I need to prove that if $\displaystyle a,b$ are integers then $\displaystyle a^2-b^2=101010$ is not possible.

My proof is this:

Since 101010 can be divided with 5 which is prime number, then one of factors $\displaystyle (a-b)(a+b)$ also must be 5.

Lets assume that $\displaystyle a-b=5$ and $\displaystyle a+b=5k$.

Because it must be $\displaystyle (a-b)(a+b)=101010$ then it must be $\displaystyle a+b=20202$.

So, we have to solve system of equations in order to see whether $\displaystyle a,b$ are integers.

$\displaystyle \begin{array}{l}

\left\{ \begin{array}{l}

a - b = 5 \\

a + b = 20202 \\

\end{array} \right. \\

\\

\left\{ \begin{array}{l}

a = 5 + b \\

a = 20202 - b \\

\end{array} \right. \\

\\

5 + b = 20202 - b \\

2b = 20202 - 5 \\

b = \frac{{20197}}{2} \\

\end{array}

$

Since $\displaystyle \frac{{20197}}{2}$ is not integer then b is not integer.

Is this proof valid?