $z^3-z(z-1)=z(z^2-z+1)=y^2$ Now, $\gcd(z,z^2-z+1)=1$ , so if the product of the two is a square, each is a square. However, notice that if $z>1$ , $z^2-2z+1<z^2-z+1<z^2$ $(z-1)^2<z^2-z+1<z^2$ Since a square cannot be between two consecutive squares, we must have that $z\le 1$ . We already determined that $z$ is a square, so $z\ge 0$ . Now, $z=0,1$ .

If $z=0$ , $x=-1$ , and $y^2=0$ . Therefore, $y=0$ , and $(-1,0)$ is a solution.

If $z=1$ , $x=0$ , and $y^2=1$ . Therefore, $y=-1,1$ , and $(0,-1)$ and $(0,1)$ are both solutions.

The answer is $\boxed{3}$ .

The equation factorises into $(x+1)(x^2+x+1)=y^2$ . We now have two cases:

Case 1: $y=0$

We then have either $x+1=0$ or $x^2+x+1=0$ For the former case we have $x = -1$ , for the latter case there are no integer solutions.

Case 2: $y\neq0$

With $x+1\neq0, x^2+x+1\neq 0$ , we note that $gcd(x+1, x^2+x+1)=1$ , so we have $x+1=m^2, x^2+x+1=n^2$ for some relatively prime integers $m, n$ .

Note also that for $x=0$ , we have $y^2=1$ , giving us $y = \pm 1$ . For negative integers $x$ ( $x \leq -1$ ), we have $(x+1)^2=x^2+2x+1 < x^2+x+1 < x^2$ For positive integers $x$ ( $x \geq 1$ )), we have $x^2 < x^2+x+1 < x^2+2x+1=(x+1)^2$ . Thus the only case where $x^2+x+1=n^2$ for integers $x, n$ is when $x=0$

Altogether we have the solutions $(x, y)=(-1,0), (0, \pm1)$ , which is a total of 3 solutions

Clearly we can factorize the polynomial on the LHS as $(x+1)(x^{2}+x+1) = y^{2}$ But we can see that $gcd(x+1,x^{2}+x+1) = 1$ .So for the RHS to be a perfect square each of these terms individually must be perfect squares themselves.Hence we have $x+1 = p^{2}$ and $x^{2}+x+1 = q^{2}$ . Now we have $x = p^{2}-1$ ..Substituting this value in the second expression we get $(p^{2}-1)^{2} +p^{2} = q^{2}$ $p^{2} = q^{2} - (p^{2}-1)^{2}$ $p^{2} = (q+p^{2}-1)(q-p^{2}+1)$ Clearly p and q are integers and if $q > 1$ , $q+p^{2}-1 > p^{2}$ Hence we must have $q=0$ or $q = 1$ . If $q=0$ ,then $(p^{2}-1)^{2}+ p^{2} = 0$ Clearly this is impossible... So $q = 1$ . Hence $p^{4}-p^{2} +1= 1$ $p^{4}-p^{2} = 0$ This correspond to 3 values of p which are 0,1,-1. This corresponds to two values of x ,i.e,x = -1 or x = 0. If x = 0,we have y = 1 or -1. If x = -1,we have y =0 . Hence there are $\boxed{3}$ ordered pairs!!!

$(x+1)(x^2+x+1) = y^2$

Since $x$ and $y$ are integers,

$(x+1)(x^2+x+1) = 1 \times y^2,$

We get $x + 1 = 1, x = 0$ and $x^2 + x + 1 = y^2, y = ± 1$

$(x+1)(x^2+x+1) = y \times y,$

We get $x + 1 = y$ and $x^2 + x + 1 = y,$ imply $x = 0, y = ±1$

$(x+1)(x^2+x+1) = y^2 \times 1,$

We get $x + 1 = y^2$ and $x^2 + x + 1 = 1,$ imply $x = -1, y = 0$

$(x+1)(x^2+x+1) = (-1) \times (-y^2),$

We get $x + 1 = -1 , x = -2$ and $x^2 + x + 1 = -y^2, y^2 = -3,$ imply this equation has no integer solution

$(x+1)(x^2+x+1) = (-y) \times (-y),$

We get $x + 1 = -y$ and $x^2 + x + 1 = -y,$ imply $x = 0, y = ±1$

$(x+1)(x^2+x+1) = (-y^2) \times (-1),$

We get $x + 1 = -y^2$ and $x^2 + x + 1 = -1,$ imply this equation has no integer solution.

Hence the solutions are $(0,1), (0,-1), (-1,0)$

CONSIDERING A FUNCTION F= x^{3}+2x^{2}+2x+1 upon differentiating the function we get a quadratic whose discriminant is less than zero.

Hence, the function F is monotonically increasing function.Also it is has a root equal to {-1}. hence for integer values of {y} the function should be a perfect square of an integer. so, we obtain (-1,0), (0,1) and (0,-1) as the integral solutions of this equation.

Let $y^2=x^3+ax^2+bx+c$

Define $D=-4a^3c+a^2b^2+18abc-4b^3-27c^2$

Then, by the Nagell-Lutz Theorem we have that a rational point $x,y$ of finite order has integer coefficients and either $y=0$ or $y|D$ .

In this particular excersise $a=2, b=2$ and $c=1$ , so that $D=-3$ . With Nagell-Lutz we only have to check $y=0, y=\pm 1$ and $y=\pm 3$ to find our points with integer coefficients.

$y=0$ : $x^3+2x^2+2x+1=0$

For $x$ to be an integer root, we must have that $x|1$ , so $x= \pm 1$ . Evaluating the expression to check if either of them are roots, we find only one point with integer coefficients; $(-1,0)$ .

$y=\pm 1$ : $x^3+2x^2+2x=0$

Thus $x=0$ or $x^2+2x+2=0$ . We can easily check that the quadric has no integer roots, so we have two points with integer coefficients; $(0,1)$ and $(0,-1)$ .

$y=\pm 3$ : $x^3+2x^2+2x-8=0$

For $x$ to be an integer root, we must have that $x|-8$ , so $x= \pm 1, x=\pm 2, x=\pm 4$ or $x=\pm 8$ . Evaluating the expression shows us that none of these are roots.

In total we have found $3$ points with integer coefficients and the Nagell-Luts Theorem guarantees us there can't be more.

x^3+2x^2+2x+1=y^2 => (x+1)(x^2+x+1)=y^2 have solution (-1,0)

x^3+2x^2+2x=y^2-1 => x(x^2+2x+2)=(y+1)(y-1) have solution (0,-1),(0,1)