Question

show that the square of any positive integer cannot be of the form 6m+2 or 6m+5​

Answer 1

Answer:

Let a be the positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 5. So, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5. (6q)2 = 36q2 = 6(6q2) = 6m, where m is any integer. (6q + 1)2 = 36q2 + 12q + 1 = 6(6q2 + 2q) + 1 = 6m + 1, where m is any integer. (6q + 2)2 = 36q2 + 24q + 4 = 6(6q2 + 4q) + 4 = 6m + 4, where m is any integer. (6q + 3)2 = 36q2 + 36q + 9 = 6(6q2 + 6q + 1) + 3 = 6m + 3, where m is any integer. (6q + 4)2 = 36q2 + 48q + 16 = 6(6q2 + 7q + 2) + 4 = 6m + 4, where m is any integer. (6q + 5)2 = 36q2 + 60q + 25 = 6(6q2 + 10q + 4) + 1 = 6m + 1, where m is any integer. Hence, The square of any positive integer is of the form 6m, 6m + 1, 6m + 3, 6m + 4 and cannot be of the form 6m + 2 or 6m + 5 for any integer m.Read more on Sarthaks.com - https://www.sarthaks.com/12814/show-that-the-square-of-any-positive-integer-cannot-be-of-the-form-6m-2-or-6m-5-for-any-integer

Answer 2

Heya mate... here is your answer

Let the square of any posititve integer be in the form 6q + 2, 6q +5

x = 6q +2

Squaring both sides

Xsquare = 36(q)square + 4 + 24q

x square = 36(q) square - 6+ 2 + 24q

xsquare = 6 [ 6(q) square - 1 + 4q] + 2

put [ 6(q) square - 1 + 4q] = m

Therefore x = 6m + 2

have a great day friend...