Prime Differences

First note that $b<a$ . If $b=1$ then $p_{a}-2\mid 2a-2$ but $p_{a}-2$ is odd, since $a-b\geq 2$ implies that $a>1$ and therefore $p_{a}$ must be odd. Hence, $p_{a}-2\mid a-1$ but this implies, since both quantities are positive, that $p_{a}\leq a+1$ which is impossible (this is easily seen by noticing $p_{3}=5$ and then seeing that $p_{k}\geq k+2$ for every positive integer $k\geq 3$ and $a\geq 3$ ).

Therefore, there are no solutions with $b=1$ so we can assume $a, b>1$ , which implies that both $p_{b}, p_{a}$ are odd. Note that there cannot be consecutive primes of the form $p, p+2, p+4, p+6$ since that implies, by the pigeonhole principle, that one of them is divisible by $3$ , and hence one of them must be $3$ . The only possibility for this is that $p=3$ , but that leads to $p+6=9$ , which is not prime, and hence this situation is impossible. Since we have that for odd consecutive primes, $p_{a}\geq p_{a-1}+2$ then, all the equalities in $p_{a}\geq p_{a-1}+2\geq p_{a-2}+4\geq p_{a-3}+6$ may not occur simultaneously, or else we have $4$ consecutive odd primes of the form $p, p+2, p+4, p+6$ , which we proved to be impossible. Hence $p_{a}>p_{a-3}+6$ and therefore, if $a-b>2$ then $p_{a}>p_{b}+2\left(a-b\right).$ Since we want that $p_{a}-p_{b}$ divides $2\left(a-b\right)$ , we need that $p_{a}\leq p_{b}+2\left(a-b\right)$ , so we conclude any solution must satisfy $a-b\leq 2$ , and hence $\boxed{a-b=2}.$ Now, this implies that we have consecutive primes of the form $p, p+2, p+4$ , implying one of them is divisible by $3$ , and hence one of them is $3$ . The only possibility for this is $3, 5, 7$ which works, and note that $7-3\mid 2\left(4-2\right)$ so $p_{2}=3$ and $p_{4}=7$ satisfy our condition.

Hence, the only solution is the ordered pair $\left(4, 2\right)$ , and thus there is $\boxed{1}$ solution. $\square$

Great problem, Hector! :)

It will be proven that $a-b= 2$ is the only value possible such that $p_{a} - p_{b} | 2(a-b)$ . Let $x = p_{a}-p_{b}$ .

(i) Consider $a-b \geq 4$ . Obviously $p_{a}- p_{b} \geq 9$ . Equality holds if and only if $(a, b)= (5, 1)$ with $2( a -b) = 8 < 9$ . Thus $a-b < 4$ .

(ii) Consider $a-b = 3$ Obviously $p_{a} -p_{b} \geq 5$ . Equality holds if and only if $(a, b) = (4, 1)$ with $2(a -b) = 6 > 5$ . Thus the only possible value of $x$ such that $x| 2(a-b)$ is $x = 6$ .

$x=6$ if and only if $p_{a}-p_{a-1}= p_{a-1}-p_{a-2}= p_{a-2}- p_{a-3} =2$ . For $k \geq 2, (2, p_{k}) =1 <=> p_{k} \equiv 1, 3, 5, 7, 9 (mod 10)$ with $p_{k} \equiv 5 (mod 10)$ if and only if $p_{k} = 5$ . If $p_{b} =5, p_{b+3=a} =13 <=> x \neq 6$ . Thus $p_{b} \neq 5$ .

For $p_{b} \geq 5, p_{b} \equiv 7 (mod 10)$ . If $p_{b} \not\equiv 7(mod 10)$ , $x > 6$ since $p_{a} \not\equiv 3 (mod 10)$ . Consider the following set $T= {{10k+ 7, 10k+ 9, 11k+ 1, 11k+ 3}}$ . By pigeon hole principle, there is $m \in T$ such that $3|m$ . Thus there is no $(a,b)$ such that $x= 6$ . Thus $x>6 \forall (a, b), a-b= 3$ .

It implies that $a-b = 2$ . If $a-b= 2$ , $x \geq 3$ . Thus $x|2(a-b)= 4$ if and only if $x= 4$ . It is easy to check that $x = 4$ if and only if $(a,b) = (4, 2)$ . Thus the only solution satisfying the above divisibility is $(a, b) = (4, 2)$ .

Because we have the constraint $a - b \geq 2$ this means we don't count twin prime pairs. I will prove that for non-twin prime pairs (except for a special exception I will get to), we have the inequality $p_a - p_b > 2(a-b)$ .

First, notice that we can completely ignore the prime $2$ because if we have an odd prime minus an even prime then that is odd and won't divide the even number $2(a-b)$ . Now consider the following numbers $p_a, p_a + 1, p_a + 2, ..., p_b$ . In this set it is clear that at most $p_a, p_a + 2, p_a + 4,..., p_b$ can be primes. This means that the distance between two primes is $p_a - p_b \geq 2(a-b)$ . This also means that equality occurs exactly when we have twin primes $p_a,p_a + 2$ or we have a twin prime triple like $p_a,p_a + 2, p_a + 4$ and we choose $p_b = p_{a+2} = p_a + 4$ . Or twin prime quadruple, etc. So let's find these twin prime triples and quadruples etc.

It is clear that if there exists a twin prime quadruple (and up) there must exist a twin prime triple so let's scout for those. We are looking for a prime $p$ such that $p+2$ and $p+4$ is prime. Let's assume that $p>3$ . Then obviously $p \equiv 1 \mod 3$ or $p \equiv 2 \mod 3$ . In the first case, we would have that $p+2 \equiv 0 \mod 3$ so $p+2$ couldn't be prime. In the second case we would have $p + 4 \equiv 0 \mod 3$ which means $p+4$ could not be prime. And if we now consider the case $p=3$ we find the unique triple $3,5,7$ . This implies there are no other triples, and because this is not a quadruple this means quadruples and quintuples (and up) don't exist.

So applying our previous deduction we know that we don't count twin primes like $3,5$ or $5,7$ but we can count $3,7$ and indeed $p_4 - p_2 = 7 - 3 = 4 = 2(4 - 2)$ . And this must be the only solution because no more twin triples exist.