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

$$beginalign
f(4xy)&=2y[f(x+y)+f(x-y)] \[4pt]
f(5)&=3
endalign$$

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,
$$2ycdot [f(x+y) + f(x-y)] = 2xcdot [f(x+y) + f(y-x)]$$
$$Leftrightarrow (x-y)cdot f(x+y) = (x+y)cdot f(x-y)$$
$$Leftrightarrow f(x+y) = fracx+yx-ycdot f(x-y)$$

If we substitute now $x=1010$ and $y=1005$, we get that
$$f(2015) = frac20155cdot 3 = 1209$$

With the provided function,

$$fleft(4xyright) = 2yleft[fleft(x + yright) + fleft(x - yright)right] tag1labeleq1$$

First, substitute $y = 0$ to get

$$fleft(0right) = 0 tag2labeleq2$$

Next, substitute $x = 0$ and use eqrefeq2 to get

$$0 = 2yleft[fleft(yright) + fleft(-yright)right] tag3labeleq3$$

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

$$fleft(yright) = -fleft(-yright) tag4labeleq4$$

In other words, $f$ is an odd function. Next, substituting $y = 1$ into eqrefeq1 gives

$$fleft(4xright) = 2left[fleft(x + 1right) + fleft(x - 1right)right] tag5labeleq5$$

Now, using $x = 1$ in eqrefeq5 gives

$$fleft(4right) = 2left[fleft(2right) + fleft(0right)right] tag6labeleq6$$

Thus, using eqrefeq2 gives

$$fleft(4right) = 2fleft(2right) tag7labeleq7$$

Similarly, using $x = 3$ in eqrefeq5 gives

$$fleft(12right) = 2left[fleft(4right) + fleft(2right)right] tag8labeleq8$$

Thus, using eqrefeq7 gives

$$fleft(12right) = 6fleft(2right) tag9labeleq9$$

Note that eqrefeq2, eqrefeq4, eqrefeq7 and eqrefeq9 all satisfy

$$fleft(nxright) = nfleft(xright) forall ; n in N tag10labeleq10$$

In particular, for eqrefeq2, $n = 0 ; forall ; x in R$; for eqrefeq4, $n = -1 ; forall ; x in R$; and for eqrefeq7 & eqrefeq9, $x = 2$ only, with $n = 2$ & $n = 6$, respectively. This doesn't prove eqrefeq10 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

$$fleft(5right) = 3 tag11labeleq11$$

gives a value of $k = frac35$ so the function would be

$$fleft(xright) = cfrac3x5 tag12labeleq12$$

To confirm this, substitute eqrefeq12 into eqrefeq5 to give on the left side

$$cfrac12x5 tag13labeleq13$$

with the right side becoming

$$2left[cfrac3left(x + 1right)5 + cfrac3left(x - 1right)5 right] = 2left[cfracleft(3x + 3 + 3x - 3right)5 right] = cfrac12x5 tag14labeleq14$$

Similarly, using eqrefeq12 in eqrefeq1 gives a left & right hand side of $frac12xy5$, 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

$$fleft(2015right) = cfrac3 times 20155 = 1209 tag15labeleq15$$

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.