# Show that the square of any positive integer cannot be of the form $6m+ 2$ or $6m + 5$ for any integer $m$.

Given:

"Square of any positive integer cannot be of the form $6m+2$ or $6m+5$ for any positive integer $m$".

To do:

We have to prove the given statement.

Solution:

According to Euclid's division lemma;

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

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

If $b\ =\ 6$, then;

• $a\ =\ 6q\ +\ r$, where $0\ \underline{< }\ r\ <\ 6$.
• So, $r\ =\ 0,\ 1,\ 2,\ 3,\ 4,\ 5$

When, $r\ =\ 0$:

$a\ =\ 6q$

Squaring on both sides, we get:

$a^2\ =\ (6q)^2$

$a^2\ =\ 36q^2$

$a^2\ =\ 6(6q^2)$

$a^2\ =\ 6m$, where $m\ =\ 6q^2$

When, $r\ =\ 1$:

$a\ =\ 6q\ +\ 1$

Squaring on both sides, we get:

$a^2\ =\ (6q\ +\ 1)^2$

$a^2\ =\ 36q^2\ +\ 12q\ + 1$

$a^2\ =\ 6(6q^2\ +\ 2q)\ +\ 1$

$a^2\ =\ 6m\ +\ 1$, where $m\ =\ 6q^2\ +\ 2q$

When, $r\ =\ 2$:

$a\ =\ 6q\ +\ 2$

Squaring on both sides, we get:

$a^2\ =\ (6q\ +\ 2)^2$

$a^2\ =\ 36q^2\ +\ 24q\ + 4$

$a^2\ =\ 6(6q^2\ +\ 4q)\ +\ 4$

$a^2\ =\ 6m\ +\ 4$, where $m\ =\ 6q^2\ +\ 4q$

When, $r\ =\ 3$:

$a\ =\ 6q\ +\ 3$

Squaring on both sides, we get:

$a^2\ =\ (6q\ +\ 3)^2$

$a^2\ =\ 36q^2\ +\ 36q\ + 9$

$a^2\ =\ 36q^2\ +\ 36q\ + 6\ +\ 3$

$a^2\ =\ 6(6q^2\ +\ 6q\ +\ 1)\ +\ 3$

$a^2\ =\ 6m\ +\ 3$, where $m\ =\ 6q^2\ +\ 6q\ +\ 1$

When, $r\ =\ 4$:

$a\ =\ 6q\ +\ 4$

Squaring on both sides, we get:

$a^2\ =\ (6q\ +\ 4)^2$

$a^2\ =\ 36q^2\ +\ 48q\ + 16$

$a^2\ =\ 36q^2\ +\ 48q\ + 12\ +\ 4$

$a^2\ =\ 6(6q^2\ +\ 8q\ +\ 2)\ +\ 4$

$a^2\ =\ 6m\ +\ 4$, where $m\ =\ 6q^2\ +\ 8q\ +\ 2$

When, $r\ =\ 5$:

$a\ =\ 6q\ +\ 5$

Squaring on both sides, we get:

$a^2\ =\ (6q\ +\ 5)^2$

$a^2\ =\ 36q^2\ +\ 60q\ + 25$

$a^2\ =\ 36q^2\ +\ 60q\ + 24\ +\ 1$

$a^2\ =\ 6(6q^2\ +\ 10q\ +\ 4)\ +\ 1$

$a^2\ =\ 6m\ +\ 1$, where $m\ =\ 6q^2\ +\ 10q\ +\ 4$

Hence, the square of any positive integer cannot be of the form $6m+2$ or $6m+5$ for any positive integer $m$.

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