# 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) ?

Apr 24, 2016

$f \left(x\right) = p {k}^{{x}^{2}}$ for constants $p \in \mathbb{R}$, $k > 0$

#### Explanation:

I will assume that by ${f}^{2} \left(0\right)$ you mean ${\left(f \left(0\right)\right)}^{2}$ rather than $f \left(f \left(0\right)\right)$ or ${f}^{\left(2\right)} \left(0\right)$

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f(x) is an even function

First note that if $f \left(0\right) = 0$ then ${\left(f \left(a\right)\right)}^{3} = 0$ for all $a$, hence $f \left(a\right) = 0$ for all $a$. So one option for $f \left(x\right)$ is the constant function $f \left(x\right) = 0$.

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

$f \left(a\right) f \left(0\right) f \left(0\right) = f \left(\sqrt{{a}^{2} + {0}^{2} + {0}^{2}}\right) f \left(0\right) f \left(0\right)$

and hence:

$f \left(a\right) = f \left(\sqrt{{a}^{2}}\right) = f \left(\left\mid a \right\mid\right)$

So we can deduce that $f \left(x\right)$ is an even function.

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Any constant function is a solution

Suppose $f \left(x\right) = k$ for all $x \in \mathbb{R}$

Then:

$f \left(a\right) f \left(b\right) f \left(c\right) = {k}^{3} = f \left(\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}\right) f \left(0\right) f \left(0\right)$

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Are there any non-constant solutions?

Suppose $f \left(0\right) = 1$ and $f \left(1\right) = k$ for some constant $k > 0$

Notice that:

$f \left(\sqrt{2}\right) = f \left(\sqrt{{1}^{2} + {1}^{2} + {0}^{0}}\right) f \left(0\right) f \left(0\right) = f \left(1\right) f \left(1\right) f \left(0\right) = {k}^{2}$

$f \left(\sqrt{3}\right) = f \left(\sqrt{{1}^{2} + {1}^{2} + {1}^{2}}\right) f \left(0\right) f \left(0\right) = f \left(1\right) f \left(1\right) f \left(1\right) = {k}^{3}$

Observing this pattern, we can define

$f \left(x\right) = {k}^{{x}^{2}}$

To find:

$f \left(a\right) f \left(b\right) f \left(c\right) = {k}^{{a}^{2}} {k}^{{b}^{2}} {k}^{{c}^{2}}$

$= {k}^{{a}^{2} + {b}^{2} + {c}^{2}}$

$= {k}^{{\left(\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}\right)}^{2}}$

$= f \left(\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}\right)$

$= f \left(\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}\right) f \left(0\right) f \left(0\right)$

Note that if $p \in \mathbb{R}$ is any constant then:

$f \left(x\right) = p {k}^{{x}^{2}}$

will also be a solution.

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