Find an expression for \({u_n}\) in terms of \(n\).

For every prime number \(p > 3\), show that \(p|{u_{p - 1}}\).

the auxiliary equation is \({\lambda ^2} - 5\lambda  + 6 = 0\)     M1

\( \Rightarrow \lambda  = 2,{\text{ }}3\)     (A1)

the general solution is \({u_n} = A \times {2^n} + B \times {3^n}\)     A1

imposing initial conditions (substituting \(n = 0,{\text{ }}1\))     M1

\(A + B = 0\) and \(2A + 3B = 1\)     A1

the solution is \(A =  - 1,{\text{ }}B = 1\)

so that \({u_n} = {3^n} - {2^n}\)     A1

[6 marks]

\({u_{p - 1}} = {3^{p - 1}} - {2^{p - 1}}\)

\(p > 3\), therefore 3 or 2 are not divisible by \(p\)     R1

hence by FLT, \({3^{p - 1}} \equiv 1 \equiv {2^{p - 1}}(\bmod p)\) for \(p > 3\)     M1A1

\({u_{p - 1}} \equiv 0(\bmod p)\)     A1

\(p|{u_{p - 1}}\) for every prime number \(p > 3\)     AG

[4 marks]