Question

How do i find a function such as `f(a)f(b)f(c) = f(sqrt(a^2+b^2+c^2))f^2(0)` ?

Answer 1

`f(x) = p k^(x^2)` for constants `p in RR`, `k > 0`

Explanation:

I will assume that by `f^2(0)` you mean `(f(0))^2` rather than `f(f(0))` or `f^((2))(0)`

`color(white)()`
f(x) is an even function

First note that if `f(0) = 0` then `(f(a))^3 = 0` for all `a`, hence `f(a) = 0` for all `a`. So one option for `f(x)` is the constant function `f(x) = 0`.

Otherwise, if we let `b = c = 0` then we find:

`f(a)f(0)f(0) = f(sqrt(a^2+0^2+0^2))f(0)f(0)`

and hence:

`f(a) = f(sqrt(a^2)) = f(abs(a))`

So we can deduce that `f(x)` is an even function.

`color(white)()`
Any constant function is a solution

Suppose `f(x) = k` for all `x in RR`

Then:

`f(a)f(b)f(c) = k^3 = f(sqrt(a^2+b^2+c^2))f(0)f(0)`

`color(white)()`
Are there any non-constant solutions?

Suppose `f(0) = 1` and `f(1) = k` for some constant `k > 0`

Notice that:

`f(sqrt(2)) = f(sqrt(1^2+1^2+0^0))f(0)f(0) = f(1)f(1)f(0) = k^2`

`f(sqrt(3)) = f(sqrt(1^2+1^2+1^2))f(0)f(0) = f(1)f(1)f(1) = k^3`

Observing this pattern, we can define

`f(x) = k^(x^2)`

To find:

`f(a)f(b)f(c) = k^(a^2)k^(b^2)k^(c^2)`

`= k^(a^2+b^2+c^2)`

`= k^((sqrt(a^2+b^2+c^2))^2)`

`= f(sqrt(a^2+b^2+c^2))`

`= f(sqrt(a^2+b^2+c^2))f(0)f(0)`

Note that if `p in RR` is any constant then:

`f(x) = p k^(x^2)`

will also be a solution.

The case `k=1` then covers the previously identified constant solution.