show that for odd positive integer to be a perfect square is should be of form 8k plus 1
Answers
Answered by
1
Let the odd positive integrity n be in the form of 4q+1, 4q+3...
n=4q+1
(n)^2= (4q+1)^2
=16q^2+8q+1
=8(2q^2+q)+1
Where k=(2q^2+q)
=8k+1
n=4q+1
(n)^2= (4q+1)^2
=16q^2+8q+1
=8(2q^2+q)+1
Where k=(2q^2+q)
=8k+1
Similar questions