1. ## Functional Equation question

Show that if:
F(xy) = F(x)F(y);
F(30) = 1;
F(10a+7) = 1 for any integer a 0< a

Then there is either simply one solution, F(x) = 1, or there are more functional equations that fit the above prerequisites. Provide justification.

I understand this is a surjective function and F(30/x) = F(7/x) = F(17/x) etc but I dont know how to identify any more solutions. I have been 'fiddling' for a while and my lack of functional equation knowledge is showing.

2. ## Re: Functional Equation question

Hey pikachu26134.

I'm not sure what results you have to use to prove this but if you don't have F(x) = 1 as the only possibility then the derivative has to be non-zero somewhere if it is differentiable.

If you can show that a non-zero derivative exists then you have proven that more solutions exist.

3. ## Re: Functional Equation question

Hi,
What do you mean by "functional"? Is this a continuous function on R? Since this is posted in the algebra forum, maybe you don't know what a continuous function is. If you do know what a continuous function is, the attachment provides a solution to the problem.

4. ## Re: Functional Equation question

Here's a hint pikachu.

Obviously, it we prove for all primes we are done since F(xy) = F(x)F(y)
From F(30)=1 we see that F(30)=F(2)F(3)F(5)=1
So F(2)=F(3)=F(5)=1 and that is the first few primes. Another small step and you can extrapolate to all primes.

Good luck!

t/n here I assumed that the function only maps to positive integers. But it is not hard to extend to rationals {use F(xy) = F(x)F(y)} and then by density of real numbers to reals.