Question

Show that the square of an odd +ve integer is of the form 8m+1.

Answer 1

Answer:

Any positive odd integer is of the form 2q+1, where q is a whole number..

Therefore, (2q+1)^2 = 4q^2+4q+1 = 4q(q+1)+1, -----(1)

q(q+1) is either 0 or even. So, it is 2m, where m is a whole number.

Therefore, (2q+1)^2= 4.2m+1 =8m+1. (From eq. 1)

Hope it helps you.✌✌

Answer 2

Answer:

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.