Define what is meant by the statement $x \equiv y(\bmod n){\text{ where }}x{\text{, }}y{\text{, }}n \in {\mathbb{Z}^ + }$ .

Hence prove that if $x \equiv y(\bmod n)$ then ${x^2} \equiv {y^2}(\bmod n)$ .

Determine whether or not ${x^2} \equiv {y^2}(\bmod n)$ implies that $x \equiv y(\bmod n)$ .

$x \equiv y(\bmod n) \Rightarrow x = y + kn,{\text{ }}(k \in \mathbb{Z})$     A1

[1 mark]

$x \equiv y(\bmod n)$

$\Rightarrow x = y + kn$     M1

${x^2} = {y^2} + 2kny + {k^2}{n^2}$     A1

$\Rightarrow {x^2} = {y^2} + (2ky + {k^2}n)n$     M1A1

$\Rightarrow {x^2} \equiv {y^2}(\bmod n)$     AG

[4 marks]

EITHER

${x^2} \equiv {y^2}(\bmod n)$

$\Rightarrow {x^2} - {y^2} = 0(\bmod n)$     M1

$\Rightarrow (x - y)(x + y) = 0(\bmod n)$     A1

This will be the case if

$x + y = 0(\bmod n){\text{ or }}x = - y(\bmod n)$     R1

so $x \ne y(\bmod n)$ in general     R1

[4 marks]

OR

Any counter example, e.g. $n = 5,{\text{ }}x = 3,{\text{ }}y = 2,$ in which case     R2

${x^2} \equiv {y^2}(\bmod n)$ but $x \ne y(\bmod n)$. (false)     R1R1

[4 marks]

While most candidates gave a correct meaning to $x \equiv y(\bmod n)$ , there were some incorrect statements, the most common being $x \equiv y(\bmod n)$ means that when x is divided by n, there is a remainder y. The true statement $8 \equiv 5(\bmod 3)$ shows that this statement is incorrect. Part (b) was solved successfully by many candidates but (c) caused problems for some candidates who thought that the result in (c) followed automatically from the result in (b).

While most candidates gave a correct meaning to $x \equiv y(\bmod n)$ , there were some incorrect statements, the most common being $x \equiv y(\bmod n)$ means that when x is divided by n, there is a remainder y. The true statement $8 \equiv 5(\bmod 3)$ shows that this statement is incorrect. Part (b) was solved successfully by many candidates but (c) caused problems for some candidates who thought that the result in (c) followed automatically from the result in (b).

While most candidates gave a correct meaning to $x \equiv y(\bmod n)$ , there were some incorrect statements, the most common being $x \equiv y(\bmod n)$ means that when x is divided by n, there is a remainder y. The true statement $8 \equiv 5(\bmod 3)$ shows that this statement is incorrect. Part (b) was solved successfully by many candidates but (c) caused problems for some candidates who thought that the result in (c) followed automatically from the result in (b).