Can this congruence be simplified? $$p(p+1) \equiv -q(q+1) \bmod pq$$
Can this be reduced to an easier format?

Mod $pq$,
$p(p+1) \equiv -q(q+1)
\iff
p^2+p = -q^2-q
\iff
p^2+q^2 +p+q =0
$
$\begin{array}\\
(p+q+1)^2
&\equiv
p^2+q^2+1+2pq+2p+2q\\
&\equiv
p^2+q^2+1+2p+2q\\
&\equiv
(p^2+q^2+p+q)+p+q+1\\
&\equiv
p+q+1\\
\end{array}
$
Therefore,
if $n = p+q+1$,
$n^2 \equiv n$
or
$n(n-1) \equiv 0
$.
If $p$ and $q$ are distinct primes
(or even just relatively prime),
then
$gcd(n-1, p) = gcd(n-1, q) = 1$,
so that
$n \equiv 0
$.
But $n < pq
$,
so that
this can not hold.
