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