Using inductive method. prove that .. 7^(2n) + 16n − 1 is divisible by 4 please help.


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Question Number 5673 by sanusihammed last updated on 23/May/16

$U s i n g i n d u c t i v e m e t h o d . p r o v e t h a t . .$ $7^{2 n} + 16 n - 1 i s d i v i s i b l e b y 4$ $p l e a s e h e l p .$
Commented by Rasheed Soomro last updated on 24/May/16

$I think 16 n is dropable at start .$ $it is unnecessary .$ $Because 4 ∣ 16 n so it is sufficient to prove$ $‘ ‘ 7^{2 n} - 1 is divisible by 4^{″}$ $∙ 7^{2 (1)} - 1 = 48 which is divisible by 4$ $∙ Let 4 ∣ 7^{2 k} - 1$ $So 7^{2 k} - 1 = 4 m where m \in Z$ $7^{2 k} = 4 m + 1$ $7^{2 k} . 7^{2} = 7^{2} (4 m + 1) = 196 m + 49$ $7^{2 (k + 1)} - 1 = 196 m + 48 = 4 (49 m + 12)$ $So 4 ∣ 7^{2 (k + 1)} - 1 if 4 ∣ 7^{2 k} - 1$ $And hence$ $for any n \in N, 4 ∣ 7^{2 n} - 1$ $since 4 ∣ 16 n$ $So finally,$ $4 ∣ 7^{2 n} + 16 n - 1 for any n \in N$
	Answered by Yozzii last updated on 24/May/16

	$L e t z (n) = 7^{2 n} + 16 n - 1 (n \in N) .$ $L e t P (n) b e t h e p r o p o s i t i o n f o r \forall n \in N,$ $z (n) i s d i v i s b l e b y 4 (o r t h a t t h e r e e x i s t s a n m \in N$ $s u c h t h a t f o r a n y n \in N, z (n) = 4 m) .$ $W h e n n = 1, z (1) = 7^{2} + 16 - 1 = 64 = 4 \times 16$ $\Rightarrow z (1) ∣ 4 \Rightarrow P (n) i s t r u e w h e n n = 1 .$ $A s s u m e n o w t h a t P (n) i s t r u e w h e n n = k, t h a t i s$ $k \in N, \exists m \in N : z (k) = 4 m .$ $F o r n = k + 1,$ $z (k + 1) - z (k) = 7^{2 (k + 1)} + 16 (k + 1) - 1 - 7^{2 k} - 16 k + 1$ $z (k + 1) - z (k) = 49 \times 7^{2 k} + 16 k + 16 - 7^{2 k} - 16 k$ $z (k + 1) - z (k) = 48 \times 7^{2 k} + 16 = 4 (12 \times 7^{2 k} + 4)$ $\Rightarrow z (k + 1) = z (k) + 4 (12 \times 7^{2 k} + 4)$ $B y t h e i n d u c t i v e h y p o t h e s i s, w e c a n w r i t e$ $z (k + 1) = 4 m + 4 (12 \times 7^{2 k} + 4)$ $z (k + 1) = 4 (m + 12 \times 7^{2 k} + 4)$ $N o w, k \in N \Rightarrow 7^{2 k} \in N . m, 4, 12 \in N$ $\Rightarrow m + 12 \times 7^{2 k} + 4 = q w h e r e q \in N .$ $H e n c e, z (k + 1) = 4 q \Rightarrow a t (n = k + 1) \in N, \exists q \in N$ $s u c h t h a t z (k + 1) = 4 q .$ $T h u s, P (k + 1) i s t r u e, a s s u m i n g P (k) i s$ $t r u e . S i n c e P (1) i s t r u e t h e n, b y t h e$ $p r i n c i p l e o f m a t h e m a t i c a l i n d u c t i o n,$ $P (n) i s t r u e f o r a l l n a t u r a l n u m b e r s n . ◻$
	Commented by sanusihammed last updated on 24/May/16

	$I r e a l l y a p p r e c i a t e t h i s . t h a n k y o u .$
	Answered by prakash jain last updated on 23/May/16

	$n = 1$ $7^{2} + 16 - 1 = 49 + 16 - 1 = 64$ $64 is divisible by 4 so statment 7^{2 n} + 16 n - 1 is$ $divisible by 4 is true for n = 1 .$ $Assume that the statement is true for n = k$ $i . e . 7^{2 k} + 16 k - 1 is divible by 4 .$ $or 7^{2 k} + 16 k - 1 = 4 m$ $7^{2 (k + 1)} + 16 (k + 1) - 1$ $= 7^{2 k} 7^{2} + 16 k + 16 - 1$ $= 7^{2 k} \cdot 49 + 16 k + 15$ $= 7^{2 k} \cdot 49 + 16 k \cdot 49 - 16 k \cdot 48 - 49 + 49 + 15$ $= 49 (7^{2 k} + 16 k - 1) - 48 \cdot 16 k + 64$ $= 49 (4 m) - 4 \cdot 12 \cdot 16 k + 4 \cdot 16$ $= 4 (49 m - 12 \cdot 16 k + 16)$ $49 m - 12 \cdot 16 k + 16 is an integer so$ $if 7^{2 k} + 16 k - 1 is divible by 4 \Rightarrow$ $7^{2 (k + 1)} + 16 (k + 1) - 1 is divisble by 4 .$
	Commented by sanusihammed last updated on 24/May/16

	$T h a n k s f o r t a k i n g y o u t i m e . i a p p r e c i a t e . t h a n k s$
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