Question

Show that the square of any positive integer is of the form $3m$ or $3m+1$ for some integer $m$ .

Answer 1

Hint: Express the positive integer in the form of $3a,3a+1$ or $3a+2$ , where $a$ is a positive integer, such that $0\le a\le 2$ . Square the number in each case and express it in the form $3m$ or $3m+1$.

Complete step-by-step answer:
Euclid’s division lemma says that any integer can be expressed in the form of $l=mq+r$ , where $l,m,q$ and $r$ are positive integers and $0\le r < m$ .
Now, let $p$ be any positive integer. So, by Euclid’s division lemma, we can write $p$ as $p=3q+r$ , where $0\le r < 3$ .
We know $p,q$ and $3$ are integers. So, $r$ must be an integer.
 $\Rightarrow r$ is an integer greater than or equal to $0$ and less than $3$.
 $\Rightarrow r$ can be equal to $0,1$ or $2$.
Case $1$ : $r=0$
When $r=0$ , $p=3q$ . So, the value of ${{p}^{2}}$ will be equal to ${{\left( 3q \right)}^{2}}=9{{q}^{2}}$ . We can write $9{{q}^{2}}$ in the form of $3m$ , where $m=3{{q}^{2}}$ .
Case $2$ : $r=1$
When $r=1$, then $p$ will be of the form $3q+1$ . So, the value of ${{p}^{2}}$ will be equal to ${{\left( 3q+1 \right)}^{2}}=9{{q}^{2}}+6q+1=3\left( 3{{q}^{2}}+2q \right)+1$ , which is of the form $3m+1$ , where $m=3{{q}^{2}}+2q$ .
Case $3$: $r=2$
 When $r=2$ , then $p$ will be of the form $3q+2$ . So, the value of ${{p}^{2}}$ will be equal to ${{\left( 3q+2 \right)}^{2}}=9{{q}^{2}}+12q+4=9{{q}^{2}}+12q+3+1=3\left( 3{{q}^{2}}+4q+1 \right)+1$ , which is of the form $3m+1$ , where \[m=3{{q}^{2}}+4q+1\] .
So, in each of the cases, we can see that the square of the positive integer $p$ is of the form $3m$ or $3m+1$ for some integer $m$ .
Hence, the square of any positive integer is of the form $3m$ or $3m+1$ for some integer $m$.

Note: The Euclid lemma helps prove many mathematical theorems. This lemma should always be remembered.