Math, asked by reachpranav, 11 months ago

Prove that the square of any odd integer is of the form 8k+1
please explain everything

Answers

Answered by great616
0

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.


reachpranav: Dude I asked you to explain
reachpranav: You just left everything blank
great616: Put k on the place of m
great616: Is this is helpful
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