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In another question I was asking if there are any different $x,y>2$ primes such that $xy+5=a(x+y)$.

Where $a=2^r-1$, and $r>2$.

I was thinking if it is able to find a Pell equation or a similar pattern of $xy+5=a(x+y)$ to say what are and how many integer solutions are there (in particular prime solutions).

Thanks.

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$xy-5=a(x+y)$ can be rewritten as $$(x-a)(y-a)=a^2+5$$ so for any fixed $a$ solving it just amounts to finding all the ways to factor $a^2+5$. So how many solutions depends on the prime factorization of $a^2+5$. I don't think there will be any formula for how many of those solutions have $x$ and $y$ prime.

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    I'm really sorry but I made a change in the question, it is not really different but it should be assumed that the -5 that I wrote is +5.2011-05-04
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    So $xy+5=a(x+y)$ can be rewritten as $$(x-a)(y-a)=a^2-5,$$ right?2011-05-05
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    true (need to spend letters)2011-05-05
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    So then what more could one do by way of an answer to your question?2011-05-06
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    If there aren't such prime pairs solve this equation, it seems very interesting. The question here if it may be unique. If when we change -5 into some other -prime will give no prime solutions. In this form of the equation we can say that for each such $a$ there are finitely (but more than 0) solutions in integers and infinitely in the union set.2011-05-06
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    So what makes you think there aren't any primes that solve the equation? $x=7$, $y=37$, $a=6$; $x=11$, $y=47$, $a=9$; $x=13$, $y=151$, $a=13$; $x=19$, $y=337$, $a=18$; $x=29$, $y=47$, $a=18$; and so on, and so on, and so forth.2011-05-06
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    Even if you insist on $a=2^r-1$ I believe there are solutions. Try $x=17179929661$, $y=4880269588100161$, $a=17179869183$.2011-05-06
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    I agree, thank you very much. It's fast seen that the number of such pairs up to n is less than ln(n). Do you think there are infinitiely many or finitely many?2011-05-07
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    My intuition says infinitely many, but my calculations show they are very rare, so I really don't know what to think.2011-05-07
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    These does not need to be common to be infinitely many..2011-05-08