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Prove that every odd integer is of the form ${4k + 1}$ or ${4k+3}$.

I have to use division algorithm to prove it but I am unable to construct a proof. Please help. Thank you! :))

Abir Mukherjee
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2 Answers2

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Every integer can be expressed as one of $$4n\equiv0\pmod2,$$ $$4n+1\equiv1\pmod2,$$ $$4n+2=2(2n+1)\equiv0\pmod2$$ and $$4n+3\equiv1\pmod2$$ where $n$ is any integer

lab bhattacharjee
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  • Yes! But I think that is what I have to prove that how odd integers are of the form ${4n +3}$. :) – Abir Mukherjee Apr 25 '14 at 14:32
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    @AbirMukherjee, You have to prove that $4n+1,4n+3$ are not divisible by $2$. See also : http://math.stackexchange.com/questions/756547/proof-about-prime-numbers – lab bhattacharjee Apr 25 '14 at 14:34
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Hint $\ 2n\!+\!1 = 4k\!+\!1 $ if $\,n=2k\,$ is even; $ $ else $\,n=2k\!+\!1\,$ is odd so $\,2n\!+\!1 =\, \ldots$

Remark $\ $This applies the division algorithm to $\,n,\,$ not $\,2n\!+\!1.\,$ If you prefer the latter, then by division we can write it in the form $\,4k\!+\!r\,$ for $\,r\in\{0,1,2,3\},\,$ and $\,4k\!+\!r\,$ odd $\,\Rightarrow\,r\,$ odd.

Bill Dubuque
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