2022HSCShort answer

Q13 (a)

3 marks

Question

Prove that for all integers $n$ with $n\ge3$ , if $2^n-1$ is prime, then $n$ cannot be even.

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Solution

Criteria	Marks
Provides correct solution	3
Uses the assumption that $n=2k$ to factor $2^n-1$ , or equivalent merit	2
Attempts to use the contrapositive, or equivalent merit	1

Sample answer:
It is equivalent to prove the contrapositive: ‘if $n$ is even, then $2^n-1$ is not prime’.

Assume $n$ is even. With $n\ge3$ , that means $n=2k$ where $k$ is an integer with $k\ge2$ .

2^n-1=2^{2k}-1=(2^k)^2-1=(2^k+1)(2^k-1)

With $k\ge2$ , $2^k+1\ge5$ and $2^k-1\ge3$ , so $2^n-1$ has two proper factors so it is not prime.

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Syllabus alignment

Proof

The Nature of ProofUse formal proof language, contradiction, counter-examples and rigorous arguments involving numbers and inequalities.

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