# 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)# ?

##### 1 Answer

#### Answer:

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

#### Explanation:

I will assume that by

**f(x) is an even function**

First note that if

Otherwise, if we let

#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

**Any constant function is a solution**

Suppose

Then:

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

**Are there any non-constant solutions?**

Suppose

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

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

will also be a solution.

The case