### 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).

Solution

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 X0, f(X0) is non-zero and put k = f ''(X0)/f(X0), 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 = - a2. Then f(Y) = A sin aY + B cos aY. But f(0) = 0, so B = 0 and we have the solution f(Y) = A sin aY. If k > 0, put k = a2. 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) = A2(sin ax cos ay + cos ax sin ay)(sin ax cos ay - cos ax sin ay) = A2(sin2ax cos2ay - cos2ax sin2ay) = A2(sin2ax - sin2ay) = f(x)2 - f(y)2, as required. Similarly for the sinh expression.