Find all functions f: Z -> Z such that f(a + b) = f(a) f(b) + f(b) for all a,b, and f(1) = k-1, where k is some integer greater than 1?

2 Answers
Jun 1, 2017

f(n)=-((1-(k-1)^n)(k-1))/(k-2)

Explanation:

For k > 2

Making b=1

f(a+1)-(k-1)f(a)=k-1 so we have the difference equation

f(n+1)-(k-1)f(n)=k-1 with solution

f(n)=C_0(k-1)^(n-1)-((1-(k-1)^n)(k-1))/(k-2)

but

f(1)=C_0-((1-(k-1))(k-1))/(k-2) = k-1 rArr C_0=0 and finally

f(n)=-((1-(k-1)^n)(k-1))/(k-2)

Jun 9, 2017

No solution exists

Explanation:

Since f(a+b)=f(a)f(b)+f(b), f(b+a)=f(b)f(a)+f(a).

Since f(a+b)=f(b+a), f(a)f(b)+f(b)=f(b)f(a)+f(a).

This means that f(a)=f(b) for all a,binZZ.

In addition, in f(a+b)=f(a)f(b)+f(b), set a=0,b=1 to get f(0+1)=f(0)f(1)+f(1), or f(0)f(1)=0.

Since f(1)=k-1,k>1, f(1)!=0. This means that f(0)=0.

But we just said that f(a)=f(b) for all a,binZZ. This means that f(x)=0 for all x inZZ. From 0=f(1)=k-1, k=1, which is not possible.

Thus, no solution exists.