Question

show that square of any odd positive integer is of the form 4q+1​

Answer 1

Answer:

4+1=5

Step-by-step explanation:

5*5=25 hence answer is 25

Answer 2

Let a be any odd integer and b = 4. Then, by Euclid’s algorithm, a = 4m + r for some integer m ≥ 0 and r = 0,1,2,3 because 0 ≤ r < 4. So, a = 4m or 4m + 1 or 4m + 2 or 4m + 3 So, a = 4m + 1 or 4m + 3 Here, a cannot be 4m or 4m + 2, as they are divisible by 2. => (4m + 1)2 = 16m2 + 8m + 1 = 4(4m2 + 2m) + 1 = 4q + 1, where q is some integer. (4m + 3)2 = 16m2 + 24m + 9 = 4(4m2 + 6m + 2) + 1 = 4q + 1, where q is some integer. Hence, The square of any odd integer is of the form 4q + 1, for some integer q.Read more on Sarthaks.com - https://www.sarthaks.com/12815/show-that-the-square-of-any-odd-integer-is-of-the-form-4q-1-for-some-integer-