# Show that the square of any positive integer cannot be of the form $3m+2$, where $m$ is a natural number.

Given: Statement "Square of any positive integer cannot be of the form $3m+2$, where $m$ is a natural number".

To prove: Here we have to prove the given statement.

Solution:

According to Euclid's lemma,

If $a$ and $b$ are two positive integers;

• $a\ =\ bq\ +\ r$, where $0\ \underline{< }\ r\ <\ b$.

If $b\ =\ 3$, then;

• $a\ =\ 3q\ +\ r$, where $0\ \underline{< }\ r\ <\ 3$.
• So, $r\ =\ 0,\ 1,\ 2$

When, $r\ =\ 0$:

$a\ =\ 3q$

Squaring on both sides, we get:

$a^2\ = (3q)^2$

$a^2\ = 9q^2$

$a^2\ = 3(3q^2)$

$a^2\ = 3m$, where $m\ =\ 3q^2$

When, $r\ =\ 1$:

$a\ =\ 3q\ +\ 1$

Squaring on both sides, we get:

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

$a^2\ = 9q^2\ +\ 6q\ + 1$

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

$a^2\ = 3m\ +\ 1$, where $m\ =\ 3q^2\ +\ 2q$

When, $r\ =\ 2$:

$a\ =\ 3q\ +\ 2$

Squaring on both sides, we get:

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

$a^2\ = 9q^2\ +\ 12q\ + 4$

$a^2\ = 9q^2\ +\ 12q\ + 3\ +\ 1$

$a^2\ = 3(3q^2\ +\ 4q\ +\ 1)\ +\ 1$

$a^2\ = 3m\ +\ 1$, where $m\ =\ 3q^2\ +\ 4q\ +\ 1$

Hence, the square of any positive number cannot be of the form $3m\ +\ 2$.