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^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.

Answer 2

SOLUTION :-

let a be any positive integer. by euclids Division lemma ,
a = bm + r , where b= 5 and r can be 0, 1, 2 ,3,4,5.

Now,
CASE : 1
where r = 0
a= 5m
a² = ( 5m )² = 25m²
a² = 5( 5m )²
a² = 5q
here ,q = 5m²

CASE : 2
a= 5m + 1
a² = ( 5m + 1 )²
= 25m² + 10m + 1
= 5( 5m² + 2m ) +1
= 5q + 1
here, q= 5m²+ 2m


CASE : 3
a = 5m + 2
a² = ( 5m + 2 ) ²
= 25m² + 20m + 4
= 5 ( 5m²+ 4m ) +4
= 5q + 4
here, q = 5m² + 4m

CASE : 4
a = 5m + 3
a² = ( 5m + 3 ) ²
= 25m² + 30m + 9
= 5 ( 5m²+ 6m ) + 9
= 5q + 9
here, q= 5m² + 6m

CASE : 5
a= 5m+ 4
a² = ( 5m+ 4 )²
= 25m²+ 40m+ 16
= 5(5m² + 8m) +16
= 5q + 16
here, q = 5m²+ 8m

as we can see from all abvove cases that square of any positive ( +ve ) integer cannot be of the form 5q + 2 , or 5q + 3 . For any integer q.