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]