# Find all functions such that $f(x^2+y^2f(x))=xf(y)^2-f(x)^2$

I am dealing with the test of the OBM (Brasilian Math Olympiad), University level, 2016, phase 2.

I hope someone can help me to discuss this test. Thanks for any help.

The question 2 says:

Find all functions $$f:\mathbb{R}\rightarrow \mathbb{R}$$ such that

## $$f(x^2+y^2f(x))=xf(y)^2-f(x)^2$$

for all $$x,y\in\mathbb{R}$$.

My attempt:

Note that $$f(0)\in\{0,-1\}$$. In fact, by taking $$x=y=0$$, we have $$f(0)=-f(0)^2$$.

Case 1 $$f(0)=0$$

By taking $$y=0$$, we have

$$f(x^2)=-f(x)^2\forall x\in\mathbb{R}$$

Particularly, $$f(1)=-f(1)^2$$, so $$f(1)\in\{0,-1\}$$.

(a) f(1)=0

By taking $$x=1$$, we have $$f(1)=f(y)^2\forall y\in\mathbb{R}$$.

So, $$f\equiv 0$$. Is trivial that it respects the statement.

(b) f(1)=-1

By taking $$x=1$$, we have $$f(1-y^2)=f(y)^2-1=-f(y^2)-1\forall y\in\mathbb{R}$$. So, to $$t\leq 0$$, we have $$f(1-x)=-f(x)-1$$.

By taking $$y=1$$, we have $$f(x^2+f(x))=x-f(x)^2=x+f(x^2) \forall x\in\mathbb{R}$$.

I could not finish this subcase

Case 2 $$f(0)=-1$$

By taking $$x=0$$,

$$f(-y^2)=-1\forall y\in\mathbb{R}$$

So, $$f(t)=-1\forall t\leq0$$.

By taking $$y=0$$,

$$f(x^2)=-x-f(x)^2 \forall x\in\mathbb{R}$$

So, $$f(t)=-\sqrt{t}-1\forall t\geq0$$.

But this function does not is correct. For instance, to $$x=y=1$$, $$f(x^2+y^2f(x))=f(1+1(-2))=f(-1)=-1$$, but $$xf(y)^2-f(x)^2=1(-2)^2-(-2)=6\not=-1$$.

My solution builds on Patrick Stevens's answer. For now on, I'm considering the case where $$f$$ isn't zero everywhere, and I'll prove that $$f(x)=-x$$ everywhere.

We already have $$f(x+1)=f(x)-1$$ for $$x\ge 0$$. But this is true for all $$x$$, here's why. Let $$t \ge 0$$ and set $$x=1$$ and $$y=\sqrt{t}$$ in the original identity, using $$f(x^2)=-f(x)^2$$. We get $$f(1-t)=-1-f(t)$$. Substitute $$t=1-s$$ to get $$f(s)=-1-f(1-s)$$ for $$s \le 1$$. Therefore, $$f(x)+f(1-x)=-1$$ for all $$x$$. Using sign-reversal and induction, we find $$f(x+n)=f(x)-n$$ for all real $$x$$ and integer $$n$$.

Let $$n$$ be an integer and $$t \ge 0$$ be a real. Set $$x=-n$$ and $$y=\sqrt{t}$$ to get $$f(n^2 + t f(n))=f(n^2) - n f(t)$$, which leads to $$f(-tn)-n^2=-n^2 - n f(t)$$, then $$f(n t) = n f(t)$$. Using sign-reversal, this is also true when $$t$$ is negative, so (replacing $$t$$ with $$x$$) $$f(n x) = n f(x)$$ for all real $$x$$ and integer $$n$$.

Replace $$x$$ with $$x/n$$ to find $$f(x/n) = n f(x/n)/n = f(nx/n)/n = f(x)/n$$. Let $$a$$ be an integer and $$b$$ be a positive integer. Then $$f((a/b)x) = f(a(x/b)) = a f(x/b) = a f(x) / b = (a/b)f(x)$$ and $$f(x + a/b) = f((bx + a)/b) = f(bx + a)/b = (f(bx) - a)/b = f(bx)/b - a/b = f(x) - a/b$$. So \begin{align} f(x+q) &= f(x)-q \\ f(qx) &= q f(x) \end{align} for all real $$x$$ and rational $$q$$.

$$f(q)=-q$$ for all rational $$q$$. Now let's show that it's true for irrational values.

We already know that $$f$$ if negative over positive values, and vice versa. Let $$x$$ be any irrational number, and let $$q < x$$ be some rational number. Then $$f(x-q)=f(x)+q$$. Since $$x-q$$ is positive, $$f(x-q)$$ is negative, and so $$f(x)<-q$$. We can choose $$q$$ to be as close as we want, so $$f(x) \le -x$$. Doing the same from the other side shows $$f(x) \ge -x$$.

• Wonderful this answer...! I just think I'm not certain about the part "So more algebra gives us $f(x+q)=f(x)-q$. If you could give us a little more details here... I'm very thanks again. Nov 8 '18 at 0:26
• I accept Patrick's answer and will give you some reward, Nov 8 '18 at 0:33

Partial progress, but not a complete answer, I'm afraid.

$$f(x^2+y^2f(x)) = xf(y)^2-f(x)^2$$

# $$f$$ has a root

Let $$y=x$$; then $$f(x^2(1+f(x)) = (x-1)f(x)^2$$. In particular, letting $$x=1$$ we obtain $$f(1+f(1)) = 0$$, so $$f$$ does have a root.

# $$f$$ is $$0$$ or has exactly the root $$0$$

Suppose $$f(x) = 0$$. Then $$f(x^2) = x f(y)^2$$ for all $$y$$, and so either $$x = 0$$ or $$f(y)^2$$ is constant as $$y$$ varies.

Suppose $$f(x) = 0$$ but $$x \not = 0$$. Then $$f(y)^2$$ is constant as $$y$$ varies; but substituting $$y = x$$ we obtain that $$f(y)^2 = 0$$ and hence $$f$$ is the constant $$0$$.

So the only possible nonzero case is that $$f$$ has exactly one root, and it is the root $$x = 0$$.

# $$f$$ is very nearly symmetric

Substitute $$y \to -y$$ to obtain the following: $$x f(y)^2-f(x)^2 = f(x^2+y^2f(x)) = x f(-y)^2-f(x)^2$$ from which $$x f(y)^2 = x f(-y)^2$$ for all $$x$$ and $$y$$; in particular, $$f(y) = \pm f(-y)$$ for all $$y$$.

# $$f$$ is odd or $$0$$

Suppose $$f(x) = f(-x)$$. Then $$x f(y) - f(x)^2 = f(x^2 + y^2 f(x)) = -x f(y) - f(-x)^2 = -x f(y) - f(x)^2$$ and so $$-x f(y) = x f(y)$$ for all $$y$$; so (since wlog $$f$$ is not the constant zero function) $$-x = x$$ and hence $$x=0$$.

So if $$f(x) = f(-x)$$ then $$x = 0$$; hence $$f(-x) = -f(x)$$ for all $$x$$.

# $$f$$ is sign-reversing or $$0$$

Note also that since $$f(x^2) = -f(x)^2$$ (by letting $$y=0$$), for every $$x > 0$$ we have $$f(x) < 0$$.

# $$f(n) = -n$$ or $$f=0$$

Substituting $$x=-1$$ gives $$f(1+y^2) = -f(y)^2-1$$ and in particular $$f(x^2+1) = f(x^2)-1$$

Therefore $$f(x+1) = f(x)-1$$ whenever $$x>0$$. This fixes the value of $$f$$ on the natural numbers: we have $$f(n) = -n$$.

We already know that the root occurs at $$x=1+f(1)$$, so $$f(1) = -1$$ (as you noted). Moreover, by letting $$x=y$$ and supposing $$f(x)=-1$$, we get $$f(0) = x-1$$ at any such $$x$$, and so $$x=1$$ is the only time $$f$$ hits $$-1$$.

• It's great, I feel we're almost there... Thank you very much. Nov 2 '18 at 21:14
• Maybe if there's some result for extends a result to $\mathbb{Z}$ to $\mathbb{R}$. Nov 2 '18 at 21:24
• The usual ways are to extend the result to $\mathbb{Q}$ by induction somehow on the denominator, and then to $\mathbb{R}$ by continuity. But neither of those steps seems easy here: showing that $f$ is continuous is a very unnatural thing to do with your problem. Nov 2 '18 at 22:06
• thank you very much. Please read Derek's response, it's very interesting ... I accept your response and give Derek some reward. Thank you. Nov 8 '18 at 0:28

Consider first the case $$x = 0$$. The equation reduces to:

$$f(y^{2}f(0)) = -f(0)^2$$

The right hand side is independent of $$y$$, leaving two possibilities: (a) $$f$$ is constant; (b) $$f(0)=0$$.

If we examine case (a), it follows that the constant is either $$0$$ or $$-1$$. Substitution into the general equation shows that only $$f = 0$$ is possible.

Case (b). Assume $$f(0)=0$$. Consider what happens when we take $$y = 0$$. The equation becomes:

$$f(x^2) = -f(x)^2$$

This has solutions of the type $$f(x) = -x^N$$ and $$f(x) = -abs(x)^N$$. In both cases we must have $$N > 0$$ to satisfy the condition $$f(0)=0$$. Now substitute both solutions into the general case, where both $$x$$ and $$y$$ are variables. It becomes quickly clear that only the first solution works and only for $$N = 1$$.

In conclusion there appear to be two solutions to the problem, namely:

$$f(x) = 0$$ and $$f(x) = -x$$

• "This has solution $f(x)=-x^N$." Why? Is this the only solution? Where is a proof? Nov 2 '18 at 7:40
• $f(x^2)=(-x^2)^N=-x^{2N}$ and $-f(x)^2+x(f(0))^2=-(-x^N)^2=-x^{2N}$, so in fact there's a solution to your first equation... But I also have doubts to the uniquess. Nov 2 '18 at 20:46
• It's great. Please, how did you get these are the only solutions for $f(x^2)=-f(x)^2$? Thank you very much. Nov 2 '18 at 21:14
• It is very difficult to prove that these are the only solutions! Perhaps an expert can point to theorems that can be used to construct such a proof. Nov 2 '18 at 21:18
• I treated your math problem as if it were a puzzle. I just tried a few ideas on a sheet of paper, using only high school math, nothing sophisticated. Nov 3 '18 at 2:08

If $$f(a)=0$$ for some $$a\ne0$$, then $$\tag{a,y}f(a^2)=af(y)^2$$ for all $$y$$, making $$|f|$$ constant and hence $$f\equiv 0$$.

Assume $$f(b)=f(-b)=c$$ for some $$b\ne0$$. Then $$\tag{b,y}f(b^2+y^2c)=bf(y)^2-c^2$$ together with $$\tag{-b,y}f(b^2+y^2c)=-bf(y)^2-c^2$$ leads to $$f\equiv 0$$.

In order to look for other solutions than the zero function, we may thus assume $$\tag1\forall x\ne0\colon f(x)\ne 0,$$ $$\tag2\forall x\ne0\colon f(x)\ne f(-x).$$ From $$\tag{1,1}f(1+f(1))=f(1)^2-f(1)^2=0$$ and $$(1)$$, we conclude $$f(0)=0$$ and $$f(1)=-1$$.

By combining $$\tag{1,y}f(1+y^2)=f(y)^2-1$$ $$\tag{1,-y}f(1+y^2)=f(-y)^2-1,$$ we see that $$f^2$$ is even, hence by $$(2)$$, $$f$$ is odd. In particular $$f(-1)=1$$. Then $$\tag{1,-1}f(1+1)=1-1=0$$ contradicts $$(1)$$.

Conclusion: The only solution is $$f\equiv 0$$.

• Thanks for your comment. I think there's something wrong at $f(1+y^2)=f(y)^2-1$ and $f(1+y^2)=f(-y)^2-1$. Should not it be $f(1\boxed{-}y^2)=f(y)^2-1=f(-y)^2-1$? Nov 8 '18 at 13:33