Show that the square of any positive integer cannot be of the form 5q+2 or 5q+3 for any integer q.
Answers
Sol: 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 = 4So, "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 + 3From 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.
Answer:
Step-by-step explanation:
Let a be the positive integer and b = 5. Then, by Euclid’s algorithm, a = 5m + r for some integer m ≥ 0 and r = 0, 1, 2, 3, 4 because 0 ≤ r < 5. So, a = 5m or 5m + 1 or 5m + 2 or 5m + 3 or 5m + 4. So, (5m)2 = 25m2 = 5(5m2) = 5q, where q is any integer. (5m + 1)2 = 25m2 + 10m + 1 = 5(5m2 + 2m) + 1 = 5q + 1, where q is any integer. (5m + 2)2 = 25m2 + 20m + 4 = 5(5m2 + 4m) + 4 = 5q + 4, where q is any integer. (5m + 3)2 = 25m2 + 30m + 9 = 5(5m2 + 6m + 1) + 4 = 5q + 4, where q is any integer. (5m + 4)2 = 25m2 + 40m + 16 = 5(5m2 + 8m + 3) + 1 = 5q + 1, where q is any integer. Hence, The square of any positive integer is of the form 5q, 5q + 1, 5q + 4 and cannot be of the form 5q + 2 or 5q + 3 for any integer q.Read more on Sarthaks.com - https://www.sarthaks.com/12813/show-that-the-square-of-any-positive-integer-cannot-be-of-the-form-5q-2-or-5q-3-for-any-integer