Question

If x, y are odd positive integers then $x^2+y^2$ must be divisible by

• A. 2
• B. 3
• C. 4
• D. 8

Answer 1

The correct option is A 2

$\Rightarrow$ Any odd positive integer is of the form $2q+1$, where $q$ is some integer.

$\Rightarrow$ Let $x=2m+1$ and $y=2n+1$, where $m$ and $n$ are some integer.

$\Rightarrow$ $x^2+y^2=(2m+1{)}^2+(2n+1{)}^2$

$\Rightarrow$ $x^2+y^2=4m^2+1+4m+4n^2+1+4n$

$\therefore$ $x^2+y^2=4(m^2+n^2+m+n)+2$

$\Rightarrow$ $x^2+y^2=4p+2$, where $p=m^2+n^2+m+n$

$\Rightarrow$ $4p$ and $2$ are even number, so $4p+2$ is also even number.

$\Rightarrow$ We know that all even numbers are divisible by 2.

$\therefore$ $x^2+y^2$ is even and divisible by $2$.