Question

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

please solve this completly
i will give you ={10}= points ​

Answer 1

Here is ur answer. Hope this helps if so mark as brainliest.

Answer 2

Answer:

Step-by-step explanation:

By using Euclid's division lemma:

a =bq+r

Let,

      b=3 ,r=0,1,2

Now,in first case

                                         (b=3 ,r=0)

a=bq+r

a=3q+0

By squaring both sides

(a)²=(3q)²

    =9q²

     =3(3q²)

                     {Putting 'm'value in (3q²)}

     =3(m)

      =3m.                  ...........(1)

In second case

                            (b=3,r=1)

a=bq+r

a=3q+1

By squaring both sides

(a)²=(3q+1)²

     =(3q)²+(1)²+2(3q)(1)

                           |(a+b)²=a²+b²+2ab

     =(9q²)+1+6q

     =3(3q²+2q²)+1

                         {Putting'm'in(3q²+2q²)}

     =3(m)+1

     =3m+1.                       ...........(2)

In third case

                           (b=3 ,r=2)

a=bq+r

a=3q+2

By squaring both sides

(a)²=(3q+2)²

    =(3q)²+(2)²+2(3q)(2)

                        |(a+b)²=a²+b²+2about

    =9q²+4+12q

    =9q²+12q+3+1

    =3(3q²+4q+1)+1

                      {Putting'm'in(3q²+4q+1)}

    =3(m)+1

    =3m+1.                ...............(3)

From (1),(2),(3) we get

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