Let $p = {2^k} + 1,{\text{ }}k \in {\mathbb{Z}^ + }$ be a prime number and let G be the group of integers 1, 2, ..., p − 1 under multiplication defined modulo p.

By first considering the elements ${2^1},{\text{ }}{2^2},{\text{ ..., }}{2^k}$ and then the elements ${2^{k + 1}},{\text{ }}{2^{k + 2}},{\text{ …,}}$ show that the order of the element 2 is 2k.

Deduce that $k = {2^n}{\text{ for }}n \in \mathbb{N}$ .

The identity is 1.     (R1)

Consider

${2^1},{\text{ }}{2^2},{\text{ }}{2^3},{\text{ ..., }}{2^k}$

${2^k} = p - 1$     R1

Therefore all the above powers of two are different     R1

Now consider

${2^{k + 1}} \equiv 2p - 2(\bmod p) = p - 2$     M1A1

${2^{k + 2}} \equiv 2p - 4(\bmod p) = p - 4$     A1

${2^{k + 3}} = p - 8$

etc.

${2^{2k - 1}} = p - {2^{k - 1}}$

${2^{2k}} = p - {2^k}$     A1

$= 1$     A1

and this is the first power of 2 equal to 1.     R2

The order of 2 is therefore 2k.     AG

Using Lagrange’s Theorem, it follows that 2k is a factor of ${2^k}$ , the order of the group, in which case k must be as given.     R2

[12 marks]

Few solutions were seen to this question with many candidates unable even to start.