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$ a \cdot (a-1) \ \mid \ n^{a}-n \ \ \ \forall \ n \in \mathbb{N} $

Find all $a$.

$$$$

Obviously $a$ has to be prime.

I found $a=2,3,7$ and $43$ as solutions.

Note that these 4 solutions are the first elements of Sylvester's Sequence. $$$$

Are there more solutions?

Bill Dubuque
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Martin Hopf
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    Apply Korselt's criterion for Carmichael numbers. – Bill Dubuque Jun 09 '19 at 18:30
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    Korselt's criterion implies $a-1$ to be squarefree. For this the next squarefree canidate for $a-1$ respectively $a$ in the sequence is: $2 \cdot 3 \cdot 7 \cdot 43 + 1 = 1807 = 13 \cdot 139$. It is not a prime! I srongly tend to say that this breaks the chain. So the above 4 solutions are the only ones. – Martin Hopf Jun 10 '19 at 16:27
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    I checked the next two numbers (5th and 6th) in the sequence and they both don't work. The sixth one is prime, but that doesn't seem to matter. I also suspect these are the only solutions, but I haven't been able to prove it yet. – QC_QAOA Jun 10 '19 at 16:54
  • $n^a-n=k×a×(a-1)$ take a=2r,2r+1 where r=0,1 then solve a quadratic equation. Or a=3r,3r+1,3r+2 where r=0,1,2, then solve elliptic curve – Guruprasad Dec 07 '24 at 19:26
  • The implication of this means if $f(a) \in \mathbb{Z}$

    $$n^a-n = af(a)$$ $a\forall ,\text{primes}$

    $$n^a-n = a(a-1)f(a)$$ $a\in (2,3,7,43)$

    $$n^a-n = a (a-1)(a-2)f(a)$$ $a=3$

    $$n^a-n = a(a-1)(a-2)(a-3)f(a)$$ no solution to this and higher

     – Aderinsola Joshua Dec 12 '24 at 12:52

1 Answers1

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For $a=1$ the problem reduces to $0\mid0$, which is true, so this is another (trivial) solution. Now let $a>1$. By Korselt's criterion, as linked in the comments, $a-1$ has to be squarefree, and for every prime $p$ dividing $a-1$, we must have $p-1\mid a-1$. We will show that the only possible values of $a-1$ are $1,2,6,42,1806$. One easily checks that $1807=13\cdot139$ doesn't satisfy the original problem, so the full list of solutions is $$\boxed{a\in\{1,2,3,7,43\}.}$$

Let $A$ be the set of all squarefree $n$ such that $p\mid n$ implies $p-1\mid n$ for all primes $p$. Call a prime $p$ an $A$-prime if $p\mid n$ for some $n\in A$. From the examples we found, we know that $2,3,7,43$ are $A$-primes. Assume there is another $A$-prime, let $p$ be the smallest one, and say $p\mid n$ for $n\in A$. Then $p-1\mid n$, so every prime factor of $p-1$ is an $A$-prime, and by minimality of $p$ must be in $\{2,3,7,43\}$. Therefore $p=1+\prod_{q\in S}q$ for some $S\subseteq\{2,3,7,43\}$. (Remember that $n$, and therefore $p-1$, is squarefree.) But a direct computation shows that no choice of $S$ yields a new prime, contradiction.

Hence $2,3,7,43$ are the only $A$-primes. Now if $43\mid n\in A$, then also $42\mid n$, so $n=2\cdot3\cdot7\cdot 43$, and similarly for the other cases. This completes the proof.

anankElpis
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