This article discovers, and PROVES, a solution of a Functional Equation that\ came up in Jonathan Bloom and Alex Burstein's article "Egge triples and unbalanced Wilf-equivalence" http://arxiv.org/abs/1410.0230 Eq. (2.5), pages 6-7. They solved it via an ingenious human argument, but we can do it faster! ------------------------------------------------------------------- Solution of a Certain Functional Equation By Shalosh B. Ekhad Theorem: Let, f(x, t), be the formal power series, in, x, t, satisfying the FUNCTIONAL equation 3 2 2 2 2 f(x, t) = (-f(x, 1) f(x, t) t x + f(x, 1) t x + f(x, 1) f(x, t) t x 2 2 - f(x, 1) t x + f(x, t) t x - f(x, 1) f(x, t) x - t x f(x, 1) - f(x, t) t x + f(x, 1) x + f(x, t) x + t - 1)/((-1 + x) (-1 + t) (t x f(x, 1) - 1)) then P=f(x,1) satisfies the ALGEBRAIC equation 2 2 + (x - 3) f(x, 1) + f(x, 1) = 0 Proof: Moving everything in the functional equation to the left, clearing denominat\ ors and taking the numerator entails the followsing algebraic relation between , f(x, t), f(x, 1), x, and, t 3 2 2 2 2 2 2 f(x, 1) f(x, t) t x - f(x, 1) t x + f(x, 1) f(x, t) t x 2 2 2 - f(x, 1) f(x, t) t x - f(x, 1) f(x, t) t x + f(x, 1) t x 2 - f(x, t) t x + f(x, 1) f(x, t) x + t x f(x, 1) - f(x, 1) x + t f(x, t) - f(x, t) - t + 1 = 0 Eliminating f(x,1) between these two equations, by taking the resultant, giv\ es the following ALGEBRAIC relation between , f(x, t), x, t 2 2 3 2 2 (2 t x - t x - t + 1) (1 - 3 f(x, t) t x + f(x, t) - 2 f(x, t) 2 2 2 2 + 6 f(x, t) x + 2 t f(x, t) + 3 f(x, t) x - f(x, t) t - 3 f(x, t) x 2 3 4 2 2 2 3 3 - 6 f(x, t) x + f(x, t) t x + 2 f(x, t) t x + 3 f(x, t) t x 2 3 3 2 3 3 2 - 2 f(x, t) t x + 2 f(x, t) t x - 3 f(x, t) t x + f(x, t) t x 2 2 2 2 4 2 2 3 - 5 f(x, t) t x + 2 f(x, t) t x + 10 f(x, t) t x - f(x, t) t x 2 2 2 3 2 2 2 3 + 4 f(x, t) t x + f(x, t) t x + 3 f(x, t) t x - 8 f(x, t) t x 2 4 3 2 2 + f(x, t) t x - 4 f(x, t) t x - 5 f(x, t) t x - 3 x - t + 3 x + 4 t x 2 3 3 2 2 2 2 + 3 t x + 3 t x - 6 t x - 3 t x + t x) = 0 To prove that the above algebraic equation for f(x,1) , namely 2 2 + (x - 3) f(x, 1) + f(x, 1) = 0 is the right one, let's substitute, t = 1, in the above relation for, f(x, t), getting 2 2 2 x (3 x - 3 x + 1) (f(x, 1) + f(x, 1) x - 3 f(x, 1) + 2) = 0 2 2 The quotient of the right sides is, (3 x - 3 x + 1) x the latter is indeed a multiple of the former, QED! This ends this article, that took, 0.185, seconds. to generate. Have a great day!