Conjecture :

For any natural number $u$, there is a natural number $t$ such that $ut + 1$ and $ut + t + 1$ are both prime. So we get a solution of the equation

$$au - b(u+1) = -1$$

with prime numbers $a$ and $b$ by setting $a := ut + t + 1 , b := ut + 1$ :

$$(ut+t+1)u - (ut+1)(u+1) = u^2t+ut+u-u^2t-ut-u-1 = -1$$

Motivation :

If the conjecture is true, then for any natural number $n$, there is a pair $(a,a+1)$ of consecutive squarefree numbers with exactly $n$ distinct prime factors.

I checked the conjecture with PARI and for $u\le10^6$, it is true. The largest number t necessary to produce the prime pair is $3420$ for the number $829123$ upto $u = 10^6$

  • $\begingroup$ The numbers are coprime and odd (if $t$ is even), so they do have a "tendency" to be prime, which means that for "small" $u$ finding such $t$ shold be easy. Thus the successful verification for $u<10^6$ is promising but may mean little in the end. $\endgroup$ – Hagen von Eitzen Jan 30 '14 at 12:12
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    $\begingroup$ I extended my search upto $u = 10^7$. $\endgroup$ – Peter Jan 30 '14 at 21:59

This is still an open problem, but it is a special case of the Hardy–Littlewood prime $k$-tuples conjecture, in this case with $k=2$. Indeed, for any positive integers $a\ne b$, we expect there to be infinitely many integers $t$ for which $at+1$ and $bt+1$ are both prime; your conjecture is the case $a=u$, $b=u+1$.


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