Question

Show that the square of any positive integer cannot be in the form of 5q + 2 or 5q + 3 , for any integer "q "

Answer 1

let a be any positive integer and b =5
by euclid division lemma
a=bm+r where 0 <_r <b
a=5m+r the possible value of r=0,1,2,3,4
if r=0 Then a=5m
a^2=(5m)^2
=25m^2 let 5m^2=q
than a^2=5q
if r=1 than a^2=(5m+1)^2=25m^2+10m+1
let 5m^2+2m=q
than a^2=5q+1
if r=2 than a^2=(5m+2)^2=5 (5m^2+4m)+4 let 5m^2+4m=q
than a^2=5q+4
if r=3 than a^2=(5m+3)^2=5 (5m^2+6m)+9
let 5m^2+6m=q
than a^2=5q+9
if r=4 than a^2=(5m+4)^2=5 (5m^2+8m)+16
let 5m^2+8m=q
than a^2=5q+16. ..........


Therefore the square of any positive integer cannot be in the form 5q+2or5q+3 for any integer q. .....

Answer 2

Let a^2 = 5 q + 1 for some integer q. a is a positive integer.

(a + 1) (a - 1) = 5 q

For a+1 =5, q = a-1 or for a-1=5 and a+1 = q, it is possible.

So a^2 can be in the form of 5q+1.

Similarly a^2 = 5q can be perfect square for q = 5.

Next for a^2 = 5 q + 4.

(a-2)(a+2) = 5 q

It is possible for a-2 = 5 and a+2=q, or vice versa.

any integer can be in the form of 5q, 5q+1, 5q+2, 5q+3, 5q+4.

For a^2 - 2 = 5q , or for a^2 = 5q +3 , q will be a fraction and not integer.

So a square cannot be in the form of 5q+2 or 5q+3.