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


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 $f(x) \ge -x$.

  • $\begingroup$ 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. $\endgroup$ Nov 8 '18 at 0:26
  • $\begingroup$ I accept Patrick's answer and will give you some reward, $\endgroup$ 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 $-1$.

  • $\begingroup$ It's great, I feel we're almost there... Thank you very much. $\endgroup$ Nov 2 '18 at 21:14
  • $\begingroup$ Maybe if there's some result for extends a result to $\mathbb{Z}$ to $\mathbb{R}$. $\endgroup$ Nov 2 '18 at 21:24
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 2 '18 at 22:06
  • $\begingroup$ thank you very much. Please read Derek's response, it's very interesting ... I accept your response and give Derek some reward. Thank you. $\endgroup$ 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 $$f(x) = -x$$

  • 2
    $\begingroup$ "This has solution $f(x)=-x^N$." Why? Is this the only solution? Where is a proof? $\endgroup$ Nov 2 '18 at 7:40
  • $\begingroup$ $f(x^2)=(-x^2)^N=-x^{2N}$ 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. $\endgroup$ Nov 2 '18 at 20:46
  • $\begingroup$ It's great. Please, how did you get these are the only solutions for $f(x^2)=-f(x)^2$? Thank you very much. $\endgroup$ Nov 2 '18 at 21:14
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    $\begingroup$ 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. $\endgroup$
    – M. Wind
    Nov 2 '18 at 21:18
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    $\begingroup$ 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. $\endgroup$
    – M. Wind
    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 $f\equiv 0$.

  • $\begingroup$ 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 $f(1\boxed{-}y^2)=f(y)^2-1=f(-y)^2-1$? $\endgroup$ Nov 8 '18 at 13:33

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