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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,

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

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

Thanks.

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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 :)2011-04-28
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    I don't know how to use latex. Those equations are equivalent to the first one (other forms to present the equation).2011-04-28
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    I fixed my question above (questions => equations). You can use LaTeX by just putting equations in dollar signs (i.e. `$y = x^2$` gives $y = x^2$).2011-04-28
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    Didn't write that p does not equal q2011-04-28
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    Actually you mean $q (p - 2^r + 1) = (2^r - 1) p - 5$ and $p (q - 2^r + 1) = (2^r - 1) q - 5$.2011-04-28
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    I'm so sorry for my bad mistakes. This is what I was about to write.2011-04-28
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    I wrote a quick script and found that $p = q = 5$ is the only solution where $p$ and $q$ are both less than 5000.2011-04-28
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    Great, but as D.Shnaks said: "10^50 is a long way from infinity". :)2011-04-28
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    I don't think that it helps, but for any solutions bot $p,q$ have to be 1 mod 4.2011-04-28
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    It is true but does not seem to be helpful..2011-04-28
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    If there is a solution, one of the primes must be 3.2011-04-28
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    How can you show such thing? This also contradicts that both primes are 1 mod 4.2011-04-28
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    There are only a couple of integer solutions less than $30000$ for each $r$ with $3\le r \le 9$ (but no primes), and no solutions for $10\le r \le 30$. Not that it proves anything.2011-04-28
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    @tomerg: Check both sides in respect to modulo 3.2011-04-28
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    I still don't see it.2011-04-28
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    @user: We have three very inconsistent comments so far. User says that both p and q must be $1 \mod{4}$, tpv says one of the primes must be 3, and MartianInvader notes that 5,5 is a solution if we allow p to equal q. Why do both have to be 1 mod 4?2011-04-28
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    Because from the equation we have: $(p+1)(q+1)=2^r(p+q)-4$ which gives p,q are not -1 mod 4. This leads that p,q are 1 mod 4.2011-04-28
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    Oh, my bad :S, sorry. Forgot the -1 in $(2^r-1)$, so what I wrote is not correct at all.2011-04-28

2 Answers 2

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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.2011-04-28
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    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.2011-04-28
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    $r=1$ also works, if you allow $q=-7$.2011-04-28
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    Still r>1 and by prime numbers I mean these are>1.2011-04-28
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    Also, $y=(2x+3)/(x-3)$ is an integer iff $x-3$ divides $9$ iff $x=-6,0,2,4,6,12$ iff $y=1,-1,-7,11,5,3$.2011-04-28
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    @lhf: How do we know that $(x-3)|9$ iff y is an integer?2011-04-28
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    $(2x+3)/(x-3)=2+9/(x-3)$.2011-04-28
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    @lhf: Oh! How nice!2011-04-28