Question

Let $p_1<p_2<p_3<p_4$ and $q_1<q_2<q_3<q_4$ be two sets of prime numbers such that $p_4-p_1=8$ and $q_4-q_1=8$. Suppose, $p_1>5$ and $q_1>5$. Prove that $30$ divides $p_1-q_1$.

Answer 1

Since $p_4-p_1=8$, and no prime is even, we observe that $\left\{\begin{array}{c}{p}_1,p_2,p_3,p_4\end{array}\right\}$ is a subset of $\left\{\begin{array}{c}{p}_1,p_1+2,p_1+4,p_1+6,p_1+8\end{array}\right\}$
Also $p_1$ is larger than $3$
If $p_1\equiv 1$ (mod $3$), then $p_1+2and{p}_1+8$ are divisible by $3$.
Hence we do not get $4$ primes in the set $\left\{\begin{array}{c}{p}_1,p_1+2,p_1+4,p_1+6,p_1+8\end{array}\right\}$.
$\therefore p_1\equiv 2$ (mod $3$) and $p_1+4$ is not a prime
We get $p_2=p_1+2,p_3=p_1+6,p_4=p_1+8$.
Consider the remainders of $p_1,p_1+2,p_1+6,p_1+8$ when divided by $5$
If $p_1\equiv 2$ (mod $5$), then $p_1+8$ is divisible by $5$ and hence is not a prime
If $p_1\equiv 3$ (mod $5$), then $p_1+2$ is divisibe by $5$
If $p_1\equiv 4$ (mod $5$), then $p_1+6$ is divisible by $5$
Hence the only possibility is $p_1\equiv 1$ (mod $5$).
Thus we see that $p_1\equiv 1$ (mod $2$),$p_1\equiv 2$ (mod $3$)$and{p}_1\equiv 1$ (mod $5$). We conclude that $p_1\equiv 11$ (mod $30$).
Similarly $q_1\equiv 11$ (mod $30$). It follows that $30$ divides $p_1-q_1$.