TITLE: Solve this functional equation: $f(x)f(y)=f(x_1)f(y_1)$ when $x^2+y^2=x_1^2+y_1^2$
QUESTION [8 upvotes]: Functional equations such as this one appear only once every several years on exams, so I feel it's hard to have a sure-fire way to approach the problem, unlike, say, solving a series convergence problem, multiple variable integration, or proving some results using basic Fourier series.
So, when I do see a solution offered for one of these problems and study the solution for a substantial amount of time, I still cannot remember how to solve these types of problems, when I come across another one.
But the question is:

Find all the real-valued continuous functions $f$ on $\mathbb R$ which satisfy $$f(x)f(y)=f(x_1)f(y_1)$$
for all $x$, $y$, $x_1$, $y_1$ such that $x^2+y^2=x_1^2+y_1^2$.

Ideally, besides offering a solution, I would love to hear about your intuition on how to solve these functional equations.
Thanks,

REPLY [7 votes]: You can linearize the problem by introducing $$g(u):=\ln\left(f(\sqrt u)\right).$$
Then with $u=x^2,v=y^2$, 
$$u+v=u'+v'\implies g(u)+g(v)=g(u')+g(v').$$
Setting $u'=0,v'=u+v$,
$$g(u)+g(v)=g(0)+g(u+v)$$
shows that the function must be affine,
$$g(u)=au+b,$$
and $$f(x)=e^{ax^2+b}=F_0\left(\frac{F_1}{F_0}\right)^{x^2}.$$

The intuitions/tricks behind this:

it is often advantageous to linearize to benefit of what we know from linear algebra and make the equations look more familiar;
products can be linearized by means of logarithms;
non-linear functions can be linearized by means of a change of variable with the function inverse;
when you have a property involving several variables, try to exploit it by assigning particular values to some of them.