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]