# Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$

Find all integer roots of: $x^2(y-1)+y^2(x-1)=1$

Obviously $(2,1)$ and $(1,2)$ are two answers. But I was unable to manipulate the equation algebraically giving a useful form for finding all other possible answers!
I also tried to view it as a quadratic equation in terms of $x$, but forming the Delta of that equation didn't help much other than if $y=1$ then certainly Delta would be a square and equation would have answer for $x$...

• For positive integer solution, you have already found all of them. Since any pair for both $x,y>1$ cannot give $1$ on the right hand side. There may be more negative integer solutions. Jul 14, 2016 at 5:28
• @mastrok What if the question asks about INTEGER answers? I edit the text of question with permission of readers. Jul 14, 2016 at 5:32
• I found that there are no negative integer solutions. Basically you just solve $y$ in terms of $x$, there are two $y(x)$. You can see that $-x-3<y_1(x)<-x-2$ and $1<y_2(x)<2,\; x\le-6$, and check that $x=-5$ is not an integer solution. Jul 14, 2016 at 5:56
• @Matrok: I agree, Did it another way, rewriting in terms of $xy$ and $x+y$ and using a discriminant argument. Jul 14, 2016 at 6:02
• @Vincenzo Oliva, thanks! how come I missed that :P Jul 22, 2016 at 6:57

Any point on the curve $x^2(y-1)+y^2(x-1)=1$ is either close to the line $x=1$, to the line $y=1$ or to the line $x+y=-2$, since:

$$x^2(y-1)+y^2(x-1)-1 = (x-1)(y-1)(x+y+2)+(x+y-3)$$

$\hspace{1.5in}$

These lines are asymptotes and the algebraic curve intersects them only at the solutions you already found. With a crude bound, we just have to test every point in the range $[-6,6]^2$ to be sure there are no other integer solutions.

• @HamidRezaEbrahimi: you're welcome. Jul 14, 2016 at 6:54
• Is it trivial that the curve doesn't cross the asymptotes outside $[-6,6]^2$? Jul 22, 2016 at 5:41

Thanks to Piquito for pointing out the negative solution.

As stated in the comments, yours are the only positive integer solutions because you'd get $LHS>1$ with larger $x,y.$ Bounding the discriminant $\Delta_x$ (resp. $\Delta_y$) with squares of polynomials of $x$ (resp. $y$) is a natural and, in this case, not a bad idea, in order to show there are no negative solutions.

WLOG let us consider the problem as a quadratic equation in $y$. We have $\Delta_x=x^4+4x^3-4x^2+4x-4,$ and since $(x^2+2x)^2=x^4+4x^3+4x^2$ we may want to try finding the smallest $m,n\in\mathbb{Z^+}$ such that for some $a$ and all $x \le a$ $$(x^2+2x-m)^2<\Delta_x<(x^2+2x-n)^2.$$ It turns out $(m,n)=(8,4)$ works with $a=-6$, and no integer $x$ fulfills $$\Delta_{x}=(x^2+2x-k)^2$$ for any $k=5,6,7.$ Therefore we only need to check $x=-1,-2,-3,-4,-5$ back in the original equation to see if there are other solutions besides yours: and there are, namely $(x,y)=(-5,2),(2,-5).$

• Hi Vincenzo: $(-5,2)$ is a solution. Regards. Jul 21, 2016 at 23:55
• You are welcome. Jul 22, 2016 at 0:10
• @Piquito: Heh. Thanks fron the old gaga. Jul 22, 2016 at 0:13
• My English is rudimentary and the translator of Google without which I could not participate in StackExchange not tell me anything about "old gaga". If you do not bother I'll thank you to tell me what that is? Jul 22, 2016 at 0:21
• Don't delete it. Just edit it (what about me that there are few days I wrote $9-1=7$. Good by. Don't answer this. Jul 22, 2016 at 0:40

It is easy to verify that $$x^2(y-1)+y^2(x-1)=1\iff2(xy)^3+(x^2+y^2)(xy)^2=(x^2+y^2+1)^2$$ It follows that $xy$ divides $x^2+y^2+1$, in other words one has $$\frac xy+\frac yx+\frac{1}{xy}\in \mathbb Z$$ The obvious solution $x=y=1$ is not solution of the proposed equation. On the other hand the number $3$ given by this $(x,y)=(1,1)$ is given also by $(x,y)=(1,2),(-5,2)$ (as expected, small multiples of $2$ and $5$ are "candidates"). It is not hard to verify that these two (and the corresponding symmetrics respect to the line $y=x$, of course) no other solution exists.

Since O.P.'s equation is symmetric respect to the line $y=x$ then the only solutions are $$(x,y)=(1,2),(2,1),(-5,2),(2,-5)$$

To pull a rabbit from a hat, let's write $y=u-2-x$. When the algebraic dust settles, the equation to solve is now

$$ux^2-u(u-2)x+(u^2-4u+5)=0$$

We see we must have $u\mid 5$, which leaves four possible values for $u$: $u=\pm1$ and $u=\pm5$. The values $u=1$ and $u=-5$ lead to quadratics $x^2+x+2=0$ and $x^2+7x-10=0$, which have no integer solutions, while $u=-1$ leads to

$$x^2+3x-10=(x+5)(x-2)=0$$

which gives $(x,y)=(-5,2)$ and $(2,-5)$. Similarly, $u=5$ leads to

$$x^2-3x+2=(x-1)(x-2)=0$$

which gives $(x,y)=(1,2)$ and $(2,1)$.

Remark: The "rabbit" here is the "$-2$" in "$u-2-x$." That is, it made sense initially to let $y=v-x$ and see what the quadratic in $x$ looked like; it turned out it implied $(v+2)\mid5$. So at that point it made sense to let $v=u-2$ and redo everything with $y=u-2-x$.

Perhaps it would be easier if you write the equation like this

$$xy(x+y)-(x+y)^2+2xy = 1$$ and now put $$a=x+y$$ and $$b=xy$$. Then you get: $$ba-a^2+2b=1\implies b ={a^2+1\over a+2}$$

Write $$c=a+2$$ and then $$b = {(c-2)^2+1\over c} = {c^2-4c+5\over c}= c-4 +{5\over c}$$

so $$c\mid 5\implies a+2\in \{-5,-1,1,5\} \implies a\in \{-7,-3,-1,3\}$$ and respectively $$b\in \{-10,-10,2,2\}$$. This should not be difficult to finsih.