5

Find all functions $f \colon \mathbb R \to \mathbb R$ such that for $\forall x, y \in \left[-\dfrac{1}{2}, +\infty\right)$, $$f(x)f(y) + f\left(\frac{1}{2} + \sqrt{xy(x + y) + \frac{1}{4}}\right) = f(xy) + f(x + y)$$

Let $P(x, y)$ be the assertion of $f(x)f(y) + f\left(\dfrac{1}{2} + \sqrt{xy(x + y) + \dfrac{1}{4}}\right) = f(xy) + f(x + y)$.

We have that for $P(0, 0)$ and $P(0, 1)$, it can be seen respectively that $\left\{ \begin{align} f(0)[f(0) - 2] &= -f(1)\\ f(0)[f(1) - 1] &= 0 \end{align} \right.$, which implies that $\left[ \begin{align} f(0) = f(1) = 0\\ f(0) = f(1) = 1 \end{align} \right.$.

And that's all I have for now, beneath are some other miscellaneous ideas that contributed nothing to my thoughts.

  • For $$P\left(\dfrac{xy}{2} - \sqrt{\left(\dfrac{xy}{2}\right)^2 - (x + y)}, \dfrac{xy}{2} + \sqrt{\left(\dfrac{xy}{2}\right)^2 - (x + y)}\right)$$, it could be obtained that $$f(x)f(y) = f\left(\dfrac{xy}{2} - \sqrt{\left(\dfrac{xy}{2}\right)^2 - (x + y)}\right) \cdot f\left(\dfrac{xy}{2} + \sqrt{\left(\dfrac{xy}{2}\right)^2 - (x + y)}\right)$$

  • For $$P\left(\frac{(m - 1) - \sqrt{m^2 - 6m + 1}}{2}, \frac{(m - 1) + \sqrt{m^2 - 6m + 1}}{2}\right)$$, we could suppose that $$f\left(\frac{(m - 1) - \sqrt{m^2 - 6m + 1}}{2}\right) \cdot f\left(\frac{(m - 1) + \sqrt{m^2 - 6m + 1}}{2}\right) = f(2(m - 1))$$

  • The same goes for $$P\left(\frac{(2m - 1) - \sqrt{4m^2 - 20m - 7}}{4}, \frac{(2m - 1) + \sqrt{4m^2 - 20m - 7}}{4}\right)$$, it happens to be that $$f\left(\frac{(2m - 1) - \sqrt{4m^2 - 20m - 7}}{4}\right) \cdot f\left(\frac{(2m - 1) + \sqrt{4m^2 - 20m - 7}}{4}\right) = f(2(2m - 1))$$

That's all for now, thanks for checking in, even if you don't have anything to add to this problem~

Arctic Char
  • 16,007
Lê Thành Đạt
  • 4,228
  • 1
    Why would anybody consider such an equation, is it just because it looks "cool" or "complicated"? –  Dec 21 '20 at 15:50
  • 1
    There is $f(1)$ in $P(a,0)$, how do you demolish that? – Zerox Dec 21 '20 at 15:51
  • Thanks for the reminder, Zerox, I initially, and incorrectly wrote $f\left(\dfrac{1}{2} - \sqrt{xy(x + y) + \dfrac{1}{4}}\right)$ instead of $f\left(\dfrac{1}{2} + \sqrt{xy(x + y) + \dfrac{1}{4}}\right)$ and must have worked on the problem before noticing the error. – Lê Thành Đạt Dec 21 '20 at 15:59
  • 1
    Note that that weird radix and can become negative!? – Hagen von Eitzen Dec 21 '20 at 16:19
  • Fortunately for $P(a,0)$ and $P(a,1)$, the argument insides the radical is always non-negative. – Divide1918 Dec 21 '20 at 16:24
  • 1
    The conditions $x,y \geq -1/2$ and $xy(x+y)+\frac{1}{4}\geq 0$ imply $xy \geq -1/2$, $x+y \geq -1$, hence it is not possible to learn anything about $f(x)$ for $x<-1$. This does not look like well formed equation. Still, solving for $f:[-1,\infty)\to \mathbb{R}$, it can be shown that $f(0)=f(1)=0$ imples $f(x)=0$ on $[-1,\infty)$ (using $P(x,0)$ and then $P(x,x)$). I don't know where $f(0)=f(1)=1$ leads. – Sil Oct 29 '22 at 05:20
  • 1
    Also $f(x)=(x-\frac{1}{2})^2+\frac{3}{4}$ satisfies the equation and is consistent with $f(0)=f(1)=1$, though I have no idea if it is the only solution in this case. – Sil Oct 29 '22 at 05:29

0 Answers0