Balkan 88/2 solution

5th Balkan 1988

Problem 2

Find all real polynomials p(x, y) such that p(x, y) p(u, v) = p(xu + yv, xv + yu) for all x, y, u, v.

Solution

Solution by Robert Israel.

Answer: The possible polynomials are p(x, y) = (x + y)^m(x - y)ⁿ for some positive integers m, n or p(x, y) = 0.

p(x, 0) p(x, 0) = p(x², 0) for all x (*). But p(x, 0) is a polynomial in x. Suppose it is a_nxⁿ + ... + a₀ with a_n non-zero. Then we must have a_n² = a_n and hence a_n = 1. Now suppose there is another non-zero term. Take a_m to be the highest such. Then the lhs of (*) has a non-zero x^m+n term and the rhs does not. Contradiction. So we must have p(x, 0) = xⁿ for some n, or p(x, 0) = 0 for all x.

If p(x, 0) = 0 for all x, then p(u, v) = p(1, 0) p(u, v) = 0, so p is identically zero. Assume then that p(x, 0) = xⁿ for some n. Then p(xu, xv) = p(x, 0) p(u, v) = xⁿ p(u, v). So p(x, y) is homogeneous of degree n. In other words, all its terms have the form a x^h y^k, where h and k are non-negative integers with sum n.

Now put g(x, y) = p( (x+y)/2, (x-y)/2). Then p(x, y) = g(x + y, x - y) and g(x, y) g(u, v) = g(xu, yv). The definition shows that g must also be homogeneous of degree n. So suppose g(x, y) = c₀yⁿ + c₁x y^n-1 + c₂x²y^n-2 + ... + c_nyⁿ. But now if we compare the expansions of g(x, y) g(u, v) and g(xu, yv) the former has terms with x and u to different powers unless all but one c_i is zero. So we must have g(x, y) = c x^hy^n-h for some h. We then require c² = c and hence c = 1. So g(x, y) = x^hy^k for some h, k. Hence p(x, y) = (x + y)^h(x - y)^k for some h, k.

5th Balkan 1988