Question

show that the square of any positive odd integer is of the form 4 m or 4 m + 1 where M is some integer​

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

Since, r = 0, 1, 2, 3

Therefore, a = 4q or 4q + 1 or 4q + 2 or 4q + 3

Since, a is an odd integer, o 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

= 4(4q2 + 2q) + 1

= 4m + 1 where m = 4q2 + 2q

Case II: When a = 4q + 3

Squaring both sides, we have,

a2 = (4q +3)2

= 16q2 + 9 + 24q

= 16 q2 + 24q + 8 + 1

= 4(4q2 + 6q + 2) +1

= 4m +1 where m = 4q2 +7q + 2

Hence, a is of the form 4m + 1 for some integer m.