# Solving functional equation $f(x)^2+f(y)^2=f(x+y)(f(f(x))+f(y))$

I was trying to find functions $f:(0,+\infty)\to(0,+\infty)$ satisfying the following functional equation $$f(x)^2+f(y)^2=f(x+y)(f(f(x))+f(y))$$ The problem is that I can't find here any reasonable substitution. The only thing I've concluded is the following $$f(f(x))-f(x)=\mathrm{const}$$ It follows from the fact that lhs is invariant under transform $(x,y)\to(y,x)$. So how could I procced from here, or may be this is a wrong way?

• Your observation definitely simplifies the problem, since you now can replace the troubling $f\bigl(f(x)\bigr)$ by $f(x)+c$ with a $c$ yet to be determined (maybe it's arbitrary). Aug 31, 2012 at 8:46
• I think $c=0$, because the only solution to this equation that I can see is constant functions Aug 31, 2012 at 8:49
• How about setting $y=0$ in the simplified functional equation $f(x)^2+f(y)^2=f(x+y)(f(x)+f(y)+c)$. This will then lead to the equation $f(x)^2+f(0)^2= f(x)^2 + f(x)( f(0)+c)$ hence $f(x)= f(0)^2/(f(0)+c)$, i.e. $f$ has to be a constant. As noticed this then in turn entails that $c=0$. Or did I miss something? Aug 31, 2012 at 13:03
• You can't put $y=0$, as it's not in the domain. Aug 31, 2012 at 13:12
• Fair point. However assuming $f$ is continous (this is of course not part of the original set of assumptions, but since this is merely a comment anyway, lets impose this additional constraint) we note that if $\lim_{y\to 0}f(y)=\infty$ then we can rewrite the functional equation as $f(x)^2/f(y) + f(y) = f(x+y)(f(x)/f(y) + 1 + c/f(y))$ for $y$ small. But in this equality the left side will converge to $\infty$ for $y\to 0$ whereas the right side converges to $f(x)$. Hence $\lim_{y \to 0}f(y)<\infty$, and this should be enough for us to extend $f$ to a function $f:[0;\infty[ \to \[0;\infty[$. Aug 31, 2012 at 13:28

Here is a solution which is not quite as elegant as I'd have liked, but requires a bit of algebraic computations: something I did in Maple.

As already noted, $$f(x+y)\Big(f\big(f(x)\big)+f(y)\Big)=f(x)^2+f(y)^2=f(x+y)\Big(f\big(f(y)\big)+f(x)\Big)$$ makes $f(f(x))-f(x)=f(f(y))-f(y)$ for all $x,y>0$ (since $f>0$), and so there is a constant $a=f(f(x))-f(x)$ for all $x$. Thus, we have $$f(x)^2+f(y)^2=f(x+y)\Big(f(x)+f(y)+a\Big).\tag{1}$$

If we set $f(x)=k\cdot g(x)$ for some constant $k>0$, we find that $g$ satisfies the same equation as $f$, but with $a$ replaced with $a/k$. Thus, by rescaling, we can without loss of generality assume that $a\in\{-1,0,1\}$.

Now comes the ugly bit. I won't include all the computations, just the main results and enough detail to explain how I got them.

Take some fixed $x$. Let $u=f(x)$. We can now use the formula $$f(x+y)=\frac{f(x)^2+f(y)^2}{f(x)+f(y)+a}\tag{2}$$ to compute $$v=f(2x)=\frac{2u^2}{2u+a}, w=f(3x)=\frac{u^2+v^2}{u+v+a}.$$ Now, we can express $f(4x)$ as either $f(2x+2x)$ or $f(x+3x)$ and require that these be equal. This gives $$\begin{split} 0=&f(x+3x)-f(2x+2x)\\ =& \frac {{u}^{2}a \left( {a}^{6}+10\,u{a}^{5}+42\,{u}^{2}{a}^{4}+100\,{ u}^{3}{a}^{3}+140\,{u}^{4}{a}^{2}+112\,{u}^{5}a+32\,{u}^{6} \right) }{ \left( 4\,{u}^{2}+2\,ua+{a}^{2} \right) \left( 16\,{u}^{4}+22\,{u}^{ 3}a+16\,{u}^{2}{a}^{2}+6\,u{a}^{3}+{a}^{4} \right) \left( 4\,{u}^{2}+ 3\,ua+{a}^{2} \right) } \end{split}$$ where we may focus on the big polynomial in the numerator which has to be zero when $u>0$ and $a\not=0$.

The big polynomial in the numerator is always positive if $a>0$: all the terms are positive. Thus, there is no possible value for $u=f(x)$. Hence, $a$ cannot be negative.

If $a<0$, as Harry Altman noted in a comment, $f(f(x))=f(x)+a$ can be iterated, replacing $x$ with $f(x)$, making $f^{n+1}(x)=f(x)+na$ ($f^k(x)$ is $f$ applied $k$ times). If $a<0$, starting with any $x$, applying $f$ enough times will make $f^{n+1}(x)=f(x)+na<0$.

My original proof was based on the same algebra as for $a>0$: have moved it to a comment at the end.

This leaves $a=0$ as the only option. We then have $f(f(x))=f(x)$, and $f(2x)=f(x+x)=f(x)$ from (2). Since $f(2x)=f(x)$, by (2) we must have $f(2x+y)=f(x+y)$ for all $x,y>0$. If we set $u=f(x)$, $v=f(y)$, this makes $$w=f(x+y)=\frac{u^2+v^2}{u+v},\quad f(2x+y)=f(x+(x+y))=\frac{u^2+w^2}{u+w},$$ which gives $$0=f(2x+y)-f(x+y)=\frac{uv(u-v)}{2u^2+uv+v^2}\implies u=v$$ which implies $f(x)=f(y)$ for all $x,y>0$.

So the only possible solution is $f(x)$ constant, which is easy to verify is a solution.

Original proof for the case $a<0$:

If $a<0$, let's assume $a=-1$, the same polynomial can be solved and has two real roots: found numerically to be $u_1=0.2580535\ldots$ and $u_2=1.887292\ldots$. Thus, for all $x$, $f(x)$ must take either of these two values. However, if we plug these values into (2), we find that if $f(x)$ and $f(y)$ takes values in $\{u_1,u_2\}$, then $f(x+y)$ does not: i.e. $f(x+y)$ then ends up taking a value other than $u_1$ or $u_2$, which is not possible.

• If $u=f(x)$, and $u$ satisfies a polynomial $P(u)=P(f(x))=0$ for all $x$, doesn't that mean the range of $f(x)$ is finite? And if it is continuous, it must be a constant function. Sep 1, 2012 at 22:31
• I can't figure out how f(2x+y)=f((x+y)+y) at that point. Sep 2, 2012 at 2:28
• @nayrb Not only is the range of $f$ finite, it's values have to be roots of $P(u)$. However, I'm not assuming that $f$ is continuous. F.ex. before finding this proof I knew that for $a=0$, $f(x)=f(rx)$ for any rational number $r$, which implies that either it is constant or everywhere discontinuous. Sep 2, 2012 at 3:27
• Note that there's an easy way to see that $a\ge 0$, without the messy algebra: If $a<0$, then iterating $f$ would eventually produce a negative number. Sep 2, 2012 at 5:19
• @Harry Good point! Will incorporate this into the answer: will make things quite a bit easier. Sep 2, 2012 at 5:33

Here's a proof that the only continuous solutions are constant. Suppose $f(x)$ were a nonconstant continuous solution; we will derive a contradiction. Since $f(x)$ is nonconstant and continuous, the range of $f(x)$ contains an interval $(a,b)$. As suggested by Norbert in the original question, we can take advantage of the $(x,y) \to (y,x)$ symmetry and we have two equations $$f(x)^2+f(y)^2=f(x+y)(f(f(x))+f(y))$$ $$f(x)^2+f(y)^2=f(x+y)(f(f(y))+f(x))$$ Hence we have $$f(f(x))+f(y) = f(f(y)) + f(x)$$ In particular, if $f(x) \neq f(y)$ are in $(a,b)$, we have $$\frac{f(f(y)) - f(f(x))}{f(y) - f(x)} = 1$$ Since $(a,b) \subset$ range$(f)$, we may take the limit of the above as $f(y)$ approaches $f(x)$, and get that $$f'(f(x)) = 1$$ This is true for any $f(x) \in (a,b)$, so we have that $f'(y) = 1$ for all $y \in (a,b)$. Hence there is a constant $c$ such that $f(y) = y + c$ on $(a,b)$. As Norbert also pointed out in his original question, there is also a constant $a$ such that we have $$f(x)^2+f(y)^2=f(x+y)(f(y)+f(x) + a)$$ So for $x,y \in (a,b)$ we have $$(x + c)^2 + (y + c)^2 = f(x + y)(x + y + 2c + a)$$ In particular, this is true for $x,y \in (a,b)$ with $x + y = a + b$, for which we have $$(x + c)^2 + (a + b - x + c)^2 = f(a + b)( a + b + 2c + a)$$ However, the left-hand side of the above is not a constant function of $x$, and the right-hand side is. This gives a contradiction and we are done.

Note that this shows that any solution whose range contains an interval is constant. So if any nonconstant $f(x)$ solves this, it will be a rather weird function.

Why not say that $$f(x) = 1$$ For all x,y

$$1^2+1^2=1*(1+1)=2$$

• Any constant function works. The question is whether any others do. Aug 31, 2012 at 16:28

This is not yet a complete answer, but comes forth with less effort than that of Einar. I do not know, whether the last move with the derivative can really be done.

We go back to the original form and use symmetry: $$\tag{1} f(x)^2+f(y)^2 = f(x+y)(f(f(x))+f(y))$$ $$\tag{2} f(x)^2+f(y)^2 = f(x+y)(f(x)+f(f(y)))$$ Subtracting and reorganizing gives (let's write $ff(x)$ for $f(f(x))$ $$\tag{3} 0 = f(x+y)(ff(x)-ff(y) -f(x)+f(y))$$ It was assumed in the definition, that $f(x)$ and thus $f(x+y)$ is not zero, so we can divide by $f(x+y)$ and rearrange $$\tag{4} f(y)-f(x) = ff(y)-ff(x)$$ Let's write $X$ for $f(x)$ now, we get $$\tag{5} Y-X = f(Y)-f(X)$$

Now we assume, $Y = X+h$ where $h \gt 0$ and write $$\tag{6} h = f(X+h)-f(X)$$ and [update] that following move was wrong $$\tag{7 wrong} 0 = {f(X+h)-f(X) \over h}$$ it must be $$\tag{7} 1 = {f(X+h)-f(X) \over h}$$

and I don't see something useful in this now....[/update].

• If $f(x)=c$, a constant, then $Y-X = f(Y)-f(X)$ only makes sense for $X=Y=c$. Thus your trick with setting $Y=X+h$ won't work, because $Y$ can only assume one value. At any rate, you could use contradiction, but the point is you need to pay attention to what are allowed values for $X$ and $Y$. Sep 1, 2012 at 22:03
• upps, my editing of the answer crossed with your comment. I'll see whether I've even messed up... Sep 1, 2012 at 22:04
• Hmm, I'm also speculating, whether using the floor, fraction function or in general taking the residue modulo something is an interesting variant because it becomes constant after the first iteration... Sep 1, 2012 at 22:29
• Wouldn't (6) imply that the limit is 1, not 0 as in (7)? Sep 2, 2012 at 1:35
• (+1) arrgh. True, what a mess. That idea was too hypnotizing, sorry... My first version of this answer was even more wrong; pehaps I should delete that whole answer... Sep 2, 2012 at 7:07