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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 $q$.