show that the square of any positive integer cannot be if the form 5q+2 or 5q+3 for some integer q
Answers
d = 5m + r,
where 0 ≤ r <5
If d = 5m, d² = 5.q,
where q is some integer and q = 5m²
If d = 5m + 1, d² = 5q + 1
If d = 5m + 2, d² = 5q + 4
If d = 5m + 3, d² = 5q + 4
If d = 5m + 4, d² = 5q + 1
Therefore, the square of any positive integer cannot be in the form of 5q + 2 or 5q + 3 for any integer "q"
hope it helps
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 5q+2 or 5q+3