prove-that-2-a-240-a-5-a-

Question Number 159556 by mnjuly1970 last updated on 18/Nov/21

p r o v e t h a t :

2 ∤ a \Rightarrow 240 ∣ a^{5} - a

Answered by floor(10²Eta[1]) last updated on 18/Nov/21

240 = 2^{4} . 3.5

∙ by fermat : a^{5} \equiv a (\mod 5) \Rightarrow 5 ∣ a^{5} - a

∙ a^{5} - a = a (a^{4} - 1) = a (a^{2} - 1) (a^{2} + 1)

= (a - 1) a (a + 1) (a^{2} + 1), but since this has

a product of three consecutive numbers

it has to be multiple of 3 \Rightarrow 3 ∣ a^{5} - a

∙ now, a^{5} - a = a (a^{2} - 1) (a^{2} + 1), since a is odd,

a = 2 k + 1 ∴ a^{5} - a = (2 k + 1) ({(2 k + 1)}^{2} - 1) ({(2 k + 1)}^{2} + 1)

= (2 k + 1) ({4 k}^{2} + 4 k) ({4 k}^{2} + 4 k + 2)

= 8 (2 k + 1) (k^{2} + k) ({2 k}^{2} + 2 k + 1), so this

is a multiple of 8 \Rightarrow if we show that

(2 k + 1) (k^{2} + k) ({2 k}^{2} + 2 k + 1) is even, {we}^{'} re done

but (2 k + 1) (k^{2} + k) ({2 k}^{2} + 2 k + 1)

= (2 k + 1) k (k + 1) ({2 k}^{2} + 2 k + 1) has a product

of two consecutive numbers, so {it}^{'} s divisible by 2

\Rightarrow 16 ∣ a^{5} - a

16.3.5 = 240 ∣ a^{5} - a

Commented by mnjuly1970 last updated on 19/Nov/21

t h a n k s a l o t . . v e r y n i c e a l i

Answered by Rasheed.Sindhi last updated on 19/Nov/21

∙ 240 ∣ a^{5} - a \Leftrightarrow {\begin{cases} 16 ∣ a^{5} - a \\ 3 ∣ a^{5} - a \\ 5 ∣ a^{5} - a \end{cases} [∵ 240 = 16.3.5]

∙ a^{5} - a = a (a - 1) (a + 1) (a^{2} + 1)

(i) T o p r o v e : 16 ∣ a^{5} - a

a \in O : a m a y b e e i t h e r 4 k + 1 o r 4 k + 3

a = 4 k + 1 :

a^{5} - a = a (a - 1) (a + 1) (a^{2} + 1)

= (4 k + 1) (4 k) (4 k + 2) (16 k^{2} + 8 k + 2)

= 16 (4 k + 1) (k) (2 k + 1) (8 k^{2} + 4 k + 1)

a = 4 k + 3 :

a^{5} - a = (4 k + 3) (4 k + 2) (4 k + 4) (16 k^{2} + 24 k + 10)

= 16 (4 k + 3) (2 k + 1) (k + 1) (8 k^{2} + 12 k + 5)

∴ 16 ∣ a^{5} - a

\begin{matrix} 16 ∣ a^{5} - a \end{matrix} \dots \dots \dots \dots \dots A

(i i) T o p r o v e : 3 ∣ a^{5} - a

a = 3 k \to 3 ∣ a

a = 3 k + 1 \to 3 ∣ a - 1

a = 3 k + 2 \to 3 ∣ a + 1

∴ 3 ∣ a^{5} - a

\begin{matrix} 3 ∣ a^{5} - a \end{matrix} \dots \dots \dots \dots \dots . . B

(i i i) T o p r o v e : 5 ∣ a^{5} - a

a = 5 k \to 5 ∣ a

a = 5 k + 1 \to 5 ∣ a - 1

a = 5 k + 2 \to 5 ∣ a^{2} + 1

a = 5 k + 3 \to 5 ∣ a^{2} + 1

a = 5 k + 4 \to 5 ∣ a + 1

∴ 5 ∣ a^{5} - a

\begin{matrix} 5 ∣ a^{5} - a \end{matrix} \dots \dots \dots \dots \dots . . C

F r o m A, B & C :

\begin{matrix} \overset{\underset{OR}{(16.3.5) ∣ a^{5} - a}}{240 ∣ a^{5} - a} \end{matrix}

Commented by mnjuly1970 last updated on 19/Nov/21

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