Question

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

Answer 1

Answer:

Step-by-step explanation:

Let take a as any positive integer and b = 3.

Then using Euclid’s algorithm we get a = 3q + r here r is remainder and value of q is more than or equal to 0 and r = 0, 1, 2 because 0 < r < b and the value of b is 3 So our possible values will 3q+0 , 3q+1 and 3q+2

Now find the square of values

Use the formula (a+b)² = a² + 2ab +b² to open the square bracket

(3q)² = 9q² if we divide by 3 we get no remainder

we can write it as 3*(3q²) so it is in form of 3m here m = 3q²

(3q+1)² = (3q)² + 2*3q*1 + 1²

=9q² + 6q +1 now divide by 3 we get 1 remainder

so we can write it as 3(3q² + 2q) +1 so we can write it in form of 3m+1 and value of m is 3q² + 2q here

(3q+2)² = (3q)² + 2*3q*2 + 2²

=9q² + 12q +4 now divide by 3 we get 1 remainder

so we can write it as 3(3q² + 4q +1) +1 so we can write it in form of 3m +1 and value of m will 3q² + 4q +1

Square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Mark me as brainliest