Question

Use Euclid's division lemma to show that the square of any positive integer is either of the form $5n,5n+1$ or $5n+4$ for some integer.

Answer 1

Let $x$ be any integer

Then,

Either $x=5m$ or $x=5m+1$ or $x=5m+2$ or, $x=5m+3$ or $x=5m+4$ for integer $x$. [ Using division algorithm]

If $x=5m$

On squaring both side and we get,

$x^2=25m^2=5\left(5m^2\right)=5n$ where $n=5m^2$

If $x=5m+1$

On squaring both side and we get,

$x^2={(5m+1)}^2$

$=25m^2+1+10m$

$=5(5m^2+2m)+1(where5m^2+2m=n)$

$=5n+1$

If $x=5m+2$

Then $x^2$

$=(5m+2{)}^2$

$=25m^2+20m+4$

$=5(5m^2+4m)+4$

$=5n+4$ [ Taking $n=5m^2+4m$]

If $x=5m+3$

Then $x^2$

$=(5m+3{)}^2$

$=25m^2+30m+9$

$=5(5m^2+6m+1)+4$

$=5n+4$ [ Taking $n=5m^2+6m+1$]

If $x=5m+4$

On squaring both side and we get,

$x^2={(5m+4)}^2$

$=25m^2+16+40m$

$=5(5m^2+8m+3)+1(where5m^2+8m+3=n)$

$=5n+1$

Hence, In each cases $x^2$ is either of the of the form $5n$ or $5n+1$ for integer $n.$.