Show that $30$ is a factor of ${n^5} - n$ for all $n \in \mathbb{N}$.

(i)     Show that ${3^{{3^m}}} \equiv 3({\text{mod}}4)$ for all $m \in \mathbb{N}$.

(ii)     Hence show that there is exactly one pair $(m,{\text{ }}n)$ where $m,{\text{ }}n \in \mathbb{N}$, satisfying the equation ${3^{{3^m}}} = {2^{{2^n}}} + {5^2}$.

METHOD 1

${n^5} - n = \underbrace {n(n - 1)(n + 1)}_{{\text{3 consecutive integers}}}({n^2} + 1) \equiv 0({\text{mod}}6)$     M1

at least a factor is multiple of 3 and at least a factor is multiple of 2     R1

${n^5} - n = n({n^4} - 1) \equiv 0({\text{mod}}5)$ as ${n^4} \equiv 1({\text{mod}}5)$ by FLT     R1

therefore, as ${\text{(5, 6)}} = 1$,     R1

${n^5} - n \equiv 0\left( {\bmod \underbrace {5 \times 6}_{30}} \right)$     A1

ie 30 is a factor of ${n^5} - n$     AG

METHOD 2

let ${\text{P}}(n)$ be the proposition: ${n^5} - n = 30\alpha$ for some $\alpha  \in \mathbb{Z}$

${0^5} - 0 = 30 \times 0$, so ${\text{P}}(0)$ is true     A1

assume ${\text{P}}(k)$ is true for some $k$ and consider ${\text{P}}(k + 1)$

${(k + 1)^5} - (k + 1) = {k^5} + 5{k^4} + 10{k^3} + 10{k^2} + 5k + 1 - k - 1$     M1

$= ({k^5} - k) + 5k\left( {\underbrace {{k^3} + 3{k^2} + 3k + 1}_{{{(k + 1)}^3}} - ({k^2} + k)} \right)$

$= ({k^5} - k) + 5k\left( {{{(k + 1)}^3} - k(k + 1)} \right)$

$= 30\alpha  + \underbrace {5k(k + 1)\left( {\underbrace {{k^2} + k + 1}_{k(k + 1) + 1}} \right)}_{{\text{multiple of 6}}}$     M1

$= 30\alpha  + 30\beta$     A1

as ${\text{P}}(0)$ is true and ${\text{P}}(k)$ true implies ${\text{P}}(k + 1)$ true, by PMI ${\text{P}}(n)$ is true for all values $n \in \mathbb{N}$     R1

Note: Award the first M1 only if the correct induction procedure is followed and the correct first line is seen.

Note: Award R1 only if both M marks have been awarded.

METHOD 3

${n^5} - n = n({n^4} - 1)$     M1

$= n({n^2} - 1)({n^2} + 1)$     A1

$= (n - 1)n(n + 1)({n^2} - 4 + 5)$     R1

$= (n - 2)(n - 1)n(n + 1)(n + 2) + 5(n - 1)n(n + 1)$     A1

each term is multiple of 2, 3 and 5     R1

therefore is divisible by 30     AG

[5 marks]

(i)     METHOD 1

case 1: $m = 0$ and ${3^{{3^0}}} \equiv 3{\text{mod}}4$ is true     A1

case 2: $m > 0$

let $N = {3^m} \geqslant 3$ and consider the binomial expansion     M1

${3^N} = {(1 + 2)^N} = \sum\limits_{k = 0}^N {}$ $\left( \begin{array}{c}N\\k\end{array} \right){2^k} = 1 + 2N + \underbrace {\sum\limits_{k = 2}^N {\left( \begin{array}{c}N\\k\end{array} \right){2^k}} }_{ \equiv ({\text{mod}}4)} \equiv 1 + 2N({\text{mod}}4)$     A1

as $\underbrace {{3^m}}_N \equiv {( - 1)^m}({\text{mod}}4) \Rightarrow 1 + 2N \equiv 1 + 2{( - 1)^m}({\text{mod}}4)$     R1

therefore $\underbrace {{3^{{3^m}}}}_{{3^N}} \equiv 1 + 2{( - 1)^m}({\text{mod}}4) \Rightarrow$ $\left\{ \begin{array}{l}\underbrace {{3^{{3^m}}}}_{{3^N}} \equiv \underbrace {1 + 2}_3({\text{mod}}4){\rm{for\,}} m {\rm{\,even}}\\\underbrace {{3^{{3^m}}}}_{{3^N}} \equiv \underbrace {1 - 2}_{ - 1 \equiv 3({\text{mod}}4)}({\text{mod}}4){\rm{for\,}} m {\rm{\,odd}}\end{array} \right.$     R1

which proves that ${3^{{3^m}}} \equiv 3({\text{mod}}4)$ for any $m \in \mathbb{N}$     AG

METHOD 2

let ${\text{P}}(n)$ be the proposition: ${3^{{3^n}}} - 3 \equiv 0({\text{mod }}4,{\text{ or }}24)$

${3^{{3^0}}} - 3 = 3 - 3 \equiv 0({\text{mod }}4{\text{ or }}24)$, so ${\text{P}}(0)$ is true     A1

assume ${\text{P}}(k)$ is true for some $k$     M1

consider ${3^{{3^{^{k + 1}}}}} - 3 = {3^{{3^k} \times 3}} - 3$     M1

$= {(3 + 24r)^3} - 3$

$\equiv 27 + 24t - 3$     R1

$\equiv 0({\text{mod }}4{\text{ or }}24)$

as ${\text{P}}(0)$ is true and ${\text{P}}(k)$ true implies ${\text{P}}(k + 1)$ true, by PMI ${\text{P}}(n)$ is true for all values $n \in \mathbb{N}$     R1

METHOD 3

${3^{{3^m}}} - 3 = 3({3^{{3^m} - 1}} - 1)$     M1A1

$= 3({3^{2k}} - 1)$     R1

$= 3({9^k} - 1)$

$= 3\underbrace {\left( {{{(8 + 1)}^k} - 1} \right)}_{{\text{multiple of 8}}}$     R1

$\equiv 0({\text{mod}}24)$     A1

which proves that ${3^{{3^m}}} \equiv 3({\text{mod}}4)$ for any $m \in \mathbb{N}$     AG

(ii)     for $m \in \mathbb{N},{\text{ }}{3^{{3^m}}} \equiv 3({\text{mod}}4)$ and, as ${2^{{2^n}}} \equiv 0({\text{mod}}4)$ and ${5^2} \equiv 1({\text{mod}}4)$ then ${2^{{2^n}}} + {5^2} \equiv 1({\text{mod}}4)$ for $n \in {\mathbb{Z}^ + }$

there is no solution to ${3^{{3^m}}} = {2^{{2^n}}} + {5^2}$ for pairs $(m,{\text{ }}n) \in \mathbb{N} \times {\mathbb{Z}^ + }$     R1

when $n = 0$, we have ${3^{{3^m}}} = {2^{{2^0}}} + {5^2} \Rightarrow {3^{{3^m}}} = 27 \Rightarrow m = 1$     M1

therefore $(m,{\text{ }}n) = (1,{\text{ }}0)$     A1

is the only pair of non-negative integers that satisfies the equation     AG

[8 marks]

Many students were able to get partial credit for part (a), although few managed to gain full marks.

There seemed to be very few good attempts at part (b), many failing at the outset to understand what was meant by ${3^{{3^m}}}$.