Question

show that the square of any positive odd integer is of the form 4m+1 for some integer m​

Answer 1

Answer:

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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

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.

Step-by-step explanation: