Question

show that the square of any integer cannot be of the form 5q plus 2 or 5q plus 3 for any integer q

Answer 1

Let a be any positive integer.

By Euclid's division lemma,

a = bm + r where b = 5

⇒ a = 5m + r

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

∴ a = 5m when r = 0

a = 5m + 1 when r = 1

a = 5m + 2 when r = 2

a = 5m + 3 when r = 3

a = 5m + 4 when r = 4

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

Case I : a = 5m

⇒ a2 = (5m)2 = 25m2

⇒ a2 = 5(5m2)

= 5q , where q = 5m2

Case II : a = 5m + 1

⇒ a2 = (5m + 1)2 = 25m2 + 10 m + 1

⇒ a2 = 5 (5m2 + 2m) + 1

= 5q + 1, where q = 5m2 + 2m

Case III : a = 5m + 2

⇒ a2 = (5m + 2)2

= 25m2 + 20m +4

= 25m2 + 20m +4

= 5 (5m2 + 4m) + 4

= 5q + 4 where q = 5m2 + 4m

Case IV: a = 5m + 3

⇒ a2 = (5m + 3)2 = 25m2 + 30m + 9

= 25m2 + 30m + 5 + 4

= 5 (5m2 + 6m + 1) + 4

= 5q + 4 where q = 5m2 + 6m + 1

Case V: a = 5m + 4

⇒ a2 = (5m + 4)2 = 25m2 + 40m + 16

= 25m2 + 40m + 15 + 1

= 5 (5m2 + 8m + 3) + 1

= 5q + 1 where q = 5m2 + 8m + 3

From all these cases, it is clear that square of any positive integer can not be of the form

5m + 2 or 5m + 3

Answer 2

sorry for the variable u may take m in place of q and q in place of m
hope it works well