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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
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This is a partial solution. But this shows, we can obtain more solutions than which are given. Suppose $x=y,$ then $$(x^2-x)^2=4x^2$$ has three solutions namely,$$x=y=0, -1, 3.$$ Now suppose $x>y>0,$ if $y=1$ then $$(x^2-1)(1-x^2)=(x+1)^2$$ which has only solution $x=-1.$
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$$ \begin{array}{l} \left( {x^2 - y} \right)\left( {y^2 - x} \right) = \left( {x + y} \right)^2 \\ \Leftrightarrow xy\left( {xy - 1} \right) = x^3 + y^3 + x^2 + y^2 \\ \Leftrightarrow xy\left( {xy - 1} \right) = x^2 \left( {1 + x} \right) + y^2 \left( {1 + y} \right) \cdots \left( * \right) \\ \left\{ \begin{array}{l} x = y \\ \left( * \right) \Leftrightarrow x^2 - 1 = 2x\left( {1 + x} \right) \\ \left( * \right) \Leftrightarrow \left( {1 + x} \right)^2 = 0 \\ \left( * \right) \Leftrightarrow \left( {x;y} \right) \in \left\{ {\left( { - 1; - 1} \right)} \right\} \\ \end{array} \right.;\left\{ \begin{array}{l} x = - y \\ \left( * \right) \Leftrightarrow x^2 + 1 = 2x^2 \\ \left( * \right) \Leftrightarrow \left( {x;y} \right) \in \left\{ {\left( {1; - 1} \right);\left( { - 1;1} \right)} \right\} \\ \end{array} \right. \\ \end{array}$$
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On the LHS the sum of the two factors is nonnegative, so $x^2-y$ and $y^2-x$ cannot be negative at the same time. Hence, if $x+y\ne0$ then both $x^2-y$ and $y^2-x$ must be positive.
If $y^2-x\ge8$ and $x^2-y\ge8$ then $$ (x+y)^2 = (x^2-y)(y^2-x) = \\ = \frac{x^2-y}2(y^2-x) +\frac{y^2-x}2(x^2-y) +0 \ge \\ \ge 4(y^2-x) +4(x^2-y) -2(x-y)^2 = \\ = 2(x+y)^2 - 4(x+y); \\ 0\le x+y \le 4. $$
Therefore, all solutions satisfy $0\le x+y\le4$, $1\le x^2-y\le 7$ or $1\le y^2-x\le 7$. It is a few simple cases and they all lead to finding the integer roots of certain polynomials with integer coefficients.
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Please don't vote. Just a pictural comment to show what the curve $(x^2-y)(y^2-x)-(x+y)^2=0$ does look like ( : it's an "elbow" ) .
$\color{red}{Red}$ is positive, $\color{green}{green}$ is negative.
The integer coordinates are yellow lines.
Picture on the right is $\color{blue}{blue}$ rectangle in picture on the left zoomed in $20 \times$ .
(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 $a = 1$.
(1) gives $(4k - 3)(9k - 2) = 25$, which has no rational solutions.
(1) gives $(4k + 4)(9k - 2) = 1$, which has no rational solutions.
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.