prove that: 2∤ a ⇒ 240∣ a^( 5) − a


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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
	$•240∣a^( 5) −a⇔ { ((16∣ a^( 5) −a)),(( 3∣ a^( 5) −a)),(( 5∣ a^( 5) −a)) :}[∵ 240=16.3.5] •a^5 −a=a(a−1)(a+1)(a^2 +1) (i)To prove:16∣ a^( 5) − a a∈O: a may be either 4k+1 or 4k+3 a=4k+1: a^5 −a=a(a−1)(a+1)(a^2 +1) =(4k+1)(4k)(4k+2)(16k^2 +8k+2) =16(4k+1)(k)(2k+1)(8k^2 +4k+1) a=4k+3: a^5 −a=(4k+3)(4k+2)(4k+4)(16k^2 +24k+10) =16(4k+3)(2k+1)(k+1)(8k^2 +12k+5) ∴ 16∣a^5 −a determinant ((( 16∣a^5 −a)))...............A (ii)To prove:3∣ a^( 5) − a a=3k→3∣a a=3k+1→3∣a−1 a=3k+2→3∣a+1 ∴ 3 ∣a^5 −a determinant (((3 ∣a^5 −a))).................B (iii)To prove:5∣ a^( 5) − a a=5k→5∣a a=5k+1→5∣a−1 a=5k+2→5∣a^2 +1 a=5k+3→5∣a^2 +1 a=5k+4→5∣a+1 ∴ 5∣a^5 −a determinant (((5∣a^5 −a))).................C From A,B & C: determinant (((240∣a^5 −a^((16.3.5)∣a^5 −a_(OR) ) )))$
	$∙ 240 ∣ a^{5} - a \Leftrightarrow {\begin{cases} 16 ∣ a^{5} - a \\ 3 ∣ a^{5} - a \\ 5 ∣ a^{5} - a \end{cases} [∵ 240 = 16.3.5]$ $∙ a^{5} - a = a (a - 1) (a + 1) (a^{2} + 1)$ $(i) T o p r o v e : 16 ∣ a^{5} - a$ $a \in O : a m a y b e e i t h e r 4 k + 1 o r 4 k + 3$ $a = 4 k + 1 :$ $a^{5} - a = a (a - 1) (a + 1) (a^{2} + 1)$ $= (4 k + 1) (4 k) (4 k + 2) (16 k^{2} + 8 k + 2)$ $= 16 (4 k + 1) (k) (2 k + 1) (8 k^{2} + 4 k + 1)$ $a = 4 k + 3 :$ $a^{5} - a = (4 k + 3) (4 k + 2) (4 k + 4) (16 k^{2} + 24 k + 10)$ $= 16 (4 k + 3) (2 k + 1) (k + 1) (8 k^{2} + 12 k + 5)$ $∴ 16 ∣ a^{5} - a$ $\begin{matrix} 16 ∣ a^{5} - a \end{matrix} . . . . . . . . . . . . . . . A$ $(i i) T o p r o v e : 3 ∣ a^{5} - a$ $a = 3 k \to 3 ∣ a$ $a = 3 k + 1 \to 3 ∣ a - 1$ $a = 3 k + 2 \to 3 ∣ a + 1$ $∴ 3 ∣ a^{5} - a$ $\begin{matrix} 3 ∣ a^{5} - a \end{matrix} . . . . . . . . . . . . . . . . . B$ $(i i i) T o p r o v e : 5 ∣ a^{5} - a$ $a = 5 k \to 5 ∣ a$ $a = 5 k + 1 \to 5 ∣ a - 1$ $a = 5 k + 2 \to 5 ∣ a^{2} + 1$ $a = 5 k + 3 \to 5 ∣ a^{2} + 1$ $a = 5 k + 4 \to 5 ∣ a + 1$ $∴ 5 ∣ a^{5} - a$ $\begin{matrix} 5 ∣ a^{5} - a \end{matrix} . . . . . . . . . . . . . . . . . C$ $F r o m A, B & C :$ $\begin{matrix} \overset{\underset{OR}{(16.3.5) ∣ a^{5} - a}}{240 ∣ a^{5} - a} \end{matrix}$
	Commented by mnjuly1970 last updated on 19/Nov/21

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