1. ## Proof validation

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

My proof is this:
Since 101010 can be divided with 5 which is prime number, then one of factors $(a-b)(a+b)$ also must be 5.
Lets assume that $a-b=5$ and $a+b=5k$.
Because it must be $(a-b)(a+b)=101010$ then it must be $a+b=20202$.
So, we have to solve system of equations in order to see whether $a,b$ are integers.

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

Is this proof valid?

2. Originally Posted by DenMac21
I need to prove that if $a,b$ are integers then $a^2-b^2=101010$ is not possible.

My proof is this:
Since 101010 can be divided with 5 which is prime number, then one of factors $(a-b)(a+b)$ also must be 5.
Lets assume that $a-b=5$ and $a+b=5k$.
101010 divisible by 5 implies one of (a-b) and (a+b) is divisible by 5, not
that one of them is 5. Also if one were 5 the other need not be divisible
by 5.

Because it must be $(a-b)(a+b)=101010$ then it must be $a+b=20202$.
So, we have to solve system of equations in order to see whether $a,b$ are integers.

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

Is this proof valid?