Show that the square of any positive odd integer is of the form 4m+1,for same integer n
Answers
Let ‘a’ be any positive integer.
Then,
According to Euclid’s division lemma,
a=bq+r
According to the question, when b = 4.
a = 4k + r, 0 < r < 4
When r = 0, we get, a = 4k
a2 = 16k2 = 4(4k2) = 4q, where q = 4k2
When r = 1, we get, a = 4k + 1
a2 = (4k + 1)2 = 16k2 + 1 + 8k = 4(4k + 2) + 1 = 4q + 1, where q = k(4k + 2)
When r = 2, we get, a = 4k + 2
a2 = (4k + 2)2 = 16k2 + 4 + 16k = 4(4k2 + 4k + 1) = 4q, where q = 4k2 + 4k + 1
When r = 3, we get, a = 4k + 3
a2 = (4k + 3)2 = 16k2 + 9 + 24k = 4(4k2 + 6k + 2) + 1
= 4q + 1, where q = 4k2 + 6k + 2
Therefore, the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.
Let ‘a’ be any positive integer.
Then,
According to Euclid’s division lemma,
a=bq+r
According to the question, when b = 4.
a = 4k + r, 0 < r < 4
When r = 0, we get, a = 4k
a2 = 16k2 = 4(4k2) = 4q, where q = 4k2
When r = 1, we get, a = 4k + 1
a2 = (4k + 1)2 = 16k2 + 1 + 8k = 4(4k + 2) + 1 = 4q + 1, where q = k(4k + 2)
When r = 2, we get, a = 4k + 2
a2 = (4k + 2)2 = 16k2 + 4 + 16k = 4(4k2 + 4k + 1) = 4q, where q = 4k2 + 4k + 1
When r = 3, we get, a = 4k + 3
a2 = (4k + 3)2 = 16k2 + 9 + 24k = 4(4k2 + 6k + 2) + 1
= 4q + 1, where q = 4k2 + 6k + 2
Therefore, the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.