2015 IMO Problems/Problem 5
Let be the set of real numbers. Determine all functions
:
satisfying the equation
for all real numbers and
.
Proposed by Dorlir Ahmeti, Albania
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Let be the set of real numbers. Determine all functions
:
satisfying the equation
for all real numbers and
.
Proposed by Dorlir Ahmeti, Albania
This problem needs a solution. If you have a solution for it, please help us out by adding it.
f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x)
(1) Put x=y=0 in the equation, We get f(0 + f(0)) + f(0) = 0 + f(0) + 0 or f(f(0)) = 0 Let f(0) = k, then f(k) = 0
(2) Put x=0, y=k in the equation, We get f(0 + f(k)) + f(0) = 0 + f(k) + kf(0) But f(k) = 0 and f(0) = k so, f(0) + f(0) = f(0)^2 or f(0)[f(0) - 2] = 0 Hence f(0) = 0, 2
Case I : f(0) = 0
Put x=0, y=x in the equation, We get f(0 + f(x)) + f(0) = 0 + f(x) + xf(0) or, f(f(x)) = f(x) Say f(x) = z, we get f(z) = z
So, f(x) = x is a solution
Case II : f(0) = 2 Again put x=0, y=x in the equation, We get f(0 + f(x)) + f(0) = 0 + f(x) + xf(0) or, f(f(x)) + 2 = f(x) + 2x
We observe that f(x) must be a polynomial of power 1 as any other power (for that matter, any other function) will make the LHS and RHS of different powers and will not have any non-trivial solutions.
Also, if we put x=0 in the above equation we get f(2) = 0
f(x) = 2-x satisfies both the above.
Hence, the solutions are f(x) = x and f(x) = 2-x.