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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!

Sufaid Saleel
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  • General case, $a,b\in\mathbb{Z}^+$, hard to do. However, a solution can be obtained via some work, when we restrict $a$ and $b$ to be primes. Try that case, and let me know if you're curious. I can supply a proof. – TBTD Mar 23 '19 at 20:00
  • @Aaron yeah!Of course!Please post your solution – Sufaid Saleel Mar 24 '19 at 01:05

1 Answers1

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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.

FredH
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  • FWIW, in this case, where one exponent is $3$, $a^3-a=(a-1)\cdot a \cdot (a+1)$. It is the product of three consecutive integers, meaning with the exception of powers of $2$, the three factors are coprime, and one of them must be divisible by $3$. That puts a lot of constraints on $b$. – Keith Backman Mar 23 '19 at 15:17
  • @KeithBackman Actually, since $b^7 - b = (b^3 - b)(b^4 + b^2 + 1)$, the factors of $b^7 - b$ have much in common with those of $a^3 - a$. One useful fact is that $7$ always divides $b^7 - b$, so $a \equiv -1,0$ or $1\pmod 7$. – FredH Mar 24 '19 at 00:01