Question

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

Answer 1

$\huge\bold{Answer :-}$

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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

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
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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
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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
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Hence, a is of the form 4m + 1 for some integer m.

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Answer 2

Answer:

Step-by-step explanation: