# Show that the square of any positive integer cannot be of the form $5q + 2$ or $5q + 3$ for any integer $q$.

Given:

"Square of any positive integer cannot be of the form $5q+2$ or $5q+3$ for any positive integer $q$".

To do:

We have to prove the given statement.

Solution:

According to Euclid's division lemma;

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

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

If $b\ =\ 5$, then;

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

When, $r\ =\ 0$:

$a\ =\ 5m$

Squaring on both sides, we get:

$a^2\ =\ (5m)^2$

$a^2\ =\ 25m^2$

$a^2\ =\ 5(5m^2)$

$a^2\ =\ 5q$, where $q\ =\ 5m^2$

When, $r\ =\ 1$:

$a\ =\ 5m\ +\ 1$

Squaring on both sides, we get:

$a^2\ =\ (5m\ +\ 1)^2$

$a^2\ =\ 25m^2\ +\ 10m\ + 1$

$a^2\ =\ 5(5m^2\ +\ 2m)\ +\ 1$

$a^2\ =\ 5q\ +\ 1$, where $q\ =\ 5m^2\ +\ 2m$

When, $r\ =\ 2$:

$a\ =\ 5m\ +\ 2$

Squaring on both sides, we get:

$a^2\ =\ (5m\ +\ 2)^2$

$a^2\ =\ 25m^2\ +\ 20m\ + 4$

$a^2\ =\ 5(5m^2\ +\ 4m)\ +\ 4$

$a^2\ =\ 5q\ +\ 4$, where $q\ =\ 5m^2\ +\ 4m$

When, $r\ =\ 3$:

$a\ =\ 5m\ +\ 3$

Squaring on both sides, we get:

$a^2\ =\ (5m\ +\ 3)^2$

$a^2\ =\ 25m^2\ +\ 30m\ + 9$

$a^2\ =\ 25m^2\ +\ 30m\ + 5\ +\ 4$

$a^2\ =\ 5(5m^2\ +\ 6m\ +\ 1)\ +\ 4$

$a^2\ =\ 5q\ +\ 4$, where $q\ =\ 5m^2\ +\ 6m\ +\ 1$

When, $r\ =\ 4$:

$a\ =\ 5m\ +\ 4$

Squaring on both sides, we get:

$a^2\ =\ (5m\ +\ 4)^2$

$a^2\ =\ 25m^2\ +\ 40m\ + 16$

$a^2\ =\ 25m^2\ +\ 40m\ + 15\ +\ 1$

$a^2\ =\ 5(5m^2\ +\ 8m\ +\ 3)\ +\ 1$

$a^2\ =\ 5q\ +\ 1$, where $q\ =\ 5m^2\ +\ 8m\ +\ 3$

Hence, the square of any positive integer cannot be of the form $5q+2$ or $5q+3$ for any positive integer $q$.

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Updated on: 10-Oct-2022

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