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

Posted by Mohankrishna Nayak 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Gaurav Seth 6 years, 10 months ago

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.

0Thank You

Sneha Singh 6 years, 10 months ago

Refer ncert

0Thank You

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CBSE > Class 10 > Mathematics