This seems not from Poland 2003...

Peter wrote: This seems not from Poland 2003... It's correct. Check it at "http://www.mimuw.edu.pl/~chel/Olimp/2etang03.html" I think we need change the problem statement as the official version : The polynomial $ W(x) = x^{4}-3x^{3}+5x^{2}-9x$ is given. Determine all pairs of different positive integers a,b satifying the equation $ W(a) = W(b)$.

Strange, that's a different paper from the one on Kalva. :

Peter wrote: Strange, that's a different paper from the one on Kalva. : What Kalva offers is the Final Round problems of PMO. The above problem is from Second Round.

suppose that $ W(a) = W(b)$ then since $ a$ and $ b$ are different we have $ (a^2 + b^2)(a + b) - 3(a^2 + ab + b^2) + 5(a + b) - 9 = 0.$ so $ (a^2 + b^2 + 5)(a + b - 3) = 3(ab - 2)$. if $ ab = 2$ then $ a = 1,b = 2$ or $ (a = 2,b = 1)$ which are solutions. if $ ab = 1$ then $ a = b = 1$ ,there are no solution in this case . now suppose $ ab > 2$. and let $ ab = x^2$.and $ a + b > 2$.($ a + b = 2$ gives $ a = b = 1$ and $ a + b = 1$ doesn't have solution) $ 3 = \frac {(a^2 + b^2 + 5)(a + b - 3)}{(ab - 2)} \ge \frac {(2ab + 5)(2\sqrt {ab} - 3)}{(ab - 2)} = \frac {(2x^2 + 5)(2x - 3)}{(x^2 - 2)}$. so after expanding (for $ x > 3/2$ ) we get $ 4x^3 - 9x^2 + 10x - 9 \le 0$. and consider the function $ f(x) = 4x^3 - 9x^2 + 10x - 9$, we have $ f'(x) = 12x^2 - 18x + 10 > 0$ so $ f$ is stricly increasing therefore $ x < 2$ (because $ f(2) > 0$).then $ ab \le 3$ so $ ab = 3$ then $ a = 1$ and $ b = 3$ (or $ a = 3$ and $ b = 1$) which not verify the equation. and if $ x \le 3/2$ then $ ab \le 9/4=2,...$ so $ ab \le 2$.

without loss of generality asume a>b. from the above post we see that $ 3 =\frac{(a^{2}+b^{2}+5)(a+b-3)}{(ab-2)} \rightarrow 3 > a+b-3 \rightarrow a+b\le 5$. Now just plug in all a,b that satisfy this and find that only 1,2 works.