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I found 8 of them and believe there is no more:

$$2+3^2=3+2^3$$ $$2+6^2=6+2^5$$ $$6+15^2=15+6^3$$ $$3+16^2=16+3^5$$ $$3+13^3=13+3^7$$ $$2+91^2=91+2^{13}$$ $$5+280^2=280+5^7$$ $$30+4930^2=4930+30^5$$

I call the solution a principal pipe of order 2. Any idea to attack this equation?

If $n=x=2$, it becomes $2+y^2=y+2^m$. It is Ramanujan and Nagell Equation, and there are only 3 solutions. If $n=2$ and $x \geq 3$, then we can use the similar way to prove that there is no other solution for $x < 10000$ with help of computer.

If $n=3$, and we assume $x$ and $y$ are prime, and $6\mid (m-1)$, then the solution is $3+13^3=13+3^7 = 2200$.

I believe there is no solution for $n>3$. I am particularly interested in solutions that both $x$ and $y$ are odd primes.

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    $\begingroup$ What about $n=1$ or $m=1$? Why do you believe there are no other solutions if $n>1$ and $m>1$? Just because you cannot find them? $\endgroup$
    – Claudio Gorodski
    Commented Feb 19, 2014 at 19:45
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    $\begingroup$ Such solutions are likely to involve consecutive powers. You might try looking at results of Pillai. Ribenboim's book on Catalan's equation has more, but I do not think it has this specific form. $\endgroup$
    – The Masked Avenger
    Commented Feb 19, 2014 at 19:59
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    $\begingroup$ Your equation is equivalent to $x(x^{m-1}-1)=y(y^{n-1}-1)$. If $p$ is a prime that divides $x-1$, then the $p$-adic valuation of $x(x^{m-1}-1)$ is $v_p(x-1)+v_p(m-1)$. If $p$ does not divide $x-1$ or $x$, then there are similar but more complicated statements that also involve the order of $x$ modulo $p$. You could probably get quite far with this kind of analysis. $\endgroup$
    – Neil Strickland
    Commented Feb 19, 2014 at 21:52
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    $\begingroup$ These are all the solutions with $2 \le n < m \le 200$ and $2 \ge x,y \le 20000$. $\endgroup$
    – Robert Israel
    Commented Feb 20, 2014 at 3:23
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    $\begingroup$ What's a pipe? What's a principal pipe? What's a principal pipe of order 2? $\endgroup$
    – Gerry Myerson
    Commented Feb 20, 2014 at 4:26

1 Answer 1

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This problem was considered in a paper of Mignotte and Petho {Publ. Math. Debrecen 1999) and subsequently in one of mine [Canad. Math. J 2001], where there is a conjecture that the equation $a^x-b^y=c$ has, for fixed positive integers $a, b$ and $c$, with $a, b \geq 2$, at most one solution in positive integer exponents $x$ and $y$, with precisely $11$ exceptions. This is provable if $c$ is suitably large or suitably small, but, as far as I know, is still open in general. This problem arose from four classical papers of Pillai.

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  • $\begingroup$ I thought it is a hard question and I am happy to get confirmed from you. This equation may not be completely solved in algebraic number theory since this equation has infinitely many solutions if $x$ and $y$ are fractional. $\endgroup$
    – Willie Wu
    Commented Feb 20, 2014 at 6:08
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    $\begingroup$ For example $$\frac{1}{2} + \frac{1}{2^b} + \frac{1}{4^b} = \frac{1}{4} + \left(\frac{1}{2} + \frac{1}{2^b}\right)^2 = \left(\frac{1}{2} + \frac{1}{2^b}\right) + \left(\frac{1}{4}\right)^b \\ \frac{1}{2} - \frac{1}{2^b} + \frac{1}{4^b} = \frac{1}{4} + \left(\frac{1}{2} - \frac{1}{2^b}\right)^2 = \left(\frac{1}{2} - \frac{1}{2^b}\right) + \left(\frac{1}{4}\right)^b$$ for any $b\geq 3$. $\endgroup$
    – Willie Wu
    Commented Feb 20, 2014 at 6:10
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    $\begingroup$ I can say with some confidence that the fact that I couldn't answer it does not make it a hard question! $\endgroup$
    – Mike Bennett
    Commented Feb 21, 2014 at 18:24

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