Proof problem : show that $\frac{n_{1}n_{2}-1}{2}$ = $\frac{n_{1}-1}{2}$ + $\frac{n_{2}-1}{2}$ mod(2)

Question

I want to show that, for odd positive integers $n_{1}$ and $n_{2}$, we have :

$\frac{n_{1}n_{2}-1}{2}$ = $\frac{n_{1}-1}{2}$ + $\frac{n_{2}-1}{2}$ mod(2)

Let $n_{1}$ = ${2m_{1} +1}$ and $n_{2}$ = ${2m_{2} +1}$

I get :

2${m_{1}m_{2}} + {m_{1}} + {m_{2}}$ $\cong$ ${m_{1} + m_{2}}$ mod(2)

My question is : how this is proving the statement ? I am not sure what to do next.

Answer 1

You can drop the $2m_1m_2$ since working mod $2.$