Question

show that the square of any positive integer cannot be of the form 5q+2 and 5q+3 for any integer q

Answer 1

Consider a be any positive integer.

By Euclid's division lemma, a = bn + r  where  b = 5

⇒ a = 5n + r

So that r can be any of 0, 1, 2, 3, 4

∴  a = 5n  when  r = 0
 a = 5n + 1  when  r = 1
 a = 5n + 2  when  r = 2
 a = 5n + 3  when  r = 3
 a = 5n + 4  when  r = 4

So, "a"  is any positive integer in the form of 5n,  5n + 1 ,  5n + 2 , 5n + 3 ,  5n + 4 for some integer n.

Case i :
 a = 5n
 ⇒  a 2 = (5n)2 = 25n 2
 ⇒  a 2 = 5(5n 2)
         = 5q , where  q = 5n 2

Case ii : 
a = 5n + 1
   ⇒  a 2 = (5n + 1)2 = 25n 2 + 10 n + 1
   ⇒  a 2 = 5 (5n 2 + 2n) + 1
           = 5q + 1,  where q = 5n 2 + 2n .

Case iii :  
a = 5n + 2
 ⇒   a 2 = (5n + 2)2
 =  25n 2 + 20n +4
 =  25n 2 + 20n +4
 =  5 (5n 2 + 4n) + 4
 =  5q + 4  where q = 5n 2 + 4n

 
Case iv:
 a = 5n + 3
 ⇒  a 2 = (5n + 3)2 = 25n 2 + 30n + 9
 = 25n 2 + 30n + 5 + 4
 = 5 (5n 2 + 6n + 1) + 4
 = 5q + 4  where  q = 5n 2 + 6n + 1
 
Case v:  
 a = 5n + 4
 ⇒  a 2 = (5n + 4)2 = 25n 2 + 40n + 16
 = 25n 2 + 40n + 15 + 1
 = 5 (5n 2 + 8n + 3) + 1
 = 5q + 1  where  q = 5n 2 + 8n + 3

From all these cases, it is clear that square of any positive integer can not be of the form 5q + 2  or  5q + 3 for any integer q.
Hope it help you...!