1. ## Proof validation

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?

2. Originally Posted by DenMac21
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 $\displaystyle a+b=5k$.
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 $\displaystyle (a-b)(a+b)=101010$ 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?