A positive integer is of the form $3q+1$, $q$ being a natural number. Can you write its square in any form other than $3m+1$, $3m$ or $3m+2$ for some integer $m$? Justify your answer.

Given: A positive integer of the form $3q\ +\ 1$.

To prove: Here we have to check if the square of $3q\ +\ 1$ is in any form other than $3m+1$, $3m$ or $3m+2$ for some integer $m$.

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, square of a positive integer of the form $3q\ +\ 1$ is always of the form $3m$ or $3m\ +\ 1$ for some integer $m$.

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Updated on: 10-Oct-2022

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