Question

prove that if x and y are both odd positive integers than x²+y² is even but not divisible by 4

Answer 1

since x and y are odd generally we can represent it as 2m+1 and 2n+1 respectively
there fore
x^2+y^2=4m^2+4n^2+2+4m+4n
=4(m^2+n^2+m+n)+2
imply..this is not divisible by 4

Answer 2

Step-by-step explanation:

we know that any odd positive integer is of the form 2q+1 for some integer q.

so, let x = 2m+1 and y = 2n+1 for some integers m and n.

•°• x²+y²= ( 2m+1)²+(2n+1)²

==> x²+y²=4(m²+n²)+4(m+n)+2

==> x²+y²= 4{(m²+n²)+(m+n)} +2

==> x²+y² = 4q +2 , where q= (m²+n²)+(m+n)

==> x²+y² is even and leaves remainder 2 when divided by 4.

==> x²+y² is even but not divisible by 4.

HENCE PROVED✔✔