Prime numbers between two multiples of numbers

I am wondering about the following question.

Do there exist infinitely many prime numbers $p$ such that there exist integers $m,n$ with $5(m+1) \geq 7n$ and $$5m < p < 7n?$$

If not, what other numbers would we have to replace $5$ and $7$ by for there to be? If so, what is the minimal set of numbers we would have to replace $5$ and $7$ by? This leads onto my second question below.

Find numbers $a,b$ with $|a-b|$ minimal such that there exist infinitely many primes $p$ with the property that there exist integers $m,n$ with $a(m+1) \geq bn$ and $$am < p < bn.$$

• Where is $n$ involved? Jun 18 '16 at 14:04
• @Bernard Thanks, that was a typo. Jun 18 '16 at 14:07
• The standard Dirichlet result gives infinitely many primes $p$ of the form $35n-1$. For such primes we have $5(7n-1)<p<7\cdot5n=5\cdot7n$ Jun 18 '16 at 14:07
• @almagest: that should be an answer. I was thinking down that line. Jun 18 '16 at 14:08
• @RossMillikan then your welcome to put it up. I am too busy on something else to think about the second part of the question. Jun 18 '16 at 14:09

Following almagest: The standard Dirichlet result gives infinitely many primes pp of the form $35n−1$. For such primes we have $5(7n−1)\lt p\lt 7⋅5n$
The same argument works with $a=6, b=7,$ primes of the form $42n-1$
• Any arithmetic progression $a+kd$ where $a,d$ are coprime and $k \in \Bbb N$ contains an infinite number of primes. Here in Wikipedia @SubhadeepDey because I stole almagest's comment for the answer by his request Jun 18 '16 at 14:16