Given that \({n^2} + 2n + 3 \equiv N(\bmod 8)\) , where \(n \in {\mathbb{Z}^ + }\) and \(0 \le N \le 7\) , prove that \(N\) can take one of only three possible values.

consider the following

     M1A1A1

we see that the only possible values so far are \(2\), \(3\) and \(6\)     R1

also, the table suggests that these values repeat themselves but we have to prove this

let \(f(n) = {n^2} + 2n + 3\) , consider

\(f(n + 4) - f(n) = {(n + 4)^2} + 2(n + 4) + 3 - {n^2} - 2n - 3\)     M1

\( = 8n + 24\)     A1

since \(8n + 24\) is divisible by \(8\),     M1

\(f(n + 4) \equiv (n)(\bmod 8)\)     A1

this confirms that the values do repeat every \(4\) values of \(n\) so that \(2\), \(3\) and \(6\) are the only values taken for all values of \(n\)     R1

[9 marks]

Solutions to this question were extremely disappointing in general. Many candidates spotted that there was a periodicity in the values of the polynomial as far as \(n = 4\) and even \(n = 8\) but then the majority simply assumed that this would continue without any need for proof. This is equivalent to noting that \({2^{2n + 1}} - 1\) is prime for \(n = 1\), \(2\) and \(3\) and simply assuming that this will be true for larger values of \(n\).