TITLE: Are $ut + 1$ and $ut + t + 1$ both prime for some t for any $u$?
QUESTION [7 upvotes]: 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 $u = 10^6$

REPLY [2 votes]: 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 $a=u$, $b=u+1$.