Question

0.67 using Euclid's cursion lemma, show
that the square of any positive
integer is either of the form an
Or Gm +1) for some integer m.​

Answer 1

Step-by-step explanation:

Use Euclid division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m.

Let a be any positive integer and b = 3.

=) a = 3q + r, r = 0 or 1 or 2.

(By Euclid's lemma)

=) a = 3q or 3q + 1 or 3q + 2 for positive integer q.

1st case,

If a = 3q :

=) a² = (3q)²

= 9q²

= 3(3q²)

= 3m, where m= 3q².

2nd case,

If a = 3q+1,

=) a² = (3q+1)²

= (3q)² + 2(3q)(1) + 1²

= 9q² + 6q + 1

= 3(3q² + 2q) + 1

= 3m + 1, where m = 3q² + 2q.

3rd case,

If a = 3q+2:

=) a² = (3q+2)²

= (3q)² + 2(3q)(2) + 2²

= 9q² + 12q + 4

= 9q² + 12q + 3 + 1

= 3(3q² + 4q + 1) + 1

= 3m + 1, where m = 3q² + 4q + 1.

Hence the square of any positive integer is either of the form 3m or 3m+1 for some integer m

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