Question

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

Answer 1

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

Let x = 2m+1 and y=2n+1

x²+ y² = (2m+1)²+(2n+1)²
x²+y² = 4m²+4m +1 +4n²+4n+1
x²+y³ = 4{(m²+n²)+(m+n)} +2
x²+y² = 4q+2

so x²+y³ is even but leaves remainder 2 when divided 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✔✔