OCR Further Additional Pure 2021 November — Question 8 12 marks

Exam BoardOCR
ModuleFurther Additional Pure (Further Additional Pure)
Year2021
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNumber Theory
TypeQuadratic residues
DifficultyHard +2.3 This Further Maths question combines recurrence relations with modular arithmetic and quadratic residues—a sophisticated proof requiring students to solve a recurrence, identify quadratic residues mod 10, compute the sequence modulo 10, and construct a proof by contradiction. The multi-stage reasoning and number theory content place it well above typical A-level questions.
Spec8.01g Second-order recurrence: solve with distinct, repeated, or complex roots8.02f Single linear congruences: solve ax = b (mod n)

8
  1. Solve the second-order recurrence system \(\mathrm { H } _ { \mathrm { n } + 2 } = 5 \mathrm { H } _ { \mathrm { n } + 1 } - 4 \mathrm { H } _ { \mathrm { n } }\) with \(H _ { 0 } = 3 , H _ { 1 } = 7\) for \(n \geqslant 0\).
    1. Write down the quadratic residues modulo 10 .
    2. By considering the sequence \(\left\{ \mathrm { H } _ { \mathrm { n } } \right\}\) modulo 10, prove that \(\mathrm { H } _ { \mathrm { n } }\) is never a perfect square.

Question 8:
AnswerMarks Guidance
8(a) Aux. Eqn. is m2 – 5m + 4 = 0  m = 1, 4
 Gen. Soln. is H = A + B  4n
n
H = 3  3 = A + B and H = 7  7 = A + 4B
0 1
 A = 53 , B = 43
(5 )
giving H = 13 + 4 n + 1
AnswerMarks
nM1
A1
M1
M1
A1
AnswerMarks
[5]1.1
1.1
1.1
1.1
AnswerMarks
1.1Use of initial terms
Solving simultaneous eqns. in A and B
AnswerMarks Guidance
(b)(i) 0, 1, 4, 5, 6, 9
[1]1.1 In any order
(ii)For n = 0, 1, 2, … 4n  4, 6, 4, 6, … (mod 10)
 5 + 4n = 9, 11, 9, 11, … (mod 10)
 9, 21, 9, 21, … (mod 10)
(5 )
 13 + 4 n  3, 7, 3, 7, … (mod 10)
since hcf(3, 10) = 1
and 3, 7 are quadratic non-residues modulo 10, so H  a square
AnswerMarks
nM1
A1
M1
A1
A1
A1
AnswerMarks
[6]3.1a
1.1
1.1
2.1
2.2a
AnswerMarks
2.4OR M1 for noting the terms of the sequence mod 10
so that final A1 can be earned also
Question 8:
8 | (a) | Aux. Eqn. is m2 – 5m + 4 = 0  m = 1, 4
 Gen. Soln. is H = A + B  4n
n
H = 3  3 = A + B and H = 7  7 = A + 4B
0 1
 A = 53 , B = 43
(5 )
giving H = 13 + 4 n + 1
n | M1
A1
M1
M1
A1
[5] | 1.1
1.1
1.1
1.1
1.1 | Use of initial terms
Solving simultaneous eqns. in A and B
(b) | (i) | 0, 1, 4, 5, 6, 9 | B1
[1] | 1.1 | In any order
(ii) | For n = 0, 1, 2, … 4n  4, 6, 4, 6, … (mod 10)
 5 + 4n = 9, 11, 9, 11, … (mod 10)
 9, 21, 9, 21, … (mod 10)
(5 )
 13 + 4 n  3, 7, 3, 7, … (mod 10)
since hcf(3, 10) = 1
and 3, 7 are quadratic non-residues modulo 10, so H  a square
n | M1
A1
M1
A1
A1
A1
[6] | 3.1a
1.1
1.1
2.1
2.2a
2.4 | OR M1 for noting the terms of the sequence mod 10
so that final A1 can be earned also
8
\begin{enumerate}[label=(\alph*)]
\item Solve the second-order recurrence system $\mathrm { H } _ { \mathrm { n } + 2 } = 5 \mathrm { H } _ { \mathrm { n } + 1 } - 4 \mathrm { H } _ { \mathrm { n } }$ with $H _ { 0 } = 3 , H _ { 1 } = 7$ for $n \geqslant 0$.
\item \begin{enumerate}[label=(\roman*)]
\item Write down the quadratic residues modulo 10 .
\item By considering the sequence $\left\{ \mathrm { H } _ { \mathrm { n } } \right\}$ modulo 10, prove that $\mathrm { H } _ { \mathrm { n } }$ is never a perfect square.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure 2021 Q8 [12]}}