P1.1aProof by deduction, exhaustion and counter-example

• B1 for choosing a value of \(n\) that gives a non-prime value, e.g. \(n = 10\) or \(n = 11\)
• B1 for showing the value is not prime, e.g. \(121 = 11 \times 11\) (or \(143 = 11 \times 13\)), with a conclusion that the claim is false

Worked solution: When \(n = 10\), \(n^2 + n + 11 = 100 + 10 + 11 = 121 = 11 \times 11\).
121 is not prime, so the claim is false.

• B1 for a correct counter-example with both facts shown, e.g. \(x = -4\): \((-4)^2 = 16 \gt 9\) but \(-4\) is not greater than 3

Worked solution: Take \(x = -4\). Then \(x^2 = 16 \gt 9\), but \(x = -4\) is not greater than 3. So the statement is false.

• B1 for “\(x \gt 3\) or \(x \lt -3\)” (accept \(|x| \gt 3\))

Worked solution: \(x^2 \gt 9 \iff |x| \gt 3\), so a true statement is: if \(x^2 \gt 9\) then \(x \gt 3\) or \(x \lt -3\).

• M1 for considering the cases \(n = 3m\), \(n = 3m + 1\) and \(n = 3m + 2\) (or \(3m - 1\)), where \(m\) is an integer
• A1 for \((3m)^2 = 9m^2 = 3(3m^2)\)
• A1 for \((3m + 1)^2 = 3(3m^2 + 2m) + 1\) and \((3m + 2)^2 = 3(3m^2 + 4m + 1) + 1\) (or \((3m - 1)^2 = 3(3m^2 - 2m) + 1\))
• A1* for a conclusion covering all cases: every integer is of one of these forms, so every square is of the form \(3k\) or \(3k + 1\)

Worked solution: Every integer \(n\) can be written as \(3m\), \(3m + 1\) or \(3m + 2\) for some integer \(m\).
\((3m)^2 = 9m^2 = 3(3m^2)\), of the form \(3k\).
\((3m + 1)^2 = 9m^2 + 6m + 1 = 3(3m^2 + 2m) + 1\), of the form \(3k + 1\).
\((3m + 2)^2 = 9m^2 + 12m + 4 = 3(3m^2 + 4m + 1) + 1\), of the form \(3k + 1\).
So the square of any integer is of the form \(3k\) or \(3k + 1\).

• M1 for using part (a): as \(a\) and \(b\) are of the form \(3m + 1\) or \(3m + 2\), \(a^2 = 3k_1 + 1\) and \(b^2 = 3k_2 + 1\)
• A1* for \(a^2 + b^2 = 3(k_1 + k_2) + 2\), which is not a multiple of 3, with a conclusion

Worked solution: Since \(a\) is not a multiple of 3, it is of the form \(3m + 1\) or \(3m + 2\), so from part (a) \(a^2 = 3k_1 + 1\). Similarly \(b^2 = 3k_2 + 1\).
Then \(a^2 + b^2 = 3(k_1 + k_2) + 2\), which leaves remainder 2 on division by 3, so it is not a multiple of 3.

• M1 for considering every integer as one of \(6k\), \(6k \pm 1\), \(6k \pm 2\), \(6k + 3\) (or \(6k, 6k + 1, \ldots, 6k + 5\))
• A1 for explaining that \(6k\) and \(6k \pm 2\) are divisible by 2 and \(6k + 3\) is divisible by 3, so a prime greater than 3 must be \(6k \pm 1\)

Worked solution: Every integer is of one of the forms \(6k, 6k \pm 1, 6k \pm 2, 6k + 3\).
\(6k = 2(3k)\) and \(6k \pm 2 = 2(3k \pm 1)\) are even, and \(6k + 3 = 3(2k + 1)\) is a multiple of 3, so none of these is a prime greater than 3.
Hence any prime greater than 3 is of the form \(6k + 1\) or \(6k - 1\), and since the prime is at least 5, \(k \ge 1\).

• M1 for expanding \((6k \pm 1)^2 - 1\) to obtain \(36k^2 \pm 12k\)
• A1 for \(12k(3k \pm 1)\)
• M1 for arguing that \(k(3k \pm 1)\) is even: if \(k\) is even it is even; if \(k\) is odd then \(3k\) is odd and \(3k \pm 1\) is even
• A1* for \(p^2 - 1 = 24 \times \frac{k(3k \pm 1)}{2}\), a multiple of 24, with a conclusion

Worked solution: \(p = 6k \pm 1\), so \(p^2 - 1 = 36k^2 \pm 12k + 1 - 1 = 12k(3k \pm 1)\).
If \(k\) is even, \(k(3k \pm 1)\) is even. If \(k\) is odd, \(3k\) is odd so \(3k \pm 1\) is even. Either way \(k(3k \pm 1) = 2j\) for an integer \(j\).
So \(p^2 - 1 = 24j\), a multiple of 24.