Show that the square of every odd integer is of the form 8m + 1


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Question Number 45439 by Tawa1 last updated on 12/Oct/18

$Show that the square of every odd integer is of the form 8 m + 1$
	Answered by tanmay.chaudhury50@gmail.com last updated on 13/Oct/18
	$(2k+1)^2 =4k^2 +4k+1 =4k(k+1)+1 now put k=1 8+1 in the form 8×1+1 put k=2 24+1 in the form 8×3+1 now assume for k=r the statement is true that is 4r(r+1)+1 in the form 8m+1 to show it[is true when k=r+1 put k=r+1 4(r+1)(r+2)+1 =4(r+1)(r+1+1)+1 =4{r(r+1)+r+r+1+1}+1 4r(r+1)+4(2r)+8+1 4r(r+1)+8(r+1)+1 form of 8m+1+8(r+1)+1$
	${(2 k + 1)}^{2} = 4 k^{2} + 4 k + 1$ $= 4 k (k + 1) + 1$ $n o w p u t k = 1 8 + 1 i n t h e f o r m 8 \times 1 + 1$ $p u t k = 2 24 + 1 i n t h e f o r m 8 \times 3 + 1$ $n o w a s s u m e f o r k = r t h e s t a t e m e n t i s t r u e$ $t h a t i s 4 r (r + 1) + 1 i n t h e f o r m 8 m + 1$ $t o s h o w i t [i s t r u e w h e n k = r + 1$ $p u t k = r + 1$ $4 (r + 1) (r + 2) + 1$ $= 4 (r + 1) (r + 1 + 1) + 1$ $= 4 {r (r + 1) + r + r + 1 + 1} + 1$ $4 r (r + 1) + 4 (2 r) + 8 + 1$ $4 r (r + 1) + 8 (r + 1) + 1$ $f o r m o f 8 m + 1 + 8 (r + 1) + 1$
	Commented by Tawa1 last updated on 13/Oct/18

	$God bless you sir$
	Answered by MJS last updated on 13/Oct/18

	$odd integer o = 2 n + 1$ $o^{2} = {(2 n + 1)}^{2}$ $n = 2 k \lor n = 2 k + 1$ $n = 2 k$ $o^{2} = {(4 k + 1)}^{2} = 16 k^{2} + 8 k + 1 =$ $= 8 (2 k^{2} + k) + 1 = 8 m + 1$ $k = \frac{n}{2} \Rightarrow m = k (2 k + 1) = \frac{n (n + 1)}{2}$ $n = 2 k + 1$ $o^{2} = {(4 k + 3)}^{2} = 16 k^{2} + 24 k + 9 =$ $= 8 (2 k^{2} + 3 k + 1) + 1 = 8 m + 1$ $k = \frac{n - 1}{2} \Rightarrow m = (k + 1) (2 k + 1) = \frac{n (n + 1)}{2}$
	Commented by Tawa1 last updated on 13/Oct/18

	$God bless you sir$
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