Prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 for some integer q.

Solution:

To Prove: that the square of any positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.

Proof: Since positive integer n is of the form of 5q or 5q + 1, 5q + 4 

If n = 5q

Then, $n^{2}=(5 q)^{2}$

$\Rightarrow \quad n^{2}=25 q^{2}$

$\Rightarrow \quad n^{2}=5(5 q)$

$\Rightarrow \quad n^{2}=5 m($ where $m=5 q)$

If n = 5q + 1

Then, $n^{2}=(5 q+1)^{2}$b

$\Rightarrow \quad n^{2}=(5 q)^{2}+10 q+1$

$\Rightarrow \quad n^{2}=25 q^{2}+10 q+1$

$\Rightarrow \quad n^{2}=5 q(5 q+2)+1$

$\Rightarrow n^{2}=5 m+1$ (where $m=q(5 q+2)$ )

If n = 5q + 2

Then, $n^{2}=(5 q+2)^{2}$

$\Rightarrow \quad n^{2}=(5 q)^{2}+20 q+4$

$\Rightarrow \quad n^{2}=25 q^{2}+20 q+4$

$\Rightarrow \quad n^{2}=5 q(5 q+4)+4$

$\Rightarrow n^{2}=5 m+4 \quad($ where $m=q(5 q+4))$

If n = 5q + 4

Then, $n^{2}=(5 q+4)^{2}$

$\Rightarrow \quad n^{2}=(5 q)^{2}+40 q+16$

$\Rightarrow \quad n^{2}=25 q^{2}+40 q+16$

$\Rightarrow \quad n^{2}=5\left(5 q^{2}+8 q+3\right)+1$

$\Rightarrow n^{2}=5 m+1$ (where $\left.m=\left(5 q^{2}+8 q+3\right)\right)$

Hence it is proved that the square of a positive integer is of the form 5q or 5q + 1, 5q + 4 for some integer q.