I encountered the following question in a math-contest
Find all continuous functions from $\mathbb{R}$ to $\mathbb{R}$ such that $$(f(x+y))^2-(f(x-y))^2=4f(x)f(y).$$
I substituted $y$ by $0$ to get that either $f$ is the zero function or the image of $0$ by $f$ is $0$. Now I found one solution (the zero function). Moving to the second case I substituted $y$ by $x$ to get $f(2x)^2=4f(x)^2$. I separated the two subcases: Either $f(2x)=2f(x)$ Or $f(2x)=-2f(x)$. Also flipping $x$ and $y$ gives us the fact that $f$ is an odd function. However I couldn't get any further. I hope you would like to help me finish the solution of this problem.
Here’s a straightforward solution. Note that you don’t need to do anything complicated with periodicity nor show the general Cauchy equation.
Letting $x=y=0$ gives $f(0)=0$.
First let’s restrict to considering $g(x)=|f(x)|$.
Let’s show by induction that $g(ky)=kg(y)$ for all nonnegative integer $k$. We already have $k=0$ and $k=1$. Letting $x=ky$ for $k\geq 2$ and applying the inductive hypothesis to $k$ and $k-1$ gives $g((k+1)y)^2=(k-1)^2f(y)+4kf(y)^2=(k+1)^2g(y)^2$, so $g((k+1)y)=(k+1)g(y)$ as desired.
Thus, for all positive rational $r=m/n$, $ g(ry)=g((m/n)y)=mg(y/n)=m/n g(y)=r g(y)$.
Thus, for all positive rational $r$, $g(r)=rg(1)$, so letting $f(1)=c,g(1)=|c|$ gives that $g$ satisfies $g(x)=|cx|$ for all positive rational $x$. Using the continuity of $g$ and the density of the positive rational numbers over the positive reals gives that $|f(x)|=g(x)=|cx|$ for all positive $x$.
If $c=0$, then $f(x)=0=cx$ for all positive $x$. Otherwise, consider $f(x)/(cx)$ for positive $x$. It is always either $1$ or $-1$, it is continuous, and it is 1 at 1 so it must be constantly 1, thus, $f(x)=cx$ for positive $x$.
Now, let’s resolve the negatives.
Substituting $(y,y)$ and $(y,-y)$ gives that $4f(y)f(y)=f(2y)^2-f(0)^2=-4f(y)f(-y)$, so $f(y)(f(y)+f(-y))=0$. Negating $y$ and adding the two equations gives that $(f(y)+f(-y))^2=0$, so $f(y)=-f(y)$. This holds for all $y$, so $f$ is odd.
Thus, for all positive $x$, $f(-x)=-f(x)=-cx=c(-x)$, so $f(y)=c y$ holds for all $y$, so all solutions are of this form. Substituting this general equation satisfies the original expression, so all equations of this form are solutions.