24th Putnam 1963 Problem B3 R is the reals. Find all f : R → R which are twice differentiable and satisfy: f(x) 2 - f(y) 2 = f(x + y) f(x - y).
Differentiate wrt x: 2 f(x) f '(x) = f '(x + y) f(x - y) + f(x + y) f '(x - y). Differentiate the result wrt y: 0 = f ''(x + y) f(x - y) - f '(x + y) f '(x - y) + f '(x + y) f '(x - y) - f(x + y) f ''(x - y). Put X = x + y, Y = x - y. Then we have f ''(X) f(Y) = f(X) f ''(Y). If f(X) = 0 for all X, then we have a solution. So suppose for some X 0 , f(X 0 ) is non-zero and put k = f ''(X 0 )/f(X 0 ), then f ''(Y) = k f(Y). Now we consider separately k = 0, k ≤ 0 and k > 0. If k = 0, then integrating gives f(Y) = AY + B. But putting y = 0 in the original relation gives immediately that f(y) = 0, so B = 0 and we have the solution f(Y) = AY. This includes the solution f(Y) = 0 noticed earlier. If k  0, put k = a 2 . Then f(Y) = A sinh aY + B cosh aY. Again B = 0 and we have the solution f(Y) = A sinh aY. It remains to check that these are solutions. It is obvious that f(Y) = AY is a solution. If f(Y) = A sin aY, then f(x + y)f(x - y) = A 2 (sin ax cos ay + cos ax sin ay)(sin ax cos ay - cos ax sin ay) = A 2 (sin 2 ax cos 2 ay - cos 2 ax sin 2 ay) = A 2 (sin 2 ax - sin 2 ay) = f(x) 2 - f(y) 2 , as required. Similarly for the sinh expression. 24th Putnam 1963 © John Scholes jscholes@kalva.demon.co.uk 5 Feb 2002