Consider the system of difference equations n>=0:

x₁(n+1)=x₁(n)+ax₂(n) x₂(n+1)=bx₁(n)

Derive two conditions regarding a and b such that if either of them holds, then the system has periodic solutions. Determine for each condition the period of the associated periodic solutions.

For your information for instance a two periodic solution means that x(n+2)=x(n) and a three periodic solution means that x(n+3)=x(n) etc. n must be a natural number.

I thought I had to do something with the eigenvalues of the system of difference equations. Which can be calculated by: (1-labda)(-labda)-ab=0, which results to labda₁=1/2-1/2root(1+4ab) and labda₂=1/2+1/2root(1+4ab). So it is a periodic solution if labda₁^{^p}=1 where p=2,3,4.... andlabda₁is unequal to 1, orlabda₂^{^p}=1 where p=2,3,4.... andlabda₂is unequal to 1, due to a theorem.

Please help me out! I'm studying for my exam on monday and I am really stuck here. Thanks in advance!