Show that the square of any odd integer is of the form 4m+1, for some integer m.

Explanation

By Euclid's division algorithm, we have a =bq+r , where `0 le t lt 4` ….(i)
On putting b=4 in Eq. (i), we get
a=4q+r, where `0 le r lt 4` i.e., r=0,1,2,3 ..(ii)
If `r=0 Rightarrow a=4q,` 4q is divisible by 2 `Rightarrow` is even
`"If" r=1 Rightarrow a =4q+1,(4q+1)` is not divisible by 2.
`"If" r=2 Rightarrow a =4q+2,2,(2q+1)` is divisible by `2 Rightarrow 2(2q+1)` is even.
`"If" r=3 Rightarrow a =4q+3,(4q+3)` is not divisible by 2.
So, for any positive integer q, `(4q+1) and (4q+3)` are odd integers,
Now, `a^(2)=(4q+1)=16q^(2)+1+8q=4(4q(2)+2q)+1` `[therefore (a+b)^(2)=a^(2)+2ab+b^(2)]`
is a square which is of the form 4m+1, where `m=(4q^(2)+2q)` is an integer.
and `a^(2)=(4q+3)^(2)=16q^(2)+9+24q=4(4q^(2)+6q+2)+1` is a square `[therefore (a+b)^(2)=a^(2)+2ab+b^(2)]`
Which is of the form 4m+1, where `m=(4q^(2)+6q+2)` is an integer
Hence, for some integer m, the square of any odd integer is of hte form 4m+1.