Question

Show that the square of any positive integer cannot be of the form $6m+2$ or $6m+5$ for any integer $m$.

Answer 1

To prove the square of any positive integer cannot be of the form $6m+2$ or $6m+5$ for any integer $m$.

Euclid's Division Lemma: For any two positive integers $a$ and $b$, there exists unique integer $q$ and satisfying $a=bq+r$, where $0\le r<b$.

If $b=6$ then $a=6q+r$ where $0\le r<6$.

So, $r=0,1,2,3,4,5$.

Case 1: When $r=0$,

$\begin{array}{rcl}a& =& 6q+0\\ & =& 6q\end{array}$

Squaring on both the sides.

$$\begin{gathered} a^2={\left(6q\right)}^2 \\ {a}^2=36q^2 \\ {a}^2=6m\mathrm{where},\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left[m=6q^2\right] \end{gathered}$$

Case 2: When $r=1$,

$\begin{array}{rcl}a& =& 6q+1\end{array}$

Squaring on both the sides.

$$\begin{gathered} a^2={\left(6q+1\right)}^2 \\ {a}^2=36q^2+1+12q \\ {a}^2=6\left(6q^2+2q\right)+1 \\ {a}^2=6m+1\mathrm{where},\left[m=6q^2+2q\right] \end{gathered}$$

Case 3: When $r=2$,

$\begin{array}{rcl}a& =& 6q+2\end{array}$

Squaring on both the sides.

$$\begin{gathered} a^2={\left(6q+2\right)}^2 \\ {a}^2=36q^2+4+24q \\ {a}^2=6\left(6q^2+4q\right)+4 \\ {a}^2=6m+4\mathrm{where},\left[m=6q^2+4q\right] \end{gathered}$$

Case 4: When $r=3$,

$\begin{array}{rcl}a& =& 6q+3\end{array}$

Squaring on both the sides.

$$\begin{gathered} a^2={\left(6q+3\right)}^2 \\ {a}^2=36q^2+9+36q \\ {a}^2=6\left(6q^2+6q+1\right)+3 \\ {a}^2=6m+3\mathrm{where},\mathbf{}\left[m=6q^2+6q+1\right] \end{gathered}$$

Case 5: When $r=4$,

$\begin{array}{rcl}a& =& 6q+4\end{array}$

Squaring on both the sides.

$$\begin{gathered} a^2={\left(6q+4\right)}^2 \\ {a}^2=36q^2+16+48q \\ {a}^2=6\left(6q^2+8q+2\right)+4 \\ {a}^2=6m+4\mathrm{where},\left[m=6q^2+8q+2\right] \end{gathered}$$

Case 6: When $r=5$,

$\begin{array}{rcl}a& =& 6m+5\end{array}$

Squaring on both the sides.

$$\begin{gathered} a^2={\left(6q+5\right)}^2 \\ {a}^2=36q^2+25+60q \\ {a}^2=6\left(6q^2+10q+4\right)+1 \\ {a}^2=6m+1\mathrm{where},\mathbf{}\left[m=6q^2+10q+4\right] \end{gathered}$$

Hence, it is showed that the square of any positive integer cannot be of the form $6m+2$ or $6m+5$ for any positive integer $m$.