show that (n^4 −n^2 ) is divisible by 12


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Question Number 90661 by Cynosure last updated on 25/Apr/20

$s h o w t h a t (n^{4} - n^{2}) i s d i v i s i b l e b y 12$
	Answered by MJS last updated on 25/Apr/20

	$n^{4} - n^{2} = n^{2} (n^{2} - 1) = n^{2} (n - 1) (n + 1) = f (n)$ $(1) n = 6 k$ $f (n) = 36 k^{2} (6 k - 1) (6 k + 1) = 12 \times 3 k^{2} (6 k - 1) (6 k + 1)$ $(2) n = 6 k + 1$ $f (n) = {(6 k + 1)}^{2} (6 k) (6 k + 2) = 12 {(6 k + 1)}^{2} k (3 k + 1)$ $(3) n = 6 k + 2$ $f (n) = {(6 k + 2)}^{2} (6 k + 1) (6 k + 3) = 12 {(3 k + 1)}^{2} (6 k + 1) (2 k + 1)$ $(4) n = 6 k + 3$ $f (n) = {(6 k + 3)}^{2} (6 k + 2) (6 k + 4) = 12 (2 k + 1) (6 k + 3) (3 k + 1) (3 k + 2)$ $(5) n = 6 k + 4$ $f (n) = {(6 k + 4)}^{2} (6 k + 3) (6 k + 5) = 12 {(3 k + 2)}^{2} (2 k + 1) (6 k + 5)$ $(6) n = 6 k + 5$ $f (n) = {(6 k + 5)}^{2} (6 k + 4) (6 k + 6) = 12 {(6 k + 5)}^{2} (3 k + 2) (k + 1)$
	Answered by JDamian last updated on 25/Apr/20

	$m = n^{4} - n^{2} = n^{2} (n^{2} - 1) = (n - 1) n^{2} (n + 1)$ $m i s t h e p r o d u c t o f t h r e e s e q u e n t i a l$ $n a t u r a l n u m b e r s . T h e r e f o r e :$ $1 m = 3 k$ $2.1 w h e n n i s e v e n \Rightarrow m = 4 k$ $2.2 w h e n n i s o d d \Rightarrow (n - 1) a n d (n + 1) a r e$ $b o t h e v e n \Rightarrow m = 4 k$ $3 F r o m a b o v e, m = 12 k$
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