Answer:

Let the positive integer be x.

Case : 1 :-

When x = 3q.

x² = (3q) ².

x² = 27q².

x² = 3(9q³).

= 3p, where p = 9q².

Case : 2 :-

When x = 3q + 1.

x² = (3q + 1)².

x² = 27q² + 1 + 6q².

x² = 3(9q³ + 2q²) + 1.

= 3p + 1, where p = 9q³ + 2q².

Hence, Proved.

Verified answer

[tex]\huge\red{\mid{\underline{\overline{\tt ANSWER }}\mid}}[/tex]

Let us consider a positive integer a

Divide the positive integer a by 3, and let r be the reminder and b be the quotient such that

a = 3b + r……………………………(1)

where r = 0,1,2,3…..

Case 1: Consider r = 0

Equation (1) becomes

a = 3b

On squaring both the side

a2 = (3b)2

a2 = 9b2

a2 = 3 × 3b2

a2 = 3p

Where p = 3b2

Case 2: Let r = 1

Equation (1) becomes

a = 3b + 1

Squaring on both the side we get

a2 = (3b + 1)2

a2 = (3b)2 + 1 + 2 × (3b) × 1

a2 = 9b2 + 6b + 1

a2 = 3(3b2 + 2b) + 1

a2 = 3p + 1

Where p = 3b2 + 2b

Case 3: Let r = 2

Equation (1) becomes

a = 3b + 2

Squaring on both the sides we get

a2 = (3b + 2)2

a2 = 9b2 + 4 + (2 × 3b × 2)

a2 = 9b2 + 12b + 3 + 1

a2 = 3(3b2 + 4b + 1) + 1

a2 = 3p + 1

where p = 3b2 + 4b + 1

∴ square of any positive integer is of form 3p or 3p+1.

Hence proved.

DON'T REPORT.