Question

Show that the square of any positive integer is of the form \[3m\] or, \[3m+1~\] for some integer \[m\].

Answer 1

Hint:
Here we will use Euclid's Division Lemma method to prove the given condition. We will first assume the integer to be any variable and then we will divide that integer by 3, and then we will write the integer in the form of quotient and remainder. Then we will take different values of remainder and we will find the square of integer for different values of remainder.

Complete step by step solution:
Let that positive integer be $a$
Now, we will divide the integer by 3 and let the remainder be $r$ and $b$ be the quotient.
We can write the integer in terms of remainder and quotient as;
$\Rightarrow a=3b+r$ ……….. $\left( 1 \right)$
Where $r$ can be 0, 1, 2, ….
Now, we will take three cases in which we take different values of $r$starting from 0.
Case 1: when $r=0$
We can write equation 1 after substituting the value of $r$ as
$\Rightarrow a=3b$
Squaring on both sides, we get
$\Rightarrow {{a}^{2}}={{\left( 3b \right)}^{2}}$
Applying exponents on the bases, we get
$\Rightarrow {{a}^{2}}=9{{b}^{2}}$
We can also write the equation as
$\Rightarrow {{a}^{2}}=3\left( 3{{b}^{2}} \right)$
Now, we will put $m$ in place of $3{{b}^{2}}$ .
$\Rightarrow {{a}^{2}}=3m$ …… $\left( 2 \right)$

Case 2: when $r=1$
We can write equation 1 after substituting the value of $r$ as
$\Rightarrow a=3b+1$
Squaring on both sides, we get
$\Rightarrow {{a}^{2}}={{\left( 3b+1 \right)}^{2}}$
Applying exponents on the bases, we get
$\Rightarrow {{a}^{2}}=9{{b}^{2}}+6b+1$
We can also write the equation as
$\Rightarrow {{a}^{2}}=3\left( 3{{b}^{2}}+2b \right)+1$
Now, we will put $m$ in place of $3{{b}^{2}}+2b$ .
$\Rightarrow {{a}^{2}}=3m+1$ ……. $\left( 3 \right)$
Case 3: when $r=2$
We can write equation 1 after substituting the value of $r$ as
$\Rightarrow a=3b+2$
Squaring on both sides, we get
$\Rightarrow {{a}^{2}}={{\left( 3b+2 \right)}^{2}}$
Applying exponents on the bases, we get
$\Rightarrow {{a}^{2}}=9{{b}^{2}}+12b+4$
We can also write the equation as
$\Rightarrow {{a}^{2}}=3\left( 3{{b}^{2}}+4b+1 \right)+1$
Now, we will put $m$ in place of $3{{b}^{2}}+4b+1$ .
$\Rightarrow {{a}^{2}}=3m+1$ ……. $\left( 4 \right)$

Hence, from equation 2, 3 and 4 we can say that the square of any positive integer is of the form \[3m\] or, \[3m+1~\].

Note:
Since we have used Euclid's division method to prove that condition. So here is the definition or explanation of this method. if we have two positive integers $a$ and $b$ , then according to Euclid’s Division Lemma method, there exist another unique integers $q$ and $r$, which satisfies the condition $a=bq+r$ where $0\le r\le b$ .