Prove-that-5-divide-n-4n-2-1-6n-2-1-for-any-natural-number-n-

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