Question

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

Answer 1

Euclid’s division lemma is used when two number’s division results into an expression of
$a = bq + r$ , where $a$ and $b$ are two positive integers, $q$ and $r$ are quotient and remainder.

Complete step-by-step answer:
Let’s consider a positive integer $a$ , whose square is to be determined.
The proof requires a number $3$ , so divide the integer $a$ by $3$ .
According to the statement of Euclid’s division lemma $a$ and $3$ are two integers such that there exists a unique $q$ and$r$ such that,
$a = 3q + r \cdots \left( 1 \right)$
Where, $q =$ quotient and $r =$ remainder when $a$ is divided by$3$.
Where $r = 0,1,2,3, \ldots$
Put $r = 0$ in equation (1),
$a = 3q \cdots \left( 2 \right)$
On squaring equation (2),
${a^2} = {\left( {3q} \right)^2} \\\ {a^2} = 9{q^2} \\\$
Taking as a common factor,
${a^2} = 3\left( {3{q^2}} \right) \\\ {a^2} = 3m \cdots \left( 3 \right) \\\$
Where, $m = 3{q^2}$

Now, put $r = 1$ in equation (1),
$a = 3q + 1 \cdots \left( 4 \right)$
Square the equation (4),
${a^2} = {\left( {3q + 1} \right)^2}$
Using the formula for ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ to expand ${\left( {3q + 1} \right)^2}$
${a^2} = {\left( {3q} \right)^2} + {1^2} + 3q \\\ {a^2} = 9{q^2} + 1 + 3q \\\$
Adjusting the terms to take out 3 as a common factor,
${a^2} = 9{q^2} + 6q + 1 \\\ {a^2} = 3\left( {3{q^2} + 2q} \right) + 1 \\\ {a^2} = 3m + 1 \cdots \left( 5 \right) \\\$
Where, $m = 3{q^2} + 2q$

Now, put $r = 2$ in equation (1),
$a = 3q + 2 \cdots \left( 6 \right)$
Square the equation (6),
${a^2} = {\left( {3q + 2} \right)^2}$
Using the formula for ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ to expand ${\left( {3q + 2} \right)^2}$
${a^2} = {\left( {3q} \right)^2} + {2^2} + 12q \\\ {a^2} = 9{q^2} + 4 + 12q \\\$
Adjusting the terms to take out 3 as a common factor ( $4$ can be written as $3 + 1$ )
${a^2} = 9{q^2} + 12q + 3 + 1 \\\ {a^2} = 3\left( {3{q^2} + 4q + 1} \right) + 1 \\\ {a^2} = 3m + 1 \cdots \left( 7 \right) \\\$
Where, $m = 3{q^2} + 4q + 1$

Put $r = 3$ in equation (1),
$a = 3q + 3 \cdots \left( 8 \right)$
Square the equation (8),
${a^2} = {\left( {3q + 3} \right)^2}$
Using the formula for ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ to expand ${\left( {3q + 3} \right)^2}$
${a^2} = {\left( {3q} \right)^2} + {3^2} + 18q \\\ {a^2} = 9{q^2} + 9 + 18q \\\$
Adjusting the terms to take out $3$ as a common factor
${a^2} = 9{q^2} + 18q + 9 \\\ {a^2} = 3\left( {3{q^2} + 6q + 3} \right) \\\ {a^2} = 3m \cdots \left( 9 \right) \\\$
Where, $m = 3{q^2} + 6q + 3$
From equation (3), (5), (7) and (9), it can be concluded that the square of any positive integer is of the form of $3m$ or $3m + 1$ .
Hence, proved.

Note: The important thing is to clearly understand the usage of the Euclid’s division lemma theorem.
It is important to adjust the equation as per the statement which has to be proved.