Factoring two RSA moduli $N_i=p_i\cdot q_i$ knowing that $p_2=p_1+2$?
The fancy name for this is factorization with an implicit hint. If the primes are unbalanced, i.e., if $\log_2 p_i &gt; 2 \log_2 q_i$, we know how to factor $n_1$ and $n_2$ quite easily. Let $k$ be the number of bits $p_1$ and $p_2$ differ by (a very small $k$, usually, for $(p, p+2)$); reduce the following lattice 

$$
\begin{pmatrix}
2^k &amp; 0 &amp; n_2 \\
0   &amp; 2^k &amp; -n_1
\end{pmatrix}
$$

using LLL or equivalent. The resulting shortest vector will be $(2^k q_1, 2^k q_2, 2q_1q_2)$. This method is described, in much more generality (i.e., in terms of shared $p_i$ bits) in Faugère, Marinier, and Renault (see also May and Ritzenhofen, Sarkar-Maitra, Kurusawa-Ueda, and Nitaj-Ariffin). 

Two recent papers make improvements to the above lattice-based method, and Theorem 1 of Lu-Peng-Zhang-Lin seems to imply that implicit factorization with balanced moduli is indeed possible. So the answer to your question would seem to be yes.