Find all real-valued functions on the reals which satisfy f( xf(x) + f(y) ) = f(x)2 + y for all x, y.
Solution
Answer: (1) f(x) = x for all x; (2) f(x) = -x for all x.
Put x = 0, then f(f(y)) = f(0)2 + y. Put y = -f(0)2 and k = f(y). Then f(k) = 0. Now put x = y = k. Then f(0) = 0 + k, so k = f(0). Put y = k, x = 0, then f(0) = f(0)2 + k, so k = 0. Hence f(0) = 0.
Put x = 0, f(f(y)) = y (*). Put y = 0, f(xf(x)) = f(x)2 (**). Put x = f(z) in (**), then using f(z) = x, we have f(zf(z)) = z2. Hence z2 = f(z)2 for all z (***). In particular, f(1) = 1 or -1. Suppose f(1) = 1. Then putting x = 1 in the original relation we get f(1 + f(y) ) = 1 + y. Hence (1 + f(y) )2 = (1 + y)2. Hence f(y) = y for all y.
Similarly if f(1) = -1, then putting x = 1 in the original relation we get f(-1 + f(y) ) = 1 + y. Hence (-1 + f(y) )2 = (1 + y)2, so f(y) = -y for all y.
Finally, it is easy to check that f(x) = x does indeed satisfy the original relation, as does f(x) = -x.
Comment. Amazingly, exactly the same question as was set 3 years later in 00/1.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002