Question

Use Euclid's division lemma to show that the square of any positive integer is of the form $3p,3p+1$.

Answer 1

Let $a$ be any positive integer and $b=3$

Then, by Euclid's division lemma,

$a=3q+r$

Since $0\le r<3$, the possible remainders are $0,1$ and $2$.

i.e., $a$ can be of the form $3q,3q+1$ or $3q+2$

Now, if $a=3q⟹a^2=9q^2=3\times 3q^2$

$a=3q+1⟹a^2=9q^2+6q+1=3(3q^2+2q)+1$

$a=3q+2⟹a^2=9q^2+12q+4=3(3q^2+4q+1)+1$

[where, $3q^2,3q^2+2q$ and $3q^2+4q+1$ can be expressed by integer ${}^{\prime }{p}^{\prime }$ for some values of $p$]

This implies that $a^2$ is of the form $3p$ or $3p+1$, where $p$ is the integer

Hence, square of any positive integer is of the form $3p$ or $3p+1$