Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

up vote 1 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.

share|improve this answer
    
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
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.