Question

Use Euclid's Division Lemma to show that the square of any positive integer is either of the form $3n$ or $3n+1$ for some integer $n.$

Answer 1

As per Euclid's Division Lemma.

If $a$ & $b$ are $2$ $+ve$ integers then $a=bq+r$ ($\because$ $0\le r<b$).

Let $a=3$, then $r=0,1,2$

$\underset{–}{r=0}$ $\underset{–}{r=1}$ $\underset{–}{r=2}$

$a=3q$ $a=3q+1$ $a=3q+2$

$a^2={9q}^2$ $a^2={9q}^2+1+6q$ $a^2={9q}^2+4+12q$

$=3\left({3q}^2\right)$ $=3({3q}^2+2q)+1$ $=3({3q}^2+4q+1)+1$

$a^2=3m$ $a^2=3m+1$ $=3m+1$

$\{m={3q}^2\}$ $\{m={3q}^2+2q\}$ $\{m={3q}^2+4q+1\}$

Therefore, the square of any positive integer is either of the form $3m$ or $3m+1.$