Here Andrei Lazanu, age 14, School No. 205, Bucharest, Romania gives another excellent solution to this problem which he has extended after doing some research on the web.

First, I calculated the first 10 polynomials that satisfy the recurrence relation given in the problem:

Pn+2(x)=xPn+1(x)-Pn(x)

where P₀(x)=0 and P₁(x)=1. I successively found:

P2(x) = x P3(x) = x2 - 1 P4(x) = x3 - 2x = x(x2 - 2) P5(x) = x4 - 3x2 + 1 P6(x) = x5 - 4x3 + 3x = x(x2 - 1)(x2 - 3) P7(x) = x6 - 5x4 + 6x2 - 1 P8(x) = x7 - 6x5 + 10x3 - 4x = x(x2 - 2)(x4 - 4x2 +2) P9(x) = x8 - 7x6 + 15x4 - 10x2 + 1 P10(x) = x(x4 - 3x2 + 1)(x4 - 5x2 + 5) .

From the examination of the expressions of the polynomials, I drew some conclusions:

(1) Odd order polynomials contain only even powers of x, including zero.

(2) Even order polynomials contain only odd powers of x.

(3) There are alternate signs of terms in each polynomial, starting with the first, of order (n-1) for P_n(x), which is positive.

I have also shown, as required in the question, that P₄(x) contains as a factor P₂(x), P₆(x) contains as factor P₃(x) (that is every root of P₃ is a root of P₆), P₈(x) contains as factor P₄(x) and P₁₀(x) contains as factor P₅(x.

P4(x) P2(x) = x2 - 2 P6(x) P3(x) = x(x2 - 3) P8(x) P4(x) = x4 - 4x2 + 2 P10(x) P5(x) = x(x4 - 5x2 + 5).

Using the defining recurrence relation we can express P₆ in terms of previous polynomials in the sequence.

P6 = xP5-P4 = (x2-1)P4-xP3 = P3P4 - P2P3 = P3(P4-P2) .

Similarly we can express P₈ in terms of previous polynomials in the sequence.

P8 = xP7-P6 = (x2-1)P6-xP5 = (x3-2x)P5-(x2-1)P4 = P4(P5-P3) .

Again we can express P₁₀ in terms of previous polynomials in the sequence.

P10 = xP9-P8 = (x2-1)P8-xP7 = (x3-2x)P7-(x2-1)P6 = (x4 - 3x2 +1)P6 - (x3 - 2x)P5 = P5P6-P4P5 = P5(P6-P4).

Editor's note: This suggests a conjecture that P_2k=P_k(P_k+1-P_k-1) where k is any natural number. This is true but the general proof is beyond the scope of school mathematics.

Andrei made further observations about the coefficients in these polynomials in the hope of finding explicit formulae for the nth order polynomial. He found, looking at Fibonacci numbers, that these polynomials are very similar to Fibonacci polynomials, which are given by the recursive relation:

Fn+2(x) = xFn+1(x) + Fn(x)

with F₀(x) = 0 and F₁(x) = 1. Using these relations, he found the first 10 Fibonacci polynomials, which are, up to the signs found in P_n(x), identical with F_n(x).

Using the explicit formula of Fibonacci polynomials from http://mathworld.wolfram.com, Andrei hoped to write correctly the explicit formula for the polynomials in the problem, as:

Pn(x)= (n-1)/2 å j=0  (-1)j Cn-j-1jxn-2j-1.

The formula works well up to n = 10. From this formula, all properties could be found easily, although Andrei was not able to demonstrate the general case that P_2n(x) is divisible by P_n(x).