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 \[m\], but not of the form \[3m + 2\].

Hint: In the given question, we have been given to prove an assertion using the given method, which is Euclid’s division lemma. To solve this question, we first need to know what the Euclid’s division lemma is. Then, we apply the Euclid's division lemma to the given question’s statement by considering all the cases and solve to get our answer.

Formula used:
We are going to use Euclid's division lemma, which is,
\[a = bq + r\], where \[0 \le r < b\]

Complete step by step solution:
In the given question, we have to prove that:
The square of any positive integer is either of the form \[3m\] or \[3m + 1\] for some integer \[m\].
First, we are going to write down Euclid's division lemma, which is,
\[a = bq + r\], where \[0 \le r < b\]
In this question, “any positive integer” is the dividend – \[a\], the divisor is the constant three – \[3\], and from the formula, the remainder – \[r\] is either of \[0,1,2\].
So, we have to consider three cases.
Let “\[a\]” be any positive integer.
When “\[a\]” is divided by \[3\], we can have three possible remainders \[\left( r \right)\] – \[0\], \[1\], or \[2\].
Case 1: \[r = 0\]
\[a = 3b + 0 = 3b\]
Squaring both sides,
\[{a^2} = {\left( {3b} \right)^2} = 9{b^2} = 3 \times 3{b^2} = 3m\], where \[m = 3{b^2}\]
So, the square in this case is of the form,
\[3m\]
Case 2: \[r = 1\]
\[a = 3b + 1\]
Squaring both sides,
\[{a^2} = {\left( {3b + 1} \right)^2} = 9{b^2} + 6b + 1 = 3\left( {3{b^2} + 2b} \right) + 1 = 3m + 1\], where \[m = 3{b^2} + 2b\]
So, the square in this case is of the form,
\[3m + 1\]
Case 3: \[r = 2\]
\[a = 3b + 2\]
Squaring both sides,
\[{a^2} = {\left( {3b + 2} \right)^2} = 9{b^2} + 12b + 4 = 9{b^2} + 12b + 3 + 1 = 3\left( {3{b^2} + 4b + 1} \right) + 1 = 3m + 1\], where \[m = 3{b^2} + 4b + 1\]
So, the square in this case is of the form,
\[3m + 1\]
Hence, proved.

Note: In the given question, we had to prove the given assertion using Euclid's division lemma. We solved it by considering the various cases of the Euclid's division lemma, calculating their values separately and then after getting a general result, combining them to prove our answer. So, it is really important that we know Euclid's division lemma, its implications and its constraints.