Question

Use Euclid's division lemma to show that the square of any positive integer is of the form 3p,3p+1.

Answer 1

By Euclids division lemma all numbers in the for 3a,3a+1,3a+2
All numbers are assumed to be b
So by squaring and both side we get b square 9n^2=3(3n^2)=3p
b square is (3a+1)^2=9a^2+6a+1=3(3a^2+2a)+1=3p+1 and similarly for other one too

Answer 2

HI

Let a be any positive integer.

Therefore, a can be 3q or 3q+1 or 3q+2

where b = 3 and r = 0,1,2

(3q)² = 9q²

= 3(3q)²

= 3m ➡️ p = 3q²

(3q+1)² = (3q)² + 2(3q)(1) + (1)²

= 9q² + 6q + 1

= 3(3q² + 2q) + 1

= 3m + 1 ➡️p = 3q² + 2q

(3q+2)² = (3q)² + 2(3q)(2) + (2)²

= 9q² + 12q + 4

= 9q² + 12q + 3 + 1

= 3(3q² + 4q + 1) + 1

= 3m + 1 ➡️p = 3q² + 4q + 1

Therefore, the square of any positive integer is of the form 3p or 3p+1.

Hope it proved to be beneficial....