Question

Prove that if x and y are both odd positive integers then x2 + y² is even
but not divisible by 4.
YT​

Answer 1

Since $\sf{x}$ and $\sf{y}$ are odd positive integers, they leave remainder 1 on division by 2. So let,

$\longrightarrow\sf{x=2m+1}$

$\longrightarrow\sf{y=2n+1}$

for some whole numbers $\sf{m}$ and $\sf{n.}$

Then,

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

$\longrightarrow\sf{x^2+y^2=4m^2+4m+1+4n^2+4n+1}$

$\longrightarrow\sf{x^2+y^2=4m^2+4n^2+4m+4n+2}$

$\longrightarrow\sf{x^2+y^2=4(m^2+n^2+m+n)+2}$

Let $\sf{m^2+n^2+m+n=p.}$ Then we see that,

$\longrightarrow\sf{x^2+y^2=4p+2\quad\quad\dots(1)}$

This means $\sf{x^2+y^2}$ leaves remainder 2 on division by 4. This implies $\sf{x^2+y^2}$ is not divisible by 4.

But from (1) we also see that,

$\longrightarrow\sf{x^2+y^2=2(2p+1)}$

Let $\sf{2p+1=q.}$ Then,

$\longrightarrow\sf{x^2+y^2=2q}$

This implies $\sf{x^2+y^2}$ is an even number.

Hence Proved!