Question

use euclids division lemma to show that the square of any poistive integer is of the form 3p. ,3p+1

Answer 1

let us take, 'x'= 3q , 3q+1, 3q+2
when, x=3q
x2 = (3q) 2
x2 = 9q2
x2 = 3(3q2)
we see that 3q2= m
so we have done the first equation 3m

when , x=3q+1
x2= (3q+1)2
[since, (a+b)2 = a2+2ab+b2]
x2= 9q+6q+1
x2= 3(3q+2q)+1
in this we see that 3q+2q= m
therefore, this satisfy the equation m+1

Answer 2

hola!
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Let x = 3q , 3q+1, 3q+2

when,

x = 3q
        
x² =  (3q)²
x² = 9q²
x²  = 3(3q²) = 3p

where,  3q2 = p

when, 

x = 3q + 1
           
x² = (3q+1)²
x² = 9q + 6q + 1
x² = 3(3q + 2q) + 1 = 3p + 1

where, 3q + 2q = p

Hence,
We prove that the square of any poistive integer is of the form 3p, 3p+1
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Hope it helps! :D