Find $p,q$ such that $p+q-pq+1=0$ where $p,q$ are primes with $p
Find $p,q$ such that $p+q-pq+1=0$ where $p,q$ are primes with $p<q$.
My try:
If we take $p=2,q=3$ the equality is satisfied.
But how can I show that this is the only solution?

$p + q-pq +1=0$ means $pq - q=p+1$ so $q=\frac {p+1}{p-1}$.
And $pq-p = q+1$ so $p =\frac {q+1}{p-1}$.
To solve $\frac {x+1}{x-1}\in \mathbb Z$, we can get $\gcd(x+1,x-1)= \gcd(x+1, (x+1)-(x-1)) = \gcd (x,2) = 1,2$  so $x-1=1,2$ (assuming $x-1 >0$) and $x+1=3,4$ and $x = 2,3$.
Alternatively we could have assigned $m = x-1$ and realized that $m|m+2$ so $m|2$ so $m = 1,2$
So $p$ and $q$ are equal to $2$ and $3$ and as $p > q$ we have $p=3; q=2$.
