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]

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