Question

prove that if x and y are odd positive integer then x²+y² squared is even but not divisible by 4

Answer 1

let, y=x+2
if, x= 1, then y= 3
x²+y²=1+9=10,
10%4 != 0,
again if, x= 5 then y= 7.
so x²+y² =74.
74%4 !=0;
Now, x and y are not connected yet both are odd integers.
let x = 5, y = 17
x²+y² = 314
314%4 !=0;

when x,y belong to N and both are odd numbers, they can be written as (2N+1)
now if x=y=2N+1 [N=1,2,3,4,.....]
x²+y²= 2×(2N+1)² = 8N² +8N +2
and For an even coefficient of N makes the whole number even and even number +2 = even. And (8N²+8N+2)%4 !=0
So if x,y are odd numbers then x²+y² is even and is not completely 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✔✔