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Prove that the square of any positive integer is of the form $3m$ or, $3m+1$ but not of the form $3m+2$.
Given: Statement "Square of any positive integer is of the form $3m$ or, $3m+1$ but not of the form $3m+2$".
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 integer is of the form $3m$ or, $3m+1$ but not of the form $3m+2$.
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