(i)     Given that $a \equiv d(\bmod n)$ and $b \equiv c(\bmod n)$ prove that

$$(a + b) \equiv (c + d)(\bmod n){\text{ .}}$$

(ii)     Hence solve the system

$$\left\{ {\begin{array}{*{20}{r}}   {2x + 5y \equiv 1(\bmod 6)} \\   {x + y \equiv 5(\bmod 6)} \end{array}} \right.$$

Show that ${x^{97}} - x + 1 \equiv 0(\bmod 97)$ has no solution.

(i)     $a \equiv d(\bmod n){\text{ and }}b \equiv c(\bmod n)$

so $a - d = pn{\text{ and }}b - c = qn$ ,     M1A1

$a - d + b - c = pn + qn$

$(a + b) - (c + d) = n(p + q)$     A1

$(a + b) \equiv (c + d)(\bmod n)$     AG

(ii)     $\left\{ {\begin{array}{*{20}{r}}   {2x + 5y \equiv 1(\bmod 6)} \\   {x + y \equiv 5(\bmod 6)} \end{array}} \right.$

adding $3x + 6y \equiv 0(\bmod 6)$     M1

$6y \equiv 0(\bmod 6){\text{ so }}3x \equiv 0(\bmod 6)$     R1

$x \equiv 0{\text{ or }}x \equiv 2{\text{ or }}x \equiv 4(\bmod 6)$     A1A1A1

for $x \equiv 0,{\text{ }}0 + y \equiv 5(\bmod 6){\text{ so }}y \equiv 5(\bmod 6)$     A1

for $x \equiv 2,{\text{ }}2 + y \equiv 5(\bmod 6){\text{ so }}y \equiv 3(\bmod 6)$     A1

If $x \equiv 4(\bmod 6),{\text{ }}4 + y \equiv 5(\bmod 6){\text{ so }}y \equiv 1(\bmod 6)$     A1

[11 marks]

Suppose x is a solution

97 is prime so ${x^{97}} \equiv x(\bmod 97)$     M1

${x^{97}} - x \equiv 0(\bmod 97)$     A1

${x^{97}} - x + 1 \equiv 1 \ne 0(\bmod 97)$

Hence there are no solutions     R1

[3 marks]

Part (a) (i) was not found difficult but using it in part (a)(ii) resulted in two or three correct lines and then abandonment of the problem.

Part (a) (i) was not found difficult but using it in part (a)(ii) resulted in two or three correct lines and then abandonment of the problem.