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Let $p_n$ be the n-th prime number starting from $p_1=2$. Find all positive integers (a,b) so that $a-b \ge 2$ and $p_a-p_b\mid 2(a-b)$.

So far I have got that for $a \ge 6 \Rightarrow p_a>2a$ but I cant use it bcs of the subtraction.

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  • $\begingroup$ @barto $a-b\geq 2$, so no twin primes. $\endgroup$ – Wojowu Jan 6 '15 at 15:27
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Let $a>b>1$. Note that $p_{b+1}-p_b\geq 2,p_{b+2}-p_{b+1}\geq 2,...,p_a-p_{a-1}\geq 2$ with $a-b$ such differences. So adding them all up we get $p_a-p_b\geq 2(a-b)$. If $p_a-p_b\mid 2(a-b)$ then $p_a-p_b\leq 2(a-b)$, so in fact $p_a-p_b=2(a-b)$, so in fact we must have $p_{k+1}-p_k=2$ for all $a\leq k<b$ (if any inequality was strict, we would get $p_a-p_b>2(a-b)$). So we have that $p_{b+2}-p_{b+1}=2$ and $p_{b+1}-p_b=2$. By looking mod 3, we see that exactly one of these is divisible by 3, and it must be $p_b$. So the only possibility, because of $a-b\geq 2$, is $a=4,b=2$.

If $b=1$, then we want $p_a-2\mid 2(a-1)$. You have shown that for $a\geq 6$ we have $p_a>2a$, so $p_a-2> 2(a-1)$ and the divisibility cannot hold. Case $a<6$ check by hand.

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Two observations to start:

  • $p_{j+1}\ge p_{j}+2$ for $j > 1$ (both are odd).
  • $p_{j+2}\ge p_{j}+6$ for $j > 2$ (as $p_{j}$, $p_{j}+2$, $p_{j}+4$, can not all be prime as one of them is divisible by $3$).

Thus, for $a \ge b +2 \ge 5$, one has $p_a - p_b \ge 2 + 2(a-b)$, and we are done.

It remains to consider the case $b=2$, and $b=1$. Using the technique given above and consideration of some special case, one will be able to finish the problem.
(If there is demand, I will complete the argument.)

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