Q. 43.9( 447 Votes )

# Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3*m* or 3*m* + 1 for some integer *m*.

Answer :

**To Prove:** Square of any number is of the form 3 m or 3 m +1**Proof:** to prove this statement from Euclid's division lemma, take any number as a divisor, in question we have 3m and 3m + 1 as the form

So,

By taking, ’ a’ as any positive integer and b = 3.

Applying Euclid’s algorithm a = bq + r.

a = 3q + r

Here, r = remainder = 0, 1, 2 and q ≥ 0 as the divisor is 3 there can be only 3 remainders, 0, 1 and 2.

So, putting all the possible values of the remainder in, a = 3q + r

a = 3q or 3q+1 or 3q+2

And now squaring all the values,

When a= 3q

Squaring both sides we get,

a^{2} = (3q)^{2}

a^{2} = 9q^{2}

a^{2} =3 (3q^{2})

a^{2} = 3 k_{1}

Where k_{1 }= 3q^{2}

When a=3q+1

Squaring both sides we get,

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

a^{2} = 9q^{2 }+ 6q + 1

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

a^{2} = 3k_{2 }+ 1

Where k_{2}_{ }= 3q^{2 }+ 2q

When a = 3q+2

Squaring both sides we get,

a^{2} = (3q + 2)^{2}a^{2} = 9q^{2 }+ 12q + 4

a^{2} = 9q^{2 }+ 12q + 3+1

a^{2} = 3( 3q^{2 }+ 4q + 1) +1

a^{2} = 3k_{3 }+ 1

_{3}

_{ }= 3q

^{2 }+ 4q + 1

Where k_{1}, k_{2} and k_{3} are some positive integers

**Hence, it can be said that the square of any positive integer is either of the form 3m or 3m+1.**

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