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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 8m + 1
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{every}\:\mathrm{odd}\:\mathrm{integer}\:\mathrm{is}\:\mathrm{of}\:\mathrm{the}\:\mathrm{form}\:\:\:\mathrm{8m}\:+\:\mathrm{1} \\ $$
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
$$\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}{k}^{\mathrm{2}} +\mathrm{4}{k}+\mathrm{1} \\ $$$$=\mathrm{4}{k}\left({k}+\mathrm{1}\right)+\mathrm{1} \\ $$$${now}\:{put}\:{k}=\mathrm{1}\:\:\:\mathrm{8}+\mathrm{1}\:\:{in}\:{the}\:{form}\:\mathrm{8}×\mathrm{1}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{put}\:{k}=\mathrm{2}\:\:\:\mathrm{24}+\mathrm{1}\:{in}\:{the}\:{form}\:\:\:\mathrm{8}×\mathrm{3}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:{now}\:{assume}\:{for}\:\:{k}={r}\:{the}\:{statement}\:{is}\:{true} \\ $$$${that}\:{is}\:\mathrm{4}{r}\left({r}+\mathrm{1}\right)+\mathrm{1}\:{in}\:{the}\:{form}\:\mathrm{8}{m}+\mathrm{1} \\ $$$$ \\ $$$${to}\:{show}\:{it}\left[{is}\:{true}\:{when}\:{k}={r}+\mathrm{1}\right. \\ $$$${put}\:{k}={r}+\mathrm{1} \\ $$$$\mathrm{4}\left({r}+\mathrm{1}\right)\left({r}+\mathrm{2}\right)+\mathrm{1} \\ $$$$=\mathrm{4}\left({r}+\mathrm{1}\right)\left({r}+\mathrm{1}+\mathrm{1}\right)+\mathrm{1} \\ $$$$=\mathrm{4}\left\{{r}\left({r}+\mathrm{1}\right)+{r}+{r}+\mathrm{1}+\mathrm{1}\right\}+\mathrm{1} \\ $$$$\mathrm{4}{r}\left({r}+\mathrm{1}\right)+\mathrm{4}\left(\mathrm{2}{r}\right)+\mathrm{8}+\mathrm{1} \\ $$$$\mathrm{4}{r}\left({r}+\mathrm{1}\right)+\mathrm{8}\left({r}+\mathrm{1}\right)+\mathrm{1} \\ $$$${form}\:{of}\:\mathrm{8}{m}+\mathrm{1}+\mathrm{8}\left({r}+\mathrm{1}\right)+\mathrm{1} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 13/Oct/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by MJS last updated on 13/Oct/18
odd integer o=2n+1 o^2 =(2n+1)^2 n=2k∨n=2k+1 n=2k o^2 =(4k+1)^2 =16k^2 +8k+1= =8(2k^2 +k)+1=8m+1 k=(n/2) ⇒ m=k(2k+1)=((n(n+1))/2) n=2k+1 o^2 =(4k+3)^2 =16k^2 +24k+9= =8(2k^2 +3k+1)+1=8m+1 k=((n−1)/2) ⇒ m=(k+1)(2k+1)=((n(n+1))/2)
$$\mathrm{odd}\:\mathrm{integer}\:{o}=\mathrm{2}{n}+\mathrm{1} \\ $$$${o}^{\mathrm{2}} =\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$${n}=\mathrm{2}{k}\vee{n}=\mathrm{2}{k}+\mathrm{1} \\ $$$$ \\ $$$${n}=\mathrm{2}{k} \\ $$$$\:\:\:\:\:{o}^{\mathrm{2}} =\left(\mathrm{4}{k}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{16}{k}^{\mathrm{2}} +\mathrm{8}{k}+\mathrm{1}= \\ $$$$\:\:\:\:\:=\mathrm{8}\left(\mathrm{2}{k}^{\mathrm{2}} +{k}\right)+\mathrm{1}=\mathrm{8}{m}+\mathrm{1} \\ $$$$\:\:\:\:\:{k}=\frac{{n}}{\mathrm{2}}\:\Rightarrow\:{m}={k}\left(\mathrm{2}{k}+\mathrm{1}\right)=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$ \\ $$$${n}=\mathrm{2}{k}+\mathrm{1} \\ $$$$\:\:\:\:\:{o}^{\mathrm{2}} =\left(\mathrm{4}{k}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{16}{k}^{\mathrm{2}} +\mathrm{24}{k}+\mathrm{9}= \\ $$$$\:\:\:\:\:=\mathrm{8}\left(\mathrm{2}{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{1}\right)+\mathrm{1}=\mathrm{8}{m}+\mathrm{1} \\ $$$$\:\:\:\:\:{k}=\frac{{n}−\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{m}=\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)=\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$
Commented by Tawa1 last updated on 13/Oct/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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