The polynomials $P_n(x,y), n=1,2,... $ are defined by \[P_1(x,y)=1, P_{n+1}(x,y)=(x+y-1)(y+1)P_n(x,y+2)+(y-y^2)P_n(x,y)\] Prove that $P_{n}(x,y)=P_{n}(y,x)$ for all $x,y \in \mathbb{R}$ and $n $.
Problem
Source: Bulgaria 1998 (round 4)
Tags: algebra, polynomial, algebra unsolved
huynguyen
10.02.2016 12:56
Sorry to revive, but it's nice to clear this topic with at least a solution.
Tintarn
10.02.2016 23:53
If I haven't made a silly mistake (...) it seems to be easily proved by induction: Denote $A(n)$ the proposition that $(y+1)P_n(x,y+2)-(x+1)P_n(y,x+2)=(y-x)P_n(x,y)$ holds for all $x,y \in \mathbb{R}$. Note that, assuming $A(n)$ to be true, we find \[P_{n+1}(x,y)-P_{n+1}(y,x)=(x+y-1)(y+1)P_n(x,y+2)-(x+y-1)(x+1)P_n(y,x+2)+(y-y^2)P_n(x,y)-(x-x^2)P_n(x,y)\]\[=(x+y-1)(y-x)P_n(x,y)+(y-x)(1-x-y)P_n(x,y)=0\]and hence $P_{n+1}$ is indeed symmetric. So it suffices to prove $A(n)$ for all $n$. We proceed by induction: $A(1)$ is trivially true. Now, assume for some $n$ we know that $A(n)$ and $A(n-1)$ are true. Hence $P_n$ is symmetric. Using these hypotheses and the recurrence relation, we find: \begin{align*}&(y+1)P_{n+1}(x,y+2)-(x+1)P_{n+1}(y,x+2)+(x-y)P_{n+1}(x,y)\\ &=(y+1)(x+y+1)(y+3)P_n(x,y+4)-(y+1)^2(y+2)P_n(x,y+2)-(x+1)(x+y+1)(x+3)P_n(y,x+4)+(x+1)^2(x+2)P_n(y,x+2)+(x-y)(x+y-1)(y+1)P_n(x,y+2)+(x-y)(y-y^2)P_n(x,y)\\ &=(y+1)(x+1)(x+y+1)P_n(y+2,x+2)+(y+1)(x+y+1)(y-x+2)P_n(x,y+2)-(y+1)^2(y+2)P_n(x,y+2)\\ &-(x+1)(y+1)(x+y+1)P_n(x+2,y+2)-(x+1)(x+y+1)(x-y+2)P_n(y,x+2)+(x+1)^2(x+2)P_n(y,x+2)+(x-y)(x+y-1)(y+1)P_n(x,y+2)+(x-y)(y-y^2)P_n(x,y)\\ &=(y+1)(x+y+1)(y-x+2)P_n(x,y+2)-(y+1)^2(y+2)P_n(x,y+2)+(x+1)^2(x+2)P_n(y,x+2)-(x+1)(x+y+1)(x-y+2)P_n(y,x+2)+(x-y)(x+y-1)(y+1)P_n(x,y+2)+(x-y)(y-y^2)P_n(x,y)\\ &=(y+1)(x-x^2)P_n(x,y+2)-(x+1)(y-y^2)P(y,x+2)+(x-y)(x+y-1)(y+1)P(x,y+2)+(x-y)(y-y^2)P_n(x,y)\\ &=(y-y^2)\left[(y+1)P_n(x,y+2)-(x+1)P(y,x+2)+(x-y)P_n(x,y) \right]\\ &=0 \end{align*}and hence $A(n+1)$ is true. Thus, we proved the induction step and hence established that $S(n)$ holds for all $n$. Done!