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Most certainly such $k$ does not exist. I am not sure whether there is a chance to prove this unconditionally, as even the situation with Sophie Germain primes (which are primes $p$ such that $2p+1$ is also prime, related to the case $k=2$ of your question) is unclear.

Conditionally, the non-existence of $k$ is an immediate consequence of the Dickson's conjecture stating that for a finite set of linear polynomials $f_1(n)=a_1n+b_1,\dotsc,f_m(n)=a_mn + b_m$ with integer coefficients and $a_1,\dotsc,a_m\ge 1$, there are infinitely many positive integers $n$ for which the values of these polynomials are all prime, unless the product $f_1(n)\dotsb f_m(n)$ has a fixed prime factor. For your problem, take $m=2$, $f_1(n)=n$, and $f_2(n)=kn+(k-1)$.

Most certainly such $k$ do not exist. I am not sure whether there is a chance to prove this unconditionally, as even the situation with Sophie Germain primes (which are primes $p$ such that $2p+1$ is also prime, related to the case $k=2$ of your question) is unclear.

Conditionally, the non-existence of $k$ is an immediate consequence of Dickson's conjecture stating that for a finite set of linear polynomials $f_1(n)=a_1n+b_1,\dotsc,f_m(n)=a_mn + b_m$ with integer coefficients and $a_1,\dotsc,a_m\ge 1$, there are infinitely many positive integers $n$ for which the values of these polynomials are all prime, unless the product $f_1(n)\dotsb f_m(n)$ has a fixed prime factor. For your problem, take $m=2$, $f_1(n)=n$, and $f_2(n)=kn+(k-1)$.

A nice reformulation of your question is as follows: is it true that every integer $k>0$ is representable as $$ k = \frac{p+1}{q+1} $$ with $p$ and $q$ both prime? Dickson's conjecture implies that, in fact, any rational $k>0$ has infinitely many such representations: write $k=u/v$ and consider the polynomials $f_1(n)=un-1$ and $f_2(n)=vn-1$.

Most certainly such $k$ does not exist. I am not sure whether there is a chance to prove this unconditionally, as even the situation with Sophie Germain primes (which are primes $p$ such that $2p+1$ is also prime, related to the case $k=2$ of your question) is unclear.

Conditionally, the non-existence of $k$ is an immediate consequence of the Dickson's conjecture stating that for a finite set of linear polynomials $f_1(n)=a_1n+b_1,\dotsc,f_m(n)=a_mn + b_m$ with integer coefficients and $a_1,\dotsc,a_m\ge 1$, there are infinitely many positive integers $n$ for which the values of these forms are all prime, unless the product $f_1(n)\dotsb f_m(n)$ has a fixed prime factor. For your problem, take $m=2$, $f_1(n)=n$, and $f_2(n)=kn+(k-1)$.

Most certainly such $k$ does not exist. I am not sure whether there is a chance to prove this unconditionally, as even the situation with Sophie Germain primes (which are primes $p$ such that $2p+1$ is also prime, related to the case $k=2$ of your question) is unclear.

Conditionally, the non-existence of $k$ is an immediate consequence of the Dickson's conjecture stating that for a finite set of linear polynomials $f_1(n)=a_1n+b_1,\dotsc,f_m(n)=a_mn + b_m$ with integer coefficients and $a_1,\dotsc,a_m\ge 1$, there are infinitely many positive integers $n$ for which the values of these polynomials are all prime, unless the product $f_1(n)\dotsb f_m(n)$ has a fixed prime factor. For your problem, take $m=2$, $f_1(n)=n$, and $f_2(n)=kn+(k-1)$.

Most certainly such $k$ does not exist, but there is no chance to prove this unconditionally, as we cannot even prove that there are infinitely many Sophie Germain primes (which are primes $p$ such that $2p+1$ is also prime, corresponding to the case $k=2$ of your question).

Conditionally, the non-existence of $k$ is an immediate consequence of the Dickson's conjecture stating that for a finite set of linear polynomials $f_1(n)=a_1n+b_1,\dotsc,f_m(n)=a_mn + b_m$ with integer coefficients and $a_1,\dotsc,a_m\ge 1$, there are infinitely many positive integers $n$ for which the values of these forms are all prime, unless the product $f_1(n)\dotsb f_m(n)$ has a fixed prime factor. For your problem, take $m=2$, $f_1(n)=n$, and $f_2(n)=kn+(k-1)$.

Most certainly such $k$ does not exist. I am not sure whether there is a chance to prove this unconditionally, as even the situation with Sophie Germain primes (which are primes $p$ such that $2p+1$ is also prime, related to the case $k=2$ of your question) is unclear.

Conditionally, the non-existence of $k$ is an immediate consequence of the Dickson's conjecture stating that for a finite set of linear polynomials $f_1(n)=a_1n+b_1,\dotsc,f_m(n)=a_mn + b_m$ with integer coefficients and $a_1,\dotsc,a_m\ge 1$, there are infinitely many positive integers $n$ for which the values of these forms are all prime, unless the product $f_1(n)\dotsb f_m(n)$ has a fixed prime factor. For your problem, take $m=2$, $f_1(n)=n$, and $f_2(n)=kn+(k-1)$.