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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).


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up vote 2 down vote accepted

$xy-5=a(x+y)$ 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. – tomerg May 4 '11 at 16:50
So $xy+5=a(x+y)$ can be rewritten as $$(x-a)(y-a)=a^2-5,$$ right? – Gerry Myerson May 5 '11 at 1:21
true (need to spend letters) – tomerg May 5 '11 at 16:15
So then what more could one do by way of an answer to your question? – Gerry Myerson May 6 '11 at 0:48
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. – tomerg May 6 '11 at 6:41

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