Given that $a \equiv b(\bmod p)$ , show that ${a^n} \equiv {b^n}(\bmod p)$ for all $n \in {\mathbb{Z}^ + }$ .

Show that ${29^{13}} + {13^{29}}$ is exactly divisible by $7$.

$a \equiv b(\bmod p) \Rightarrow a = b + pN,N \in \mathbb{Z}$      M1

${a^n} = {(b + pN)^n} = {b^n} + n{b^{n - 1}}pN \ldots$     M1A1

$= {b^n} + pM$ where $M \in \mathbb{Z}$     A1

this shows that ${a^n} \equiv {b^n}(\bmod p)$     AG

[4 marks]

$29 \equiv 1(\bmod 7) \Rightarrow {29^{13}} \equiv {1^{13}} \equiv 1(\bmod 7)$     M1A1

$13 \equiv - 1(\bmod 7) \Rightarrow {13^{29}} \equiv {( - 1)^{29}} \equiv - 1(\bmod 7)$     A1

therefore ${29^{13}} + {13^{29}} \equiv 1 + ( - 1) \equiv 0(\bmod 7)$     M1A1

this shows that ${29^{13}} + {13^{29}}$ is exactly divisible by $7$     AG

[5 marks]