Question

Use Euclid's Division Lemma to show that the square of any positive integer is either $5n,5n+1or5n+4$ for some integer.

Answer 1

Let us consider a positive numbers $a$.

As per the given data:

$a=5nora=5n+1ora=5n+2ora=5n+3ora=5n+4$ for integer $a$.

We know that According to Euclid’s Division Lemma

$a=bq+r${ condition for $ris(0\le r<b)$}

$a=5q+r—————\left(i\right)b=5$

so $r$ is an integer which lies in between $0and5$

Hence $r$ can be either $0,1,2,3,4$

So, now we can have five cases which we will take one by one.

Case I- When $r=0$, the equation $\left(1\right)$ becomes

$a=5q$

Now, squaring both the sides, we get

$$\begin{gathered} a^2={\left(5q\right)}^2 \\ {a}^2=25q^2 \\ {a}^2=5\left(5q^2\right) \\ {a}^2=5nwheren=5q^2 \end{gathered}$$

Case II- When $r=1$, the equation $\left(1\right)$ becomes

$a=5q+1$

Now, squaring both the sides, we get

$$\begin{gathered} a^2={\left(5q+1\right)}^2 \\ {a}^2=25q^2+1+10q \\ {a}^2=25q^2+10q+1 \\ {a}^2=5\left(5q^2+2q\right)+1 \\ {a}^2=5n+1wheren=5q^2+2q \end{gathered}$$

Case III- When $r=2$, the equation $\left(1\right)$ becomes

$a=5q+2$

Now, squaring both the sides, we get

$$\begin{gathered} a^2={\left(5q+2\right)}^2 \\ {a}^2=25q^2+4+20q \\ {a}^2=25q^2+20q+4 \\ {a}^2=5\left(5q^2+4q\right)+4 \\ {a}^2=5n+4wheren=5q^2+4q \end{gathered}$$

Case IV- When $r=3$, the equation $\left(1\right)$ becomes

$a=5q+3$

Now, squaring both the sides, we get

$$\begin{gathered} a^2={\left(5q+3\right)}^2 \\ {a}^2=25q^2+9+30q \\ {a}^2=25q^2+30q+5+4 \\ {a}^2=5\left(5q^2+6q+1\right)+4 \\ {a}^2=5n+4wheren=5q^2+6q+1 \end{gathered}$$

Case V- When $r=4$, the equation $\left(1\right)$ becomes

$a=5q+4$

Now, squaring both the sides, we get

$$\begin{gathered} a^2={\left(5q+4\right)}^2 \\ {a}^2=25q^2+16+40q \\ {a}^2=25q^2+40q+15+1 \\ {a}^2=5\left(5q^2+6q+3\right)+1 \\ {a}^2=5n+1wheren=5q^2+6q+3 \end{gathered}$$

Therefore, In each case, $a$ is either of the form $5n,5n+1or5n+4$ for integer $n$