# All wrapped in functions

Find all functions $$f:\mathbb{R}\rightarrow\mathbb{R}$$ such that $$f(x)f\big(f(x)+y\big)=f\big(x^2\big)+f(xy)$$ for all $$x,y\in\mathbb R$$

Problem by me

Most elegant solution wins!

Not sure this rates as elegant but it is certainly short:

At $$x=0,y=-f(0)$$ we have $$f(0)^2=2f(0)$$, a quadratic equation with solutions $$0,2$$.

Case 1: $$f(0)\ne 0$$

substitute $$x=0$$ to get $$2f(y+2)=4$$ for any $$y$$, therefore $$f\equiv 2$$.

Case 2: $$f(0)=0$$

Case 2a: there is another zero $$f(z)=0,z\ne 0$$

substitute $$x=z$$ to get $$f(z^2)=f(zy)$$ for any $$y$$, i.e. $$f$$ must be constant $$f\equiv 0$$.

Case 2b: $$f(0)=0$$ is the only zero

Then, letting $$y$$ equal $$0$$ and $$-f(x)$$, respectively, we can make good use of $$f(x)f(f(x))=f(x^2)=-f(-xf(x))$$ With $$x=1$$ this yields $$f(f(1))=1$$ and $$f(-f(1))=-f(1)$$. With $$x=f(1)$$ we now get $$f(f(1)^2)=f(1)$$ and with $$x=-f(1)$$, finally, $$f(1)=1$$ and $$f(-1)=-1$$.

We now go back to the original equation. Comparing $$x=1,y=t$$ with $$x=-1,y=-t$$ we get $$f(-z)=-f(z)$$ for any $$z$$. Finally, setting $$y=-x$$ we get $$f(f(x)-x)=0$$, i.e. $$f(x)=x$$ for any $$x$$.

• This is similar to my solution. You get the checkmark! Jul 26 '20 at 3:42
• Hi @CulverKwan, thanks for the entertaining puzzle. I'm new to this section. Do you ever disclose your own solution? I'd like to see whether I missed any shortcuts. Jul 26 '20 at 3:53

First, let's get rid of all constant solutions. If $$f(x)=c$$ is a solution, then the equation gives

$$c^2=2c\implies c\in \{0,2\}$$. So from now on let's assume $$f$$ is non-constant.

If some $$x$$ satisfies $$f(x)=0$$, then plugging that in,

$$f(xy)=-f(x^2)$$, and since $$f$$ is non-constant, $$x$$ must be zero. Also, putting $$x=0$$ gives $$f(0)f(f(0)+y)=2f(0)$$, and again $$f(0)\ne 0$$ leads to $$f$$ being constant, so in fact $$f(x)=0\iff x=0$$.

Plugging $$y=0$$ now gives

$$f(x^2)=f(x)f(f(x))$$, so the equation can be written $$f(x)f(f(x)+y)=f(x)f(f(x))+f(xy)$$. Call this alternate version $$(\star)$$.

Now we claim that

$$f$$ is injective. Indeed, suppose $$f(a)=f(b)$$ with $$a\ne b$$. As seen already, $$a,b$$ are non-zero. Now substituting $$x\mapsto a$$ and $$x\mapsto b$$ in $$(\star)$$, we see that all the terms except $$f(xy)$$ have the same value in both cases. Thus $$f(ay)=f(by)$$ for all reals $$y$$. This shows that $$f$$ takes the same value on any two reals with ratio $$a/b$$.

Now plugging $$x=1$$ in the original equation gives $$f(1)f(f(1)+y)=f(1)+f(y)$$. Using this equation for $$y\mapsto ay$$ and $$y\mapsto by$$, we see that in fact $$f(f(1)+ay)=f(f(1)+by)$$, and given any $$c\ne a/b$$, one can choose $$y$$ so that the numbers $$f(1)+ay$$ and $$f(1)+by$$ have ratio $$c$$. Thus given any ratio, one can find two reals with that ratio so that $$f$$ has the same value on them.

And thus, to prove injectivity,

Note that given any non-zero $$x_1,x_2$$, we can find $$c,d$$ so that $$c/d=x_1/x_2$$ and $$f(c)=f(d)$$. Using the same kind of argument as on $$a,b$$, this would give $$f(cy)=f(dy)$$, so in particular $$f(x_1)=f(x_2)$$, so in fact $$f$$ is constant on non-zero reals. Because of $$f(x^2)=f(x)f(f(x))$$, the only possibility (remembering $$f$$ is non-constant) is $$f(x)=1$$ for non-zero $$x$$ and $$0$$ for $$x=0$$; this function, however, doesn't work! So $$f$$ has to be injective.

Phew! Now we are ready for the finish. Indeed, plug in

$$y=-f(x)$$; we get $$f(x^2)=-f(-xf(x))$$. Putting $$x\mapsto -x$$ gives $$f(-xf(x))=f(xf(-x))$$, and by injectivity this gives $$f(-x)=-f(x)$$ (well, as long as $$x\ne 0$$, but this holds for $$x=0$$ anyway so we are good). This says $$f$$ is an odd function, so in fact $$f(x^2)=f(xf(x))$$. Again, injectivity allows us to "cancel out" the outer $$f$$'s, giving $$f(x)=x$$ (again, for non-zero $$x$$, but $$f(0)=0$$ is already known).

Thus these are all such $$f$$:

The constant functions $$0$$ and $$2$$, and the identity function.

• Any more elegant solutions? No need to prove injective. Jun 19 '20 at 4:26
• This is awesome! Jun 19 '20 at 21:47
• Does this proof touch on the case when f never takes on the value 0? Jun 19 '20 at 22:34
• @phenomist the second spoiler block shows that can never happen for non-constant $f$. Jun 20 '20 at 3:01

Not an answer, but a step toward an answer:

Let $$x$$ be a number such that $$f(x)=0$$. Then the definition becomes $$0=f(x^2)+f(xy)$$. Thus $$f(xy)$$ is constant as a function of $$y$$ (for this $$x$$) and thus either $$x=0$$ or $$f$$ is constant. We wind up with two choices: $$f\equiv 0$$; or $$f(x)=0\Rightarrow x=0$$ (which latter includes the case that $$f$$ is never $$0$$).

(The examples already found by other answerers —

$$f\equiv 0$$, $$f\equiv 2$$, and $$f(x)=x$$

— all meet this criterion.)

• Good progress, I am using this method! Try finishing it. Jun 19 '20 at 4:26

I'm not sure about "all functions", but I've found 2:

$$f(x) = 0$$

and

$$f(x) = 2$$

Along with the solutions provided by Steve, I've found that:

$$f(x) = x$$

Is also a valid solution

$$f(x)f(f(x)+y) = f(x)f(x+y) = x(x+y) = x^2+xy$$

$$f(x^2)+f(xy) = x^2+xy$$ (Note this doesn't apply for $$f(x) = cx$$ where $$c\ne1$$, as you result with $$c^3x^2 + c^2xy = cx^2+cxy$$