7
$\begingroup$

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 $u = 10^6$

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ I extended my search upto $u = 10^7$. $\endgroup$ – Peter Jan 30 '14 at 21:59
2
$\begingroup$

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 $a=u$, $b=u+1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.