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Show that the square of an odd positive ...

Show that the square of an odd positive integer is of the form `8q+1,` for some integer `qdot`

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Let a = 2n +1 be any odd integer , `n in I` (non -negative)
`therefore " " a^(2) = (2n+1)^(2)`
`= 4n^(2) + 4n +1`
Since , `n in I` , so two cases arise here .
Case I : Let n is an even number say n = 2p , `p in I` .
= `16p^(2) + 8p +I`
= `8(2p^(2) +p) +1`
= 8k +1 `" " ....(1)`
Case II : Let n is an odd number say n = 2r + 1 , `r in I. `
`therefore " " a^(2) = 4(2r +1)^(2) + 4(2r+1) +1`
= `4(4r^(2) + 4r +1) + 8r + 4 + 1`
= `16r^(2) + 24r + 8 +1`
= `8(2r^(2) + 3r +1) + 1`
= `8k +1`
From equations (1) and (2) , we can say that square of any odd number is always of the form 8k +1 . Hence Proved .
In other words , we can say that `a^(2)` will leave a remainder 1 when divided by 8 . Hence Proved.
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