Question

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

Answer 1

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.