Question
Wed June 13, 2012 By:
 

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

Expert Reply
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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