Prove that the square of any odd integer is of the form 8k+1
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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:
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