# Constructing a functional equation that has given solution set.

Motivation was that all the functional equations that appear in Olympiad have mainly one solution. for example, $xf(x^2)f(f(y))+f(yf(x))=f(xy)(f(f(x^2))+f(f(y^2)))$

or

$f(x+y)^2=2f(x)f(y)+max(f(x^2+y^2),f(x^2)+f(y^2))$

You can find infinitely many of them in AOPS site.

Let's consider following functional equations.

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

this clearly has infinitely many non-trivial solutions, that is, $f(x)=x^a$

My question is, does my following conjecture hold?

Conjecture: we call functional equation GOOD if $f(x)=x$ is a solution, and it is consist of algebraic operation of $f$. If there are finitely many non-constant solutions. Then the equation's only non-trivial solution is $f(x)=x$.

My question could be unclear, so basically I'm asking the following.

Can you make a functional equation that only $f(x)=x, f(x)=x^2$ is solution?

More generally, is it possible to construct a functional equation that the solution set is given set?

• Inequalities? These are all equations (equalities). – Robert Israel Jan 19 '18 at 8:30
• mistakes were made! sorry – Simo Ryu Jan 19 '18 at 8:32
• $f(x)=x^2$ is a functional equation. – Professor Vector Jan 19 '18 at 10:24
• @Somos That has no solutions except when $x = x^2$ (i.e. $x=0$ or $1$). – Robert Israel Jan 19 '18 at 21:50

Your conjecture is not correct. For an example possibly more interesting than the trivial cases, like the one suggested by @RobertIsrael, you can take the following functional equation: $$g \big( x g ( y ) - x y \big) + x y \big( 2 - y + x y ^ 2 \big) = ( 1 - y ) g ( x y ) + x \big( 1 + x y ^ 2 \big) g ( y )$$ The only solutions of this equation are given by $g ( x ) = x$ and $g ( x ) = 2 x - x ^ 2$. To show this, define $f ( x ) = g ( x ) - x$ and you'll have: $$f \big( x f ( y ) \big) = ( 1 - y ) f ( x y ) + x ^ 2 y ^ 2 f ( y )$$ You can see here for a proof showing that the only solutions to this equation are $f ( x ) = 0$ and $f ( x ) = x - x ^ 2$.
Try $$(f(x)-x)(f(y)-y^2) = 0$$