Difference between revisions of "2015 IMO Problems/Problem 5"

Revision as of 19:31, 7 November 2016

Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f$ : $\mathbb{R}\rightarrow\mathbb{R}$ satisfying the equation

$f(x+f(x+y))+f(xy) = x+f(x+y)+yf(x)$

for all real numbers $x$ and $y$ .

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 $\mathbb{R}$ be the set of real numbers. Determine all functions $f$ : $\mathbb{R}\rightarrow\mathbb{R}$ satisfying the equation

$f(x+f(x+y))+f(xy) = x+f(x+y)+yf(x)$

for all real numbers $x$ and $y$ .

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.

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Difference between revisions of "2015 IMO Problems/Problem 5"

Revision as of 19:31, 7 November 2016

@@ Line 19: / Line 19: @@
-All forms of a function can be expressed in a major form as
+f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x)
-as
-<math>f(r) = ar^t + c</math>
+(1) Put x=y=0 in the equation,
-using this,  <math>f(x+f(x+y))+f(xy) = a(x + a(x+y)^t + c)^t + c +a(xy)^t + c</math>
+We get f(0 + f(0)) + f(0) = 0 + f(0) + 0
-and     <math>x+f(x+y)+yf(x) =x + a(x+y)^t + c + ay(x)^t +yc</math>
+or f(f(0)) = 0
-and for both expressions to be equal,
+Let f(0) = k, then f(k) = 0
-<math>t</math> has to be 1,
-<math>c</math> has to be 0,
+(2) Put x=0, y=k in the equation,
-and <math>a</math> has to be 1.
+We get f(0 + f(k)) + f(0) = 0 + f(k) + kf(0)
-therefore the function that can satisfy the equation in the question is <math>f(r) = r^1 + 0</math>
+But f(k) = 0 and f(0) = k
-which is <math>f(r) = r</math>.
+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.
 [[Category:Olympiad Algebra Problems]]
 [[Category:Functional Equation Problems]]