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Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m or 3m$+$1 for some integer m.
Given :
The given positive integer is 'm'.
To do :
We have to show that the square of any positive integer is either of the form 3m or 3m$+$1 for some integer m.
Solution :
By Euclid's division algorithm,
$a = b q + r$, where $0 \leq r < b$.
Let a be the positive integer and b $=$ 3.
Now, $a = 3 q + r$, where $0 \leq r < 3$.
The possibilities of r is 0 , 1 , 2.
When $r = 0$,
$a = 3 q + r$
$a = 3 q + 0$
$a = 3 q $
Squaring on both sides,
$a^2 = (3 q)^2$
$a^2 = 9 q^2$
$a^2 = 3(3 q^2)$
$a^2 = 3 m$ ; where $m = 3 q^2$.
When $r = 1$
$a = 3 q + r$
$a = 3 q + 1$
Squaring on both sides,
$a^2 = (3 q + 1)^2$
$a^2 = (3 q)^2 + 1^2 + 2 (3q) (1)$ $[(a+b)^2 = a^2 + b^2 + 2ab]$
$a^2 = 9 q^2 + 1 + 6q$
$a^2 = 9 q^2 + 6q + 1$
$a^2 = 3(3 q^2 + 2 q) + 1$
$a^2 = 3 m + 1$ ; where $m = 3 q^2 + 2 q$.
When $r = 2$
$a = 3 q + r$
$a = 3 q + 2$
Squaring on both sides,
$a^2 = (3 q + 2)^2$
$a^2 = (3 q)^2 + 2^2 + 2 (3q) (2)$ $[(a+b)^2 = a^2 + b^2 + 2ab]$
$a^2 = 9 q^2 + 4 + 12q$
$a^2 = 9 q^2 + 12 q + 4$
$a^2 = 9 q^2 + 12 q + 3 + 1$
$a^2 = 3(3 q^2 + 4 q + 1) + 1$
$a^2 = 3 m + 1$ ; where $m = 3 q^2 + 4 q + 1$.
Hence, the square of any positive integer is either of the form 3m or 3m$+$1 for
some integer m.
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