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

Revision as of 12:07, 14 May 2017

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

Solution:

$f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x)$ for all real numbers $x$ and $y$ .

(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 $1$ : $f(0) = 0$

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 $2$ : $f(0) = 2$ 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 $\boxed{\color{red}{f(x) = x}}$ and $\boxed{\color{red}{f(x) = 2-x}}$ .

@@ Line 7: / Line 7: @@
 Proposed by Dorlir Ahmeti, Albania
-[b][i][size = 100]Solution:[/b][/i][/size]
+Solution:
 <math>f(x+f(x+y)) + f(xy) = x + f(x+y) + yf(x)</math> for all real numbers <math>x</math> and <math>y</math>.

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

Revision as of 12:07, 14 May 2017