Prove that 5 divide n(4n^2 + 1)(6n^2 + 1) for any natural number n


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Question Number 158240 by HongKing last updated on 01/Nov/21

$Prove that 5 divide$ $n ({4 n}^{2} + 1) ({6 n}^{2} + 1)$ $for any natural number n$
	Answered by Rasheed.Sindhi last updated on 01/Nov/21

	$A (n) = n ({4 n}^{2} + 1) ({6 n}^{2} + 1)$ $n i s p o s s i b l y e q u a l t o o n e o f t h e f o l l o w i n g :$ $5 m, 5 m + 1, 5 m + 2, 5 m + 3, 5 m + 4$ $n = 5 m :$ $A (5 m) = 5 m (4 {(5 m)}^{2} + 1) (6 {(5 m)}^{2} + 1)$ $O b v i o u s l y 5 ∣ A (5 m)$ $n = 5 m + 1 :$ $F o r n = 5 m + 1$ $T h e f a c t o r {4 n}^{2} + 1 o f A (n)$ $= 4 {(5 m + 1)}^{2} + 1 = {100 m}^{2} + 40 m + 5$ $= 5 ({20 m}^{2} + 8 m + 1)$ $∴ 5 ∣ A (5 m + 1)$ $n = 5 m + 2 :$ $F o r n = 5 m + 2 t h e f a c t o r {6 n}^{2} + 1$ $= 6 {(5 m + 2)}^{2} + 1 = {150 m}^{2} + 120 m + 25$ $= 5 ({30 m}^{2} + 24 m + 5)$ $∴ 5 ∣ A (5 m + 2)$ $n = 5 m + 3 :$ $T h e f a c t o r {6 n}^{2} + 1 o f A (n)$ $= 6 {(5 m + 3)}^{2} + 1 = {150 m}^{2} + 180 m + 55$ $= 5 ({30 m}^{2} + 36 m + 11)$ $∴ 5 ∣ A (5 m + 3)$ $n = 5 m + 4 :$ $T h e f a c t o r {4 n}^{2} + 1 o f A (n)$ $= 4 {(5 m + 4)}^{2} + 1 = {100 m}^{2} + 160 + 65$ $= 5 ({20 m}^{2} + 32 m + 13)$ $∴ 5 ∣ A (5 m + 4)$ $HENCE$ $5 ∣ A (n) in all possible cases$
	Commented by HongKing last updated on 01/Nov/21

	$thank you so much dear S er cool$
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