Answer:$(x,y)=(0,1),(0,-1),(-1,0)$. $y^2=(x+1)(x^2+x+1)$ If $d|x+1$ and $d|x^2+x+1$, then $d|x^2+2x+1-x^2-x-1=x$ so $d|x+1-x=1$ gives us that $d=1$. $gcd(x+1,x^2+x+1)=1$. Both $x+1$ and $x^2+x+1$ should be perfect squares. $x+1=a^2$ and $x^2+x+1=b^2$ $\implies x=a^2-1\implies (a^2-1)^2+a^2=b^2\implies (a^2)^2\geq a^4-a^2+1=b^2$ if $a^2 \geq 1$.$(*)$ Let $b=a^2-k$ such that $k \geq 0$. \[a^4-a^2+1=b^2=a^4-2a^2k+k^2\]\[2a^2k-a^2=k^2-1\]\[a^2(2k-1)=k^2-1\]We have $2k-1|k^2-1$ so $2k-1|k^2-1-2k+1=k(k-2)$ and because of $gcd(2k-1,k)=1$ if $k$ is not $0$, we can say that $2k-1|k-2\implies 2k-1|2k-1-2k+4=3\implies 2k-1=1,3 \implies k=1,2$ If $k=0$, then $a^2=1$ If $k=1$, then $a^2=1$ If $k=2$, then $a^2=1$ If we put $1$ on $a^2$, we get \[1=1^4-1^2+1=b^2\]So $x+1=1 \implies \boxed{x=0,y=1}\boxed{x=0,y=-1}$ $(*):$If $a=0$, then $\boxed{x=-1,y=0}$

Note that $y^2=(x+1)(x^2+x+1)$ and $gcd(x+1, x^2+x+1)=1$. So, $x+1, x^2+x+1$ are both perfect squares or one of them is zero. But $0 \not = x^2+x+1 >x^2, <(x+1)^2$ for $x>0$ and $(x+1)^2<x^2+x+1<x^2$ for $x<-1$, so $x=0, y=1;-1$ and $x=-1, y=0$ are only solutions.

$y^2=x^3+2x^2+2x+1$ $\implies y^2=2x(2x+1)+(x+1)(x^3-x+1)$ $\implies y^2=(x+1)(x^3+x+1)$ Now using Euclidean reduction, we see that, $gcd(x+1,x^3+x+1) = gcd(x+1,x^3)$ but also $x+1| x^3+1$, so $gcd(x+1,x^3)=1$, Now either $x+1=x^3+x+1$ or one of the two factors is $0$, Computing all these cases we find that the final possible pairs are, $(0,1);(0,-1),(-1,0)$

Jishnu4414l wrote: $y^2=x^3+2x^2+2x+1$ $\implies y^2=2x(2x+1)+(x+1)(x^3-x+1)$ $\implies y^2=(x+1)(x^3+x+1)$ No. Because $x^3+2x^2+2x+1 \not = (x+1)(x^3+x+1)$.

Rewrite equation like this $y^2=(x+1)(x^2+x+1)$ , and we use gcd . GCD($x+1$;$x^2+x+1$)=1 , then both of the equation need to be perfect square , let's look for $x^2+x+1$ , we can see that , $x^2<x^2+x+1<(x+1)^2$ , and there is some special cases , $x^2=x^2+x+1$ and $(x+1)^2=x^2$ and from this equations we get these solutions : (0;1) , (0;-1) and (-1;0)

Euler... wrote: Rewrite equation like this $y^2=(x+1)(x^2+x+1)$ , and we use gcd . GCD($x+1$;$x^2+x+1$)=1 , then both of the equation need to be perfect square , let's look for $x^2+x+1$ , we can see that , $x^2<x^2+x+1<(x+1)^2$ , and there is some special cases , $x^2=x^2+x+1$ and $(x+1)^2=x^2$ and from this equations we get these solutions : (0;1) , (0;-1) and (-1;0) $x^2+x+1>x^2$ is not true for negative $x$. We can see it by trying $-2$ Similarily $x^2+x+1<x^2+2x+1$ is not true which can be seen by trying $-1$.

Notice that $y^2=x^3+2x^2+2x+1=(x+1)(x^2+x+1)$. Since gcd$(x+1,x^2+x+1)=1$, we have that both $x+1$ and $x^2+x+1$ are squares. Thus let $a^2=x+1$ and $b^2=x^2+x+1$. $b^2=x^2+x+1 \implies (2b)^2-(2x+1)^2=3 \implies (2b+2x+1)(2b-2x-1)=3$. Consider all cases $(2b+2x+1,2b-2x-1)=(\pm 1, \pm 3)$ or $(\pm 3, \pm 1)$. Solving yields $x=0,-1$. In either case, $x+1$ is a square. So the only solutions are $(x,y)=(0, \pm 1),(-1,0)$.

My solution is virtually the same as @bin_sherlo's. Sorry man. I didn't mean it. I claim that all integer pairs that satisfy this equation are $\boxed{(x,y) \in \{(-1, 0), (0, -1), (0, 1)\}}.$ We have that $y^2 = (x + 1)(x^2 + x + 1)$, and as $\gcd(x + 1, x^2 + x + 1) = 1$, we have that $x + 1 = a^2$ and $x^2 + x + 1 = b^2$ for relatively prime integers $a, b$. Hence we have that \[ \begin{cases} x + 1 = a^2 \\ x^2 + x + 1 = b^2 \end{cases} \implies \begin{cases} (a^2 - 1)^2 + a^2 = b^2 \leq a^4 \implies b \leq a^2. \end{cases} \]Now set $b = a^2 - k$. We then have that \begin{align*} (a^2 - 1)^2 + a^2 &= (a^2 - k)^2 \\ \iff a^2(2k - 1) &= k^2 - 1\\ \implies 2k - 1&\mid k^2 - 1 \\ \implies 2k - 1 &\mid 2(k^2 - 1) - (2k - 1) = k - 2 \\ \implies 2k - 1 &\mid 2(k - 2) - (2k - 1) = -3 \implies 2k - 1 \mid 3. \end{align*}Hence we have that $k = 1,2 \implies a = 0, 1 \implies \boxed{x = -1, 0}$. Hence $\boxed{y = 0, \pm 1}$ for each corresponding $x$, yielding us our claimed solutions, as desired. $\blacksquare$

y²=x³+2x²+2x+1. y²=(x+1)(x²+x+1). y² ≥0, x²+x+1=(x+1/2)²+3/4>0. ∴ x ≥-1. By Euclid's algorithm, gcd (x+1, x²+x+1)=gcd (1, x+1)=1. Thus, x+1 and x²+x+1 are both perfect squares. x ≥1 ---->x² <x²+x+1<(x+1)² ∴ x=-1 or 0 case 1)x=-1: y²=-1+2-2+1=0. --->y=0 case 2)x=0: y²=0+0+0+1=1. --->y=1,-1. Therefore, the answer is (-1,0),(0,1),(0,-1).