Question

Show that square of every positive integer is of the form 5p or 5p 1 or 5p 4 where p is some integer

Answer 1

Step-by-step explanation:

Let x be any positive integer and y = 5.

Using Euclid's Division Lemma,we have

x = yq + r , 0 ≤ r < b

⇒x = 5q + r , where r =0,1,2,3 and 4.

So, x can be = 5q, (5q + 1), (5q + 2), (5q + 3) and (5q + 4).

Case:1

Taking x = 5q

On squaring both sides, We get,

x² = 25q²

⇒x² = 5(5q²)

⇒x² = 5p   [ let 5q² = p for some integer p]

Case:2

Taking x = (5q + 1)

On squaring both sides, We get

⇒x² = (5q+1)²

⇒x² = (5q)² + 2×5q×1 + 1²

⇒x² = 25q² + 10q + 1

⇒x² = 5(5q² + 2) + 1

⇒x² = 5p + 1   [Let 5q²+2 = p for some integer p]

Case: 3

Taking x = (5q + 2)  

On  squaring both sides ,We get

⇒x² = (5q+2)²

⇒x² = 25q² + 20q + 4

⇒x² =  5(5q² + 4q ) + 4

⇒x² = 5p + 4     [Let 5q²+4q = p for some integer p)]

Case: 4

Taking x = (5q + 3)

On squaring both sides , We get

⇒x² = 25q²+30q + 9

⇒x² = 5 (5q² + 6q + 1) +4

⇒x² = 5p + 4           [Let 5q² + 6q + 1 = p for some integer p]

Case: 5

Taking x = (5q + 4)

On squaring both sides , We get

⇒x² = 25q²+40q + 16

⇒x² = 5 (5q² + 8q + 3) + 1

⇒x² = 5p + 1           [Let 5q² + 8q + 3 = p for some integer p]

From the above,the square of any positive integer is of the form 5p, (5p + 1) and (5p + 4) where p is some integer.  (Hence Proved)

________________________________

//Hope This will Helped You//

//Please Mark it as Brainliest//

#BAL   #ANSWERWITHQUALITY

Answer 2

Complete question is Show that square of every positive integer is of the form 5p or 5p+1 or 5p+4 where p is some integer.

Step-by-step explanation:

Let x be any positive integer.

then,$x = 5p$or$x = 5p + 1$or$x = 5p + 4$for integer x.

If $x = 5p$

${x}^{2} = {5p}^{2} = 25 {p}^{2} = 5 \times 5 \times {p}^{2} = 5n$

Where $n = 5 {p}^{2}$

If $x = 5p + 1$

where $n = 5 {p}^{2} + 2p$

If $x = 5p + 4$

${x}^{2} = {(5p + 4)}^{2} = 25 {p}^{2} + 40p + 16 = 5(5 {p}^{2} + 8p + 3) + 1 = 5n + 1$

Where $n = 5 {p}^{2} + 8p + 3$

In each three cases ${x}^{2}$is either of the form $5p$or $5p + 1$or$5p + 4$and for interger p. [proved]