MathDB

8

Part of 2014 Iran MO (3rd Round)

Problems(1)

Recursive Polynomial

Source: Iran 3rd round 2014 - final exam problem 8

9/16/2014
The polynomials kn(x1,…,xn)k_n(x_1, \ldots, x_n), where nn is a non-negative integer, satisfy the following conditions k0=1k_0=1 k1(x1)=x1k_1(x_1)=x_1 kn(x1,…,xn)=xnkn−1(x1,…,xn−1)+(xn2+xn−12)kn−2(x1,…,xn−2)k_n(x_1, \ldots, x_n) = x_nk_{n-1}(x_1, \ldots , x_{n-1}) + (x_n^2+x_{n-1}^2)k_{n-2}(x_1,\ldots,x_{n-2}) Prove that for each non-negative nn we have kn(x1,…,xn)=kn(xn,…,x1)k_n(x_1,\ldots,x_n)=k_n(x_n,\ldots,x_1).
algebrapolynomialalgebra unsolved