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

$$f(4xy)=2y[f(x+y)+f(x-y)]$$

$$f(5)=3$$

Find the value of $f(2015)$.

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

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 by $y$ 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} imply that

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

This suggests 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 is

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

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

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

Thus, eqref{eq12} gives a solution to eqref{eq1}. I'm not quite sure how to prove it's unique (note it's been almost 30 years since I've done very much higher level math, with this being one reason this solution is likely more wordy than necessary), but I assume it is. As such, we 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.