it might be a silly question but i need help please
given RSA system , where $n=pq , p\ and \ q \ are \ primes $ , $ v_0,v_1,v_2, v_3 \ are \ known $
$p^p \equiv v_0 \mod q$
$q^q \equiv v_1 \mod p$
$q^p + p^q \equiv v_2 \mod pq$
$(p+q)^{p+q} \equiv v_3 \mod pq$
Mathmatical facts:
$p^{q-1} \equiv 1 \mod q$
is obtained using $\phi(q)=q-1$ , Fermat: $p^{\phi(q)} \equiv 1 \mod q$
$q^{p-1} \equiv 1 \mod p$
$q^p + p^q \equiv p+q \mod pq$
is obtained using $(p^{q} \equiv p \mod pq) + (q^{p} \equiv q \mod pq)$
$(p+q)^{p+q} \equiv p^{p+q}+ q^{p+q} \mod pq$
Edit : The solution from https://ctf-wiki.github.io/ctf-wiki/crypto/asymmetric/rsa/rsa_theory/#2018-national-security-week-pure-math
is $q=\frac{v_3-v_0*v2}{v_1-v_0}$ where you can rewrite as
$q=\frac{p^{p+q}+ q^{p+q} -p^p*(q^p + p^q)}{q^q-p^p}= \frac{p^{p+q}+ q^{p+q} -p^{p+q}-(pq)^p }{q^q-p^p}= \frac{q^{p}*(q^q-p^p) }{q^q-p^p} = q^p = q $
I can prove this backward how is calculated but what is the fundamental idea that used to bring the above equation?
