Standard +0.8 This requires proving a general statement about prime numbers using algebraic factorization. Students must recognize that 2^n - 1 factors as (2^(n/2) - 1)(2^(n/2) + 1) when n is even, then show both factors are greater than 1. While the factorization is a standard difference of squares pattern, applying it correctly in a proof context and ensuring all logical steps are complete makes this moderately challenging for A-level.
6 Shona makes the following claim.
" \(n\) is an even positive integer greater than \(2 \Rightarrow 2 ^ { n } - 1\) is not prime"
Prove that Shona's claim is true.
\(2^{2k} - 1\) or \(4^k - 1\) (where \(k\) is an integer \(> 1\)) \(= (2^k)^2 - 1 = (2^k-1)(2^k+1)\)
M1, A1
or \(2^{2k+2}-1\) or \(2^{2k+4}-1\); Allow \(2^{2n}-1\); \(= (2^{k+1}-1)(2^{k+1}+1)\) oe or \((2^{k+2}-1)(2^{k+2}+1)\) oe; Induction method: Assume \(2^k-1\) is \(\div\) by 3 (\(k\) even) M1; Let \(2^k - 1 = 3p\) (\(p\) integer)
\((2^k+1) > 1\) and \(k > 1\), hence \((2^k - 1) > 1\); Hence \((2^k-1)(2^k+1)\) is the product of two integers, both \(> 1\), and hence \(2^n - 1\) is not prime
M1, A1
Both statements needed; \(2^{k+2} - 1 = 4\times2^k - 1 = 4\times2^k - 4 + 3 = 4(2^k-1)+3 = 4\times3p+3\) which is \(\div\) by 3; When \(k=2\): \(2^2 - 1 = 3\) so \(\div\) by 3 A1; Hence true for all even \(n\). Claim true A1
# Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2^{2k} - 1$ or $4^k - 1$ (where $k$ is an integer $> 1$) $= (2^k)^2 - 1 = (2^k-1)(2^k+1)$ | M1, A1 | or $2^{2k+2}-1$ or $2^{2k+4}-1$; Allow $2^{2n}-1$; $= (2^{k+1}-1)(2^{k+1}+1)$ oe or $(2^{k+2}-1)(2^{k+2}+1)$ oe; **Induction method:** Assume $2^k-1$ is $\div$ by 3 ($k$ even) M1; Let $2^k - 1 = 3p$ ($p$ integer) |
| $(2^k+1) > 1$ and $k > 1$, hence $(2^k - 1) > 1$; Hence $(2^k-1)(2^k+1)$ is the product of two integers, both $> 1$, and hence $2^n - 1$ is not prime | M1, A1 | Both statements needed; $2^{k+2} - 1 = 4\times2^k - 1 = 4\times2^k - 4 + 3 = 4(2^k-1)+3 = 4\times3p+3$ which is $\div$ by 3; When $k=2$: $2^2 - 1 = 3$ so $\div$ by 3 A1; Hence true for all even $n$. Claim true A1 |
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6 Shona makes the following claim.\\
" $n$ is an even positive integer greater than $2 \Rightarrow 2 ^ { n } - 1$ is not prime"\\
Prove that Shona's claim is true.
\hfill \mbox{\textit{OCR H240/02 2019 Q6 [4]}}