The Problem I liked to be in exam was the one I have written below but I arrived at it only three days before exam and it was too late!!because it had to be tested by other people and there was no time for this and also in the team selection meeting the problem you saw on exam was selected so the change was not easy... The Problem is This one: Prove that there exist only finite pairs of positive integers $(x,y)$ such that : $ (y^{3}+xy-3)(x^{2}+3x-y)=(x^{3}-xy+1)(y^{2}+x+3y) $

goodar2006 wrote: Find all integer numbers $x$ and $y$ such that: \[(y^3+xy-1)(x^2+x-y)=(x^3-xy+1)(y^2+x-y).\] Proposed by Mahyar Sefidgaran Sketch. Case 1. $x^2+x-y = 0$ or $y^2+x-y = 0$ ... Case 2. $x^2+x-y \neq 0$ and $y^2+x-y \neq 0$. The equation becomes: (1) $ y-x+2 = \frac{x-y+1}{x^2+x-y} + \frac{x-y+1}{y^2+x-y}$. Case 2.1. $x \geq y+2$. Then LHS of (1) is $\leq 0$ but RHS is $> 0$, which is impossible. Case 2.2. $x=y+1$ ... Case 2.3. $x=y$ ... case 2.4. $ x < y $. Case 2.4.1. $|x^2+x-y| \leq 1$ or $|y^2 +x-y| \leq 1$ which leads to $x^2+x-y = \pm 1$ or $y^2 +x-y = \pm 1$ ... Case 2.4.2 $|x^2+x-y| \geq 2$ and $|y^2 +x-y| \geq 2$. We get: $|\text{RHS of (1)}| \leq |x-y+1|/2 + |x-y+1|/2 = y -x-1 $ $|\text{LHS of (1)}| = y-x+2 > y-x-1$ , which is impossible.

Here's a (long, but relatively straightforward) proof that $(y^3+xy-3)(x^2+3x-y)=(x^3-xy+1)(y^2+x+3y)$ has finitely many integer solutions. Of course, it suffices to show that for some $M>2$ (to be determined later), there exist no solutions $(x,y)$ with \[|x-y|,|x|,|y|,|x^2+3x-y|,|y^2+x+3y|,|3x-y|,|x+3y|\ge M.\]Suppose otherwise, and consider such a solution $(x,y)$. Then we have \begin{align} x-y &= \frac{3x^2-1}{x^2+3x-y}-\frac{3y^2+3}{y^2+3y+x} \\ x-y &= \frac{3x+9y-3}{y^2+3y+x}-\frac{9x-3y+1}{x^2+3x-y}. \end{align}We now consider a few cases. Case 1: $x^2+3x-y,y^2+3y+x<0$, so $x^2+y^2+4x+2y+2M\le 0$, a contradiction for sufficiently large $M$. Case 2: $y^2+3y+x<0<x^2+3x-y$, whence from (1) we have $0+0<x-y<-y^2-4y$, which is absurd for sufficiently large $M$. Case 3: $x^2+3x-y<0<y^2+3y+x$, so $y\ge x^2+3x+1$. But then \begin{align*} x^2+2x+1\le y-x &=\frac{3x^2-1}{y-x^2-3x}+\frac{3y^2+3}{y^2+3y+x} \\ &\le\frac{3x^2-1}{M}+\frac{3y^2+3}{y^2+3(x^2+3x+1)+x}\le\frac{3x^2-1}{M}+3, \end{align*}a clear contradiction for large enough $M$ (we use the fact that $3(x^2+3x+1)+x\ge1$ for large $|x|$). Case 4: $x^2+3x-y,y^2+3y+x>0$. Subcase 4.1: $3x-y,3y+x>0$. Then $0<(3x^2-1)/(x^2+3x-y)\le 3$ and $0<(3y^2+3)/(y^2+3y+x)\le3$, a contradicting (1) whenever $M>6$. Subcase 4.2: $3x-y>0>3y+x$. Then from (2), we have $0+0>x-y\implies 0>5x-5y=2(3x-y)-(3y+x)>0$, contradiction. Subcase 4.3: $3x-y<0<3y+x$. Then from (2), we have $0+0<x-y\implies 0<5x-5y=2(3x-y)-(3y+x)<0$, contradiction. Subcase 4.4: $3x-y,3y+x<0$, so $y<-x/3\implies 0<(3x^2-1)/(x^2+3x-y)<(3x^2-1)/(x^2+10x/3)<4$ for sufficiently large $|x|$. (Note that $10x=3(3x-y)+(3y+x)<0\implies x<0$.) Substituting into (1), we get \[x-y<4-\frac{3y^2+3}{y^2+3y+x}<4,\]so taking $M$ large enough, we must have $x-y<0$; from the other direction in (1), we get \[0<x-y+\frac{3y^2+3}{y^2+3y+x}\implies 0<(y-x)(y^2+3y+x)<3y^2+3.\]But $y-x\ge M$ and $y^2+3y+x\ge M$, so by the concavity of $f(x)=(y-x)(y^2+3y+x)$ we know that $f$ attains its minimum on the boundary, whence $3y^2+3>f(x)\ge M(y^2+4y-M)\ge 4(y^2+4y-4)$, a contradiction for sufficiently large $M\ge4$ (observe that $y^2+4y=(y-x)+(y^2+3y+x)\ge 2M$).

We can use the Runge's method. The curve $(y^{3}+xy-1)(x^{2}+x-y)=(x^{3}-xy+1)(y^{2}+x-y)$ may be approximating by the line $y-x+2=0$ and the parabols $y^2+x-y+1=0$, $x^2+x-y+1=0$.