Show that square of any positive integer is of the form $ 5p, 5p+1, 5p+4 $.

Given: 

Square of any positive integer is of the form $5p$, $5p\ +\ 1$, $5p\ +\ 4$.

To prove: 

We have to show that the square of any positive integer is of the form $5p$, $5p\ +\ 1$, $5p\ +\ 4$.

Solution:

Let 'a' be an integer such that $a\ =\ 5m\ +\ r$.

According to Euclid's division algorithm:

$a\ =\ bm\ +\ r$, where $0\ \underline{< }\ r\ <\ b$.

Here, b = 5. So,

$a\ =\ 5m\ +\ r$, where $0\ \underline{< }\ r\ <\ 5$.

Now we need to consider all cases of r.

When r = 0:

Let us consider p to be equal to $m^2$.

When $r\ =\ 0$, we can conclude that $a\ =\ 5m$.

$a\ =\ 5m$

$a^2\ =\ ( 5m )^2$

$a^2\ =\ 5 \left( 5m^{2}\right)$

$a^2\ =\ 5p$

When r = 1:

Let us consider p to be equal to $5m^2\ +\ 2m$.

$a\ =\ 5m\ +\ 1$

$a^2\ =\ ( 5m\ +\ 1 )^2$

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

$a^2\ =\ 5( 5m^2\ +\ 2m )\ +\ 1$

$a^2\ =\ 5p\ +\ 1$

When r = 2:

Let us consider p to be equal to $5m^2\ +\ 4m$.

$a\ =\ 5m\ +\ 2$

$a^2\ =\ ​( 5m\ +\ 2 )^2$

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

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

$a^2\ =\ 5p\ +\ 4$

When r = 3:

Let us consider p to be equal to $5m^2\ +\ 6m\ +\ 1$.

$a\ =\ 5m\ +\ 3$

$a^2\ =\ ​(5m\ +\ 3)^2$

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

$a^2\ =\ 25m^2\ +\ 30m\ +\ 5\ +\ 4$

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

$a^2\ =\ 5p\ +\ 4$.

When r = 4:

Let us consider p to be equal to $5m^2\ +\ 8m\ +\ 3$.

$a\ =\ 5m\ +\ 4$

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

$a^2\ =\ 25m^2\ +\ 40m\ +\ 15\ +\ 1$

$a^2\ =\ 5 ( 5m^2\ +\ 8m\ +\ 3 )\ +\ 1$

$a^2\ =\ 5p\ +\ 1$

Hence, the square of any positive integer is of the form 5p or 5p $+$ 1 or 5p $+$ 4.

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Updated on: 10-Oct-2022

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