Question

Prove that the square of any positive integer us
of the form 3m or 3m +1 but not of the
form 3 m +2.​

Answer 1

Answer:

here is your answer

Step-by-step explanation:

Let a and b any odd positive integer.

We apply the division lemma with

a and b=3

since 0is equal/<r<3,

the possible remainders are 0, 1 and 2

I.e can be 3q,3q+1or 3q+2where q is the quotient

Now

a^2=(3q)^2=9q^2

which can be written in the form

=3(3q^2)

=3m,since m =3q^2

it is divisible by 3

Again,

a^2=(3q+1)^2

=9 q^2+6q+1

=3m+1 (since m=3q^2+2q,3(3q^2+2q) is divisible by 3)

Lastly

a^2=(3q+2)^2

=9q^2+12q+4

=(9q^2+12q+3)+1

=3(3q^2+4q+1)+1

which can be written in the form

3m+1,since

3(3q^2+4q+1)is divisible by 3

therefore

The square of any positive integer is either of the form 3mor 3m +1

for some integer m .

I hope it will help

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