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Are there only finite quadruples of non-negative integers $(a,b,c,d)$ that satisfy the following equation:

$$2^a + 3^b = 2^c + 3^d \quad ?$$

with $a \neq c$.

I found these:

  • $5 = 2^2 + 3^0 = 2^1 + 3^1$
  • $11 = 2^3 + 3^1 = 2^1 + 3^2$
  • $17 = 2^4 + 3^0 = 2^3 + 3^2$
  • $35 = 2^5 + 3^1 = 2^3 + 3^3$
  • $259 = 2^8 + 3^1 = 2^4 + 3^5$
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    $\begingroup$ Is there anything stopping you from taking $c=a$ and $d=b$? $\endgroup$ – Nick Peterson Mar 27 '14 at 15:14
  • $\begingroup$ I think you mean that $a \neq c$ and $b \neq d$? $\endgroup$ – Indrayudh Roy Mar 27 '14 at 15:26
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    $\begingroup$ OEIS says there are no more below $10^{4000}$ It seems very likely you have them all. $\endgroup$ – Ross Millikan Mar 27 '14 at 15:27
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    $\begingroup$ You probably just want to say $a \neq c$. Your first solution violates $a+b \neq c+d$ $\endgroup$ – Ross Millikan Mar 27 '14 at 15:29
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    $\begingroup$ @Lucian, I do not think that is right, since some prime factors of $2^{x}-1$ may be present in $3^{y}-1$. The correct conclusion would be $2^{c}||(3^{y}-1)$ and $3^{b}||(2^{x}-1)$. $\endgroup$ – Indrayudh Roy Mar 27 '14 at 16:05

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