Question
Wed June 13, 2012 By:

# Show that square of any odd integer is of form 4m+1 for some integer m.

Thu June 14, 2012
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.

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