Question

show that the square of any positive integer is of the form flame or 3M + 1 for some integer m

Answer 1

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

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

Answer 2

Step-by-step explanation:

let ' a' be any positive integer and b = 3.

we know, a = bq + r , 0 <  r< b.

now, a = 3q + r , 0<r < 3.

the possibilities of remainder = 0,1 or 2

Case I - a = 3q

a² = 9q² .

= 3 x ( 3q²)

= 3m (where m = 3q²)

Case II - a = 3q +1

a² = ( 3q +1 )²

=  9q² + 6q +1

= 3 (3q² +2q ) + 1

= 3m +1 (where m = 3q² + 2q )

Case III - a = 3q + 2

a² = (3q +2 )²

= 9q² + 12q + 4

= 9q² +12q + 3 + 1

= 3 (3q² + 4q + 1 ) + 1

= 3m + 1 ( where m = 3q² + 4q + 1)

From all the above cases it is clear that square of any positive integer ( as in this case a² ) is either of the form 3m or 3m +1.

Hence, it is solved .