How prove this diophantine equation $(x^2-y)(y^2-x)=(x+y)^2$ have only three integer solution? HAPPY NEW YEAR To Everyone! (Now Beijing time  00:00 (2015))

Let $x,y$ are integer numbers,and such $xy\neq 0$,
Find this diophantine equation all solution
$$(x^2-y)(y^2-x)=(x+y)^2$$

I use Wolf found this equation only have two  nonzero integer solution $(x,y)=(-1,1),(-1,-1)$,see wolf
$$\Longleftrightarrow x^3+y^3+x^2+y^2=xy(xy-1)$$
But How prove it?
and I found sometimes,and my problem almost similar with 2012 IMO shortlist:2012 IMO shortlist
(Apologies for the rather long solution.)



Write $x = ka$, $y = kb$, with $a, b$ relatively prime.
The equation $(x^2 - y)(y^2 - x) = (x + y)^2$ becomes
$$
\newcommand{\divs}{\; \mid \;}
(ka^2 - b)(kb^2 - a) = (a + b)^2 \tag{1}
$$
Or, rearranged,
$$
(ka + 1)(k b + 1) ab = (a + b)(kb^2 + b + ka^2 + a) \tag{2}
$$
Taking (2) modulo $a$, we get $kb^3 + b^2 \equiv 0$, and
dividing by $b^2$ (which is a unit mod $a$) we get $kb + 1 \equiv 0$.
So $a \divs kb + 1$ and likewise $b \divs ka + 1$.
Let $c = \text{gcd}(ka + 1, kb + 1)$,
$ka + 1 = bcA$, $kb + 1 = acB$ for relatively prime integers $A, B$.
Rewrite (2) as
$$
(bcA)(acB) ab = (a + b)(abcB + abcA)
$$
i.e.
$$
abc A B = (a + b)(A + B). \tag{3}
$$
Since $a$ and $b$ are relatively prime to $a + b$,
$$
ab \divs A + B \tag{4}
$$
Since $A$ and $B$ are relatively prime to $A + B$,
$$
AB \divs a + b \tag{5}
$$
Recall the following integer properties:


If $m \divs n$, then $|m| \le |n|$ OR $n = 0$.
$|m + n| \le |mn| + 1$ for all $m,n \ne 0$.
(Follows from triangle inequality and $(|m| - 1)(|n| - 1) \ge 0$.)


From these and (4), (5) we get that
$$
|a + b| \le |ab| + 1 \le |A + B| + 1 \le |AB| + 2 \le |a + b| + 2
$$
unless $A + B = 0$ or $a + b = 0$ or one of $A, B, a, b$ is $0$.

Case 1: $\boldsymbol{A + B = 0}$

Since $A$ and $B$ are relatively prime, we conclude WLOG $A = 1$, $B = -1$.
From (3), $abc = 0$.
But $a, b \ne 0$, so $c = 0$.
So $ka + 1 = kb + 1 = 0$.
So $x = ka = -1$ and $y = kb = -1$,
and in this case we get the solution $(x,y) = \boxed{(-1, -1)}$.

Case 2: $\boldsymbol{a + b = 0}$

Since $a$ and $b$ are relatively prime, we conclude WLOG $a = 1$, $b = -1$.
From (1),
$(k - 1)(k + 1) = 0$,
so either $k = 1$ or $k = -1$,
giving the solutions $(x,y) = (ka, kb) = (k, -k) = \boxed{(1, -1), (-1, 1)}$.

Case 3: $\boldsymbol{abAB = 0}$

From (3) this case implies $0 = (a + b)(A + B)$,
so this case is encompassed in the first two.

Case 4: $\boldsymbol{|a + b| \le |ab| + 1 \le |A + B| + 1 \le |AB| + 2 \le |a + b| + 2}$

From the inequality, we conclude that
$$
|a + b| \le |a| + |b| \le |ab| + 1 \le |a + b| + 2 \le |a| + |b| + 2
$$
Therefore,
$$
|ab| + 1 = |a| + |b| + r
$$
where $r \in \{0, 1, 2\}$.
In particular,
$$
(|a| - 1)(|b| - 1) \in \{0, 1, 2\}
$$
We know $a, b \ne 0$.
The first subcase is if $|a| = 1$ or $|b| = 1$ (WLOG $|a| = 1$).
Otherwise, $(|a|, |b|) = (2, 2), (2, 3), \text{ or } (3, 2)$,
Since $a$ and $b$ are relatively prime,
the first of these is impossible.
Otherwise we may assume $|a| = 2$.
By flipping the sign of $k$, we may assume $a$ is positive.
So we are left with three subcases: $a = 2$ and $b = 3$,
or $a = 2$ and $b = -3$, or $a = 1$.


(Case 4a) $a = 2$, $b = 3$


(1) gives
$(4k - 3)(9k - 2) = 25$,
which has no rational solutions.


(Case 4b) $a = 2$, $b = -3$


(1) gives
$(4k + 4)(9k - 2) = 1$,
which has no rational solutions.


(Case 4c) $a = 1$


From (4), $b \divs A + B$.
By the inequality we are still assuming,
$|A + B| + 1 =$ $|a + b|$, $|a + b| + 1$, or $|a + b| + 2$.
This implies $A + B = \pm(a + b)$
or $A + B \in \pm(a + b) \pm 1$.
With $a = 1$, we have
$A + B = b + 1, -b - 1, b + 2, b, - b, -b - 2$.

(Case 4c-i)
If $A + B = b + 1$ or $-b - 1$, then from (4) $b \divs A + B$ so $b = \pm 1$.
$a = 1$ and $b = -1$ implies $a + b = 0$, which we have already covered.
So $a = b = 1$.
Then (2) gives $(k + 1)^2 = 2(2k + 2)$,
so $k = -1$ or $k = 3$.
If $k = -1$ then $(x,y) = (-1, -1)$ which we already found.
If $k = 3$ then we have the solution $(x,y) = \boxed{(3,3)}$.

(Case 4c-ii)
If $A + B = b + 2$ or $-b - 2$, then from (4) $b \divs A + B$ so $b \divs 2$.
$b = \pm 1$ is covered in case 4c-i.
So assume $b = \pm 2$.
Then $A + B = 0$ or $\pm 4$.
If $A + B = 0$ this is covered by Case 1.
So assume $|A + B| = 4$.
Rewriting the inequality assumed in Case 4,
$$
|b + 1| \le |b| + 1 \le |A + B| + 1 \le |AB| + 2 \le |b + 1| + 2
$$
The middle term is $|A + B| + 1 = 5$, so $5 \le |b + 1| + 2$,
so $b = 2$.
Returning to (1), we get
$$
(k - 2)(4k - 1) = 9
$$
so $4k^2 - 9k - 7 = 0$,
which has no integer solutions.

(Case 4c-iii)
If $A + B = b$ or $-b$,
then write
$A + B = \frac{ka+1}{bc} + \frac{kb+1}{ac}$
so that $ka^2 + a + kb^2 + b = abc(A + B) = \pm b^2 c$.
Then from (2) we get
$$
(k + 1)(bk + 1) = \pm (b + 1) b^2 c
$$
Since no factors of $b$ are shared by $bk + 1$, $b^2 \divs k+1$.
But we also have $k + 1 + kb^2 + b = \pm b^2 c$,
implying $b^2 \divs  b$.
Since $b \ne 0$, $b \in \{-1, 1\}$.
So $(a,b)$ is one of $(1, -1)$ or $(1, 1)$.
But $(a,b) = (1, -1)$ satisfies $a + b = 0$, so was covered by Case 2.
And $(a,b) = (1,1)$ was covered in Case 4c-i.
Note that your equation is basically 

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