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

Latest revision as of 23:24, 14 February 2024

Problem

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 -- fallacy

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. -- fallacy

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 1: / Line 1: @@
+==Problem==
 Let <math>\mathbb{R}</math> be the set of real numbers. Determine all functions <math>f</math>:<math>\mathbb{R}\rightarrow\mathbb{R}</math> satisfying the equation
@@ Line 7: / Line 9: @@
 Proposed by Dorlir Ahmeti, Albania
-{{solution}}
+==Solution==
-Let <math>\mathbb{R}</math> be the set of real numbers. Determine all functions <math>f</math>:<math>\mathbb{R}\rightarrow\mathbb{R}</math> satisfying the equation
+<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>.
+(1) Put <math>x=y=0</math> in the equation,
+We get<math> f(0 + f(0)) + f(0) = 0 + f(0) + 0</math>
+or <math>f(f(0)) = 0</math>
+Let <math>f(0) = k</math>, then <math>f(k) = 0</math>
+(2) Put <math>x=0, y=k</math> in the equation,
+We get <math>f(0 + f(k)) + f(0) = 0 + f(k) + kf(0)</math>
+But <math>f(k) = 0</math> and <math>f(0) = k</math>
+so, <math>f(0) + f(0) = f(0)^2</math>
+or <math>f(0)[f(0) - 2] = 0</math>
+Hence <math>f(0) = 0, 2</math>
+Case <math>1</math> : <math>f(0) = 0</math>
+Put <math>x=0, y=x</math> in the equation,
+We get <math>f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)</math>
+or, <math>f(f(x)) = f(x)</math>
+Say <math>f(x) = z</math>, we get <math>f(z) = z</math>
+So, <math>f(x) = x</math> is a solution -- fallacy
+Case <math>2</math> : <math>f(0) = 2</math>
+Again put <math>x=0, y=x</math> in the equation,
+We get <math>f(0 + f(x)) + f(0) = 0 + f(x) + xf(0)</math>
+or, <math>f(f(x)) + 2 = f(x) + 2x</math>
-<math>f(x+f(x+y))+f(xy) = x+f(x+y)+yf(x)</math>
+We observe that <math>f(x)</math> must be a polynomial of power <math>1</math> as any other power (for that matter, any other function) will make the <math>LHS</math> and <math>RHS</math> of different powers and will not have any non-trivial solutions. -- fallacy
-for all real numbers <math>x</math> and <math>y</math>.
+Also, if we put <math>x=0</math> in the above equation we get <math>f(2) = 0</math>
-Proposed by Dorlir Ahmeti, Albania
+<math>f(x) = 2-x</math> satisfies both the above.
-{{solution}}
+Hence, the solutions are <math>\boxed{\color{red}{f(x) = x}}</math> and <math>\boxed{\color{red}{f(x) = 2-x}}</math>.
+==See Also==
-All forms of a function can be expressed in a major form as
+{{IMO box|year=2015|num-b=4|num-a=6}}
-as
-<math>f(r) = ar^t + c</math>
-using this,  <math>f(x+f(x+y))+f(xy) = a(x + a(x+y)^t + c)^t + c +a(xy)^t + c</math>
-and     <math>x+f(x+y)+yf(x) =x + a(x+y)^t + c + ay(x)^t +yc</math>
-and for both expressions to be equal,
-<math>t</math> has to be 1,
-<math>c</math> has to be 0,
-and <math>a</math> has to be 1.
-therefore the function that can satisfy the equation in the question is <math>f(r) = r^1 + 0</math>
-which is <math>f(r) = r</math>.
 [[Category:Olympiad Algebra Problems]]
 [[Category:Functional Equation Problems]]

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

Latest revision as of 23:24, 14 February 2024

Problem

Solution

See Also