Let $L_n$ be the $n$th Lucas number and $p$ a prime number.
I noticed something with Lucas Number : it seems than $(L_{n+1}+2)\equiv0 \mod n$ is right only when $n$ is a prime $p$ and only if $p^2$ has digits in nondecreasing order.
For example :
$(L_{5+1}+2) = 18+2 = 20$ and $20\equiv0 \mod 5$ and $5^2 = 25$. $25$ has digits in nondecreasing order
$(L_{7+1}+2) = 47+2 = 49$ and $49\equiv0 \mod 7$ and $7^2 = 49$. $49$ has digits in nondecreasing order
$(L_{11+1}+2) = 322+2 = 324$ and $324\equiv5 \mod 11$ and $11^2 = 121$. $121$ has not digits in nondecreasing order
$(L_{13+1}+2) = 843+2 = 845$ and $845\equiv0 \mod 13$ and $13^2 = 169$. $169$ has digits in nondecreasing order
$(L_{17+1}+2) = 5778+2 = 5780$ and $5780\equiv0 \mod 17$ and $17^2 = 289$. $289$ has digits in nondecreasing order
$(L_{19+1}+2) = 15127+2 = 15129$ and $15129\equiv5 \mod 19$ and $13^2 = 361$. $361$ has not digits in nondecreasing order
$(L_{37+1}+2) = 87403803+2 = 87403805$ and $87403805\equiv0 \mod 37$ and $37^2 = 1369$. $1369$ has digits in nondecreasing order
I tested with the sequence https://oeis.org/A028865 and it works until 337 for me. (I didn't check higher)
Is there a way to explain that ? I don't know how to start for proving it.
The primes $p$ for which $L_{p+1}\equiv-2\bmod p$ end with $2, 3, 5$ or $7$. The primes in the OEIS sequence, whose squares must end with $4, 5,$ or $9$ for the digit ordering to work, meet this criterion, but there are also other primes for which $L_{p+1}\equiv-2\bmod p$. For instance, $p=23$ would satisfy the modular relation but $23^2=\color{blue}{52}9$.
The next unlisted primes ending in $3$ or $7$ are $43,47,53$ and these also appear to hold the modular relation, but their squares again do not have all digits sorted.