Find all positive integer $a,b$ such that $a^3-b^7=a-b$ 
Find all positive integer $a,b$ such that $a^3-b^7=a-b$.

I found a pair $(a,b)=(13,3)$.I failed to find anything helpful. Please help me.
Thank you in advance!
 A: Apart from the trivial solution $(a,b) = (1,1)$, there is likely nothing more to find.  I cannot offer you a proof, but there are computational results, conjectures and a proof of a closely-related result.
Rewriting the equation as $a^3 - a = b^7 - b$, it can be seen as saying that some integer $n$ can be written as $m^k - m$ in two different ways with positive integers $m,k$ (in particular, with $k=3$ and $k=7$).  Your example corresponds to $n = 2184 = 13^3 - 13 = 3^7 - 3$.
Without requiring any particular values of $k$, only $8$ positive values of $n$ with this property are known; they are listed in OEIS sequence A057896.  There you will also find a claim by Giovanni Resta that the next such $n$, if one exists, must be greater than $10^{24}$.
Dana Mackenzie has called integers $n$ which can be written as $m^k - m$ in two different ways doubly absurd numbers.  In the paper $2184$: An Absurd (and Adsurd) Tale, Mackenzie notes that Mike Bennett conjectured that there were no further doubly absurd numbers in $2001$.
Mackenzie also proves (Theorem $1$ in the cited paper) that there are no further solutions to $a^3 - a = b^k - b$ with $k$ odd and greater than $3$ and with $a$ being prime apart from the one you have found.  That does not settle your question, but it does show that solutions have been sought before and not found, and if there are any to find, they will require that $a$ is greater than $10^8$ and composite.
