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We let $p\neq q$ be odd prime numbers and $r$ be integer $>2$. Are there such $p,q$ satisfying $pq=(2^r-1)(p+q)-5$?

This is clear from here that,


and $p(q-2^r+1)=(2^r-1)q-5$.


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The question is clear, but what do the last two equations represent? Also, please use LaTeX formatting - it's easier on the eyes :) – Hans Parshall Apr 28 '11 at 16:35
I don't know how to use latex. Those equations are equivalent to the first one (other forms to present the equation). – tomerg Apr 28 '11 at 16:38
I fixed my question above (questions => equations). You can use LaTeX by just putting equations in dollar signs (i.e. $y = x^2$ gives $y = x^2$). – Hans Parshall Apr 28 '11 at 16:41
Didn't write that p does not equal q – tomerg Apr 28 '11 at 16:43
Actually you mean $q (p - 2^r + 1) = (2^r - 1) p - 5$ and $p (q - 2^r + 1) = (2^r - 1) q - 5$. – Robert Israel Apr 28 '11 at 16:44
up vote 3 down vote accepted

Over at tomerg's other, closely-related question The form $xy+5=a(x+y)$ and its solutions with $x,y$ prime I found $p=17179929661$, $q=4880269588100161$, $r=34$ is a solution.

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Blockquote EDIT: After writing, the original problem was edited to read $p(q - 2^r - 1)...$ and so this is all beautiful speculation on a different problem. Had I checked his derivation, I might have noticed. But I didn't.

Firstly, I wonder - is there a significance to this question? I've no idea. But I put together some scratch work real quick and found a solution - so there's that.

I note firstly that at most one can be even (considering the equation mod 2 gives this). So I thought, what if p were 2?

Then we have $$ 2(q + 2^r - 1) = (2^r - 1)1 - 5$$ $$ 2^{r+1} + 3 = (2^r - 3)q$$ $$ \frac{2^{r+1} + 3}{2^r - 3} = q $$

And if $ r = 2$, we get that $ q = 11$. Although I don't have it yet, I suspect this is the only solution for q for $p = 2$. So there is at least one answer.

Without having a better intuition for the problem (as I don't really know if there's anything special about this equation, if it means something in particular, etc.), I don't see a better method of attack than this sort of play.

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@mixedmath: the question requires $p$ and $q$ be odd. – Zev Chonoles Apr 28 '11 at 16:59
Okay, I responded to the original, unedited post. This is too bad. It's still fun, though not applicable to this question. But upon checking, even then they were to be odd. I really just outdid myself on this one. – mixedmath Apr 28 '11 at 17:02
$r=1$ also works, if you allow $q=-7$. – lhf Apr 28 '11 at 17:22
Still r>1 and by prime numbers I mean these are>1. – tomerg Apr 28 '11 at 17:24
Also, $y=(2x+3)/(x-3)$ is an integer iff $x-3$ divides $9$ iff $x=-6,0,2,4,6,12$ iff $y=1,-1,-7,11,5,3$. – lhf Apr 28 '11 at 17:34

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