Question

Show that the square of an odd positive integer is of the form 8m + 1, where m is some whole number. ​

Answer 1

Let a be any positive integer and b = 4.

Then, by Euclid's algorithm a = 4q + r for some integer q  0 and 0  r < 4

Thus, r = 0, 1, 2, 3

Since, a is an odd integer, so a = 4q + 1 or 4q + 3

Case I: When a = 4q + 1

Squaring both sides, we have, a2 = (4q + 1)2

a2 = 16q2 + 1 + 8q

   = 8(2q2 + q) + 1

    = 8m + 1, where m = 2q2 + q

Case II: When a = 4q + 3

Squaring both sides, we have,

a2 = (4q +3)2

   = 16q2 + 9 + 24q

   = 16 q2 + 24q + 8 + 1

    = 8(2q2 + 3q + 1) +1

    = 8m +1 where m = 2q2 + 3q + 1

Hence, a is of the form 8m + 1 for some integer m.