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If we consider the Leyland numbers of the form $n^6+6^n$, how to prove that there is no prime for odd $n>1$? I found some attempts on here and here you can find a general related question, while here you have a list of Leyland prime numbers.

The case $n^4+4^n$ is pretty easy to solve by completing the square (Sophie Germain's identity), cf. here and AoPS, so is it possible to use something similar?

Naively I tried to use the factorisation of $a^6+27b^6$ but I'm going nowhere with that.

Bill Dubuque
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user967210
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    Please edit to include your efforts. Note that this is clear, save for the case $7,|,n$. – lulu Apr 02 '25 at 20:02
  • @lulu thank you for pointing that out. I tried to use naively the factorisation of $a^6+27b^6$ but with no results. – user967210 Apr 02 '25 at 20:09
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    I think this is probably false. According to factordb, most such numbers have small factors, but some don't: e.g. $6^{259} + 259^6$ factors as P32 $\cdot$ P170. That doesn't look like an algebraic factorization either, so I suspect there's nothing stopping you from finding a prime if you look far enough. – Ravi Fernando Apr 02 '25 at 22:47
  • @RaviFernando well I linked a page with the actually known Leyland primes, they computed the results for large $n$ but found none – user967210 Apr 02 '25 at 23:42
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    @user967210 Possibly relevant forum thread (from 2014, on the Leyland factoring forum): https://www.mersenneforum.org/node/14039. It looks like the user henryzz reached the same conclusion as me. – Ravi Fernando Apr 03 '25 at 00:48
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    Now posted to MO, https://mathoverflow.net/questions/490792/is-n66n-composite-for-all-n1 – Gerry Myerson Apr 10 '25 at 00:08

1 Answers1

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Too long for a comment: We may consider the last digit of powers:

$A=6^n+n^6$

$n$ may end to following digit:

$n\equiv (1, 2, 3, 4, 5, 6, 7, 8, 9)\bmod 10$

Corresponding the last digits for $n^6$ are:

$n^6\equiv(1, 4, 9, 6, 5, 6, 9, 4, 1)\bmod 10$

Considering the fact that the last digit of any power of $6$ is $6$, the last digits of $A$ are:

$A\equiv [7, 10(\equiv 0), 15(\equiv 5), 12(\equiv 2), 11(\equiv 1), 12(\equiv 2), 15(\equiv 5), 10(\equiv 0), 7]\bmod 10 $

That is if the last digit of $n$ are numbers $2, 3, 4, 6, 7, 8 $ then the last digit of $A$ are $0, 5, 2, 2, 5, 0$ which means $A$ is composite. Now we have to concentrate on when the last digit of $A$ are $7$ and $1$ .

1): Last digit $7$ represent when the last digit of $n$ is $1$ or $9$.

2): Last digit $1$ represent when the last digit of $n$ is $5$

Numerical experiment shows:

$n=11\rightarrow A=364568617= 52081231\times 7$

$n=9\rightarrow A=10607137=1515591\times 7$

$n=5\rightarrow A=23401=3343\times 7$

For $n=5$ we may write:

$(7-1)^5+(7-2)^6\equiv -1\bmod 7+2^6\bmod 7 \equiv 63 \bmod \equiv 0\bmod 7$

For $n=11$ we may write :

$(7-1)^{11}\equiv -1\bmod 7$

$11^6\equiv 1\bmod 7$

$\Rightarrow A\equiv 0\bmod 7$

For $n=9$ we may write :

$(7-1)^{9}\equiv -1\bmod 7$

$9^6\equiv 1\bmod 7$

$\Rightarrow A\equiv 0\bmod 7$

For $n=7$ we have:

$6^7+7^6=79517\times 5$

We may write:

$(5+1)^7\equiv 1\bmod 5$

$(7+2)^6\equiv 2^6=64\equiv -1\bmod 5$

$\Rightarrow A\equiv 0\bmod 5$

This holds for all odd powers of $7$:

$(7^{2k+1})^6=(7^6)^{2k+1}\equiv (-1)^{2k+1}\equiv -1\bmod 5$

$(5+1)^{7^{2k+1}}\equiv 1\bmod 5$

So: $A\equiv 0\bmod 5$

sirous
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    Please state explicitly in the answer what you claim to have proved. $\ \ $ – Bill Dubuque Apr 05 '25 at 17:53
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    Like Bill said, it's hard to decipher what you're claiming here. But it looks like you're saying that when $A = n^6 + 6^n$ is congruent to $1 \pmod{10}$ (i.e. $n \equiv 5 \pmod{10}$), it's always divisible by $7$. That is false; the first counterexample is $n=35$. (As lulu hinted in the comments, we have $7 \mid A$ if and only if $n$ is odd and not divisible by $7$.) – Ravi Fernando Apr 05 '25 at 21:07
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    Yes, as in covering congruences, it is trivial to deduce that $,A,$ has a factor in $,2,3,5,7,$ except when $!\bmod 210!:, \pm n,\equiv, 35,49,91,,$ i.e. $,n\equiv 35,49,91,119,161,175,$ [if my mental arithmetic is correct]. That's the easy part. But the least prime factors of $,A,$ get big really fast in these remaining cases, e.g. $,9007,11117,$ for $,n=35,49.\ \ $ – Bill Dubuque Apr 06 '25 at 19:14
  • The prior congruences arise by using little Fermat to infer $$\begin{align} 2\mid A&\iff n\equiv 0\ \pmod{2}\ 3\mid A&\iff n\equiv 0\ \pmod{3}\ 5\mid A&\iff n\equiv \pm 2!!!\pmod 5\ 7\mid A&\iff n\not\equiv 0\ \pmod7 \end{align}\qquad$$ – Bill Dubuque Apr 06 '25 at 23:10
  • If all fail then $,n\equiv 1\pmod{2}$ and $,n\equiv 0\pmod{7},,$ i.e. $,n\equiv 7\pmod{14},,$ and $,n\equiv \pm1=:j \pmod{!3},,$ $,n\equiv \pm1,0 =: k\pmod{5},,$ so by CRT we get $6$ roots $(j,k)\bmod 210$ $$\begin{align}(-1,0) \cong 35;&\ \ \ \ \ \ \ (1,\ 0)\cong -35\equiv 175,,\text{ its negative}\ (1,-1)\cong 49;&\ \ \ \ (-1,\ 1)\cong -49\equiv 161\ (1,1)\cong 91;&\ \ (-1,-1)\cong -91\equiv 119\end{align}\qquad$$ – Bill Dubuque Apr 06 '25 at 23:21
  • Not true. As I wrote above $,5\mid A!\iff! n\equiv \pm2\pmod{!5}.,$ Proof $ $ By Fermat: $!\bmod 5!:\ n^5\equiv n,$ so $,n^6\equiv n^2,$ so $,0\equiv n^6+6^n\equiv n^2+1\equiv n^2-4!\iff! n\equiv \pm2.\ \ $ – Bill Dubuque Apr 07 '25 at 08:18