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Suppose the function $f:Bbb RtoBbb R$ satisfies the following conditions:

$$begin{align}
f(4xy)&=2y[f(x+y)+f(x-y)] \[4pt]
f(5)&=3
end{align}$$

Find the value of $f(2015)$.

I have tried to find some other hiding condition, like $f(0)=0,$ but which is useless.

Firstly, it’s trivial that
$$f(0) = f(4cdot 0cdot 0) = 2cdot 0 cdot [f(0) + f(0)] = 0$$

Next, we see that
$$0 = f(0) = f(4cdot 0 cdot y) = 2ycdot [f(y) + f(-y)]text{,}$$
which implies that $f(-y) = -f(y)$ for all $yneq 0$.

Afterwards, one notices that
$$f(4xy) = f(4yx)$$
Hence,
$$2ycdot [f(x+y) + f(x-y)] = 2xcdot [f(x+y) + f(y-x)]$$
$$Leftrightarrow (x-y)cdot f(x+y) = (x+y)cdot f(x-y)$$
$$Leftrightarrow f(x+y) = frac{x+y}{x-y}cdot f(x-y)$$

If we substitute now $x=1010$ and $y=1005$, we get that
$$f(2015) = frac{2015}{5}cdot 3 = 1209$$

With the provided function,

$$fleft(4xyright) = 2yleft[fleft(x + yright) + fleft(x - yright)right] tag{1}label{eq1}$$

First, substitute $y = 0$ to get

$$fleft(0right) = 0 tag{2}label{eq2}$$

Next, substitute $x = 0$ and use eqref{eq2} to get

$$0 = 2yleft[fleft(yright) + fleft(-yright)right] tag{3}label{eq3}$$

Thus, for all values of $y$ other than $0$, dividing both sides by $2y$ gives

$$fleft(yright) = -fleft(-yright) tag{4}label{eq4}$$

In other words, $f$ is an odd function. Next, substituting $y = 1$ into eqref{eq1} gives

$$fleft(4xright) = 2left[fleft(x + 1right) + fleft(x - 1right)right] tag{5}label{eq5}$$

Now, using $x = 1$ in eqref{eq5} gives

$$fleft(4right) = 2left[fleft(2right) + fleft(0right)right] tag{6}label{eq6}$$

Thus, using eqref{eq2} gives

$$fleft(4right) = 2fleft(2right) tag{7}label{eq7}$$

Similarly, using $x = 3$ in eqref{eq5} gives

$$fleft(12right) = 2left[fleft(4right) + fleft(2right)right] tag{8}label{eq8}$$

Thus, using eqref{eq7} gives

$$fleft(12right) = 6fleft(2right) tag{9}label{eq9}$$

Note that eqref{eq2}, eqref{eq4}, eqref{eq7} and eqref{eq9} all satisfy

$$fleft(nxright) = nfleft(xright) forall ; n in N tag{10}label{eq10}$$

In particular, for eqref{eq2}, $n = 0 ; forall ; x in R$; for eqref{eq4}, $n = -1 ; forall ; x in R$; and for eqref{eq7} & eqref{eq9}, $x = 2$ only, with $n = 2$ & $n = 6$, respectively. This doesn't prove eqref{eq10} always works, but it suggests that $f$ is a linear multiple of $x$, i.e, $fleft(xright) = kx text{ for a non-zero constant } k in R$, which from the given condition of

$$fleft(5right) = 3 tag{11}label{eq11}$$

gives a value of $k = frac{3}{5}$ so the function would be

$$fleft(xright) = cfrac{3x}{5} tag{12}label{eq12}$$

To confirm this, substitute eqref{eq12} into eqref{eq5} to give on the left side

$$cfrac{12x}{5} tag{13}label{eq13}$$

with the right side becoming

$$2left[cfrac{3left(x + 1right)}{5} + cfrac{3left(x - 1right)}{5} right] = 2left[cfrac{left(3x + 3 + 3x - 3right)}{5} right] = cfrac{12x}{5} tag{14}label{eq14}$$

Similarly, using eqref{eq12} in eqref{eq1} gives a left & right hand side of $frac{12xy}{5}$, confirming it's also a solution of the original equation. I'm not quite sure offhand how to prove it's unique, but I assume it is. We thus get a final answer of

$$fleft(2015right) = cfrac{3 times 2015}{5} = 1209 tag{15}label{eq15}$$

Note: The solution could involve just "guessing" the function being a constant multiple of the argument, and then showing this works to determine the final answer, but I thought it might be useful for people to see how one can approach solving this problem somewhat systematically as I did. In particular, I tried using small constant values for $x$ and $y$, checking what information this offered, and then seeing if I could leverage this when using other values. For example, I started with $x = 0$ and $y = 0$, then used $y = 1$, $x = 1$ and $x = 3$. Each time, I used previous details to help simplify and learn more about the new values, until I saw the pattern which I then confirmed worked.