Let a be any odd integer and b = 4.
According to Euclid’s algorithm,
a = 4m + r for some integer m ≥ 0
And r = 0,1,2,3 because 0 ≤ r < 4.
So, we have that,
a = 4m or 4m + 1 or 4m + 2 or 4m + 3 So, a = 4m + 1 or 4m + 3
We know that, a cannot be 4m or 4m + 2, as they are divisible by 2.
(4m + 1)2 = 16m2 + 8m + 1
= 4(4m2 + 2m) + 1
= 4q + 1, where q is some integer and q = 4m2 + 2m.
(4m + 3)2 = 16m2 + 24m + 9
= 4(4m2 + 6m + 2) + 1
= 4q + 1, where q is some integer and q = 4m2 + 6m + 2
Therefore, Square of any odd integer is of the form 4q + 1, for some integer q.
Hence Proved.