Question

show that the square of any positive integer is either of the form 4q or 4q+1 for some integer q.​

Answer 1

Solution:

Since any positive integers 'n' in the form 2m or, 2m + 1

If n = 2m, then

n^2 = 4m^2 = 4q

where, q = m^2

If n = 2m + 1, then

n^2 = ( 2m + 1 )^2

n^2 = 4m^2 + 4m + 1

n^2 = 4m ( m + 1 ) + 1

n^2 = 4q + 1

where, q = m ( m + 1 )

Thus, it proves that the square of any positive integer is either of the form 4q or 4q+1.