Prove that the number $14 641$ is the fourth power of an integer in any base greater than $6$.

For $a,b \in \mathbb{Z}$ the relation $aRb$ is defined if and only if $\frac{a}{b} = {2^k}$ , $k \in \mathbb{Z}$ .

  (i)     Prove that $R$ is an equivalence relation.

  (ii)     List the equivalence classes of $R$ on the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

$14641$ (base $a > 6$ ) $= {a^4} + 4{a^3} + 6{a^2} + 4a + 1$ ,     M1A1

$= {(a + 1)^4}$     A1

this is the fourth power of an integer     AG

[3 marks]

(i)     $aRa$ since $\frac{a}{a} = 1 = {2^0}$ , hence $R$ is reflexive     A1

$aRb \Rightarrow \frac{a}{b} = {2^k} \Rightarrow \frac{b}{a} = {2^{ - k}} \Rightarrow bRa$

so R is symmetric     A1

$aRb$ and $bRc \Rightarrow \frac{a}{b} = {2^m}$, $m \in \mathbb{Z}$ and $bRc \Rightarrow \frac{b}{c} = {2^n}$ , $n \in \mathbb{Z}$     M1

$\Rightarrow \frac{a}{b} \times \frac{b}{c} = \frac{a}{c} = {2^{m + n}}$ , $m + n \in \mathbb{Z}$    A1

$\Rightarrow aRc$ so transitive     R1

hence $R$ is an equivalence relation     AG

(ii)     equivalence classes are {1, 2, 4, 8} , {3, 6} , {5, 10} , {7} , {9}     A3

Note: Award A2 if one class missing, A1 if two classes missing, A0 if three or more classes missing.

[8 marks]

This was not difficult but a surprising number of candidates were unable to do it. Care with notation and logic were lacking.

The question was at first straightforward but some candidates mixed up the properties of an equivalence relation with those of a group. The idea of an equivalence class is still not clearly understood by many candidates so that some were missing.