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Hint: Use closure property with addition and subtraction over integers.

We have given that If $x$ and $y$ are positive integers. Let $x = 2h$for$h \in \mathbb{Z}$and$y = 2k$for$k \in \mathbb{Z}$(Here$\mathbb{Z}$ represents the set of integers).Since,$x - y$ is even so we can also write it as multiple of $2$. That is $x - y = 2m$ for some $m \in \mathbb{Z}$.

Now consider,

$\begin{gathered}

{x^2} - {y^2} \\= (x + y)(x - y){\text{ [Using, }}{a^2} - {b^2} = (a + b)(a - b){\text{]}} \\

{\text{ = }}(2h + 2k)(2m){\text{ [Using, }}x = 2h,y = 2k,x - y = 2m{\text{]}} \\

= 2 \times 2(h + k)(m){\text{ [2 is taken out from both brackets]}} \\

{\text{ = 4}}(h + k)(m) \\

\end{gathered} $

Since, ${x^2} - {y^2}$ can be expressed out as the multiple of $4.$so, it can be divisible by $4$ and the multiple will be $m(h + k){\text{ where, }}m,h,k \in \mathbb{Z}$. Hence Proved.

Note: In number theories one needs to visualise the numbers in order to solve it. For example, if a problem says something is multiple of $4.$then immediately visualise that number as $4k{\text{ for some }}k \in \mathbb{Z}$.

We have given that If $x$ and $y$ are positive integers. Let $x = 2h$for$h \in \mathbb{Z}$and$y = 2k$for$k \in \mathbb{Z}$(Here$\mathbb{Z}$ represents the set of integers).Since,$x - y$ is even so we can also write it as multiple of $2$. That is $x - y = 2m$ for some $m \in \mathbb{Z}$.

Now consider,

$\begin{gathered}

{x^2} - {y^2} \\= (x + y)(x - y){\text{ [Using, }}{a^2} - {b^2} = (a + b)(a - b){\text{]}} \\

{\text{ = }}(2h + 2k)(2m){\text{ [Using, }}x = 2h,y = 2k,x - y = 2m{\text{]}} \\

= 2 \times 2(h + k)(m){\text{ [2 is taken out from both brackets]}} \\

{\text{ = 4}}(h + k)(m) \\

\end{gathered} $

Since, ${x^2} - {y^2}$ can be expressed out as the multiple of $4.$so, it can be divisible by $4$ and the multiple will be $m(h + k){\text{ where, }}m,h,k \in \mathbb{Z}$. Hence Proved.

Note: In number theories one needs to visualise the numbers in order to solve it. For example, if a problem says something is multiple of $4.$then immediately visualise that number as $4k{\text{ for some }}k \in \mathbb{Z}$.

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