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#?

Question

873 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-01T15:01:36

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

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)#

Answer 2 · 2017-06-09T02:45:50

No solution exists

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.