A massively palindromic prime is when for in all bases (in which it has more than $2$ digits) it is a palindrome:
Edit : My example was broken because of a silly miscalculation, I haven't even found one yet.
Do there exist any palindromic primes? How many massively palindromic primes exist? If infinite, how large percentage of all primes? If none exist, can we maybe calculate some bound of number of bases a prime is palindromic in?
I am very new in number theory and don't have many ideas how to start proving this.
I will prove this theorem.
Theorem: There is no massively palindromic number larger than $7$.
Proof
For $n>105$, it is obvious that $2(\lfloor\sqrt{n}\rfloor-2)^2>2(\sqrt{n}-3)^2=2n-12\sqrt n+18>n$. Therefore, there exists at least $3$ bases, $\lfloor\sqrt{n}\rfloor$, $\lfloor\sqrt{n}\rfloor-1$ and $\lfloor\sqrt{n}\rfloor-2$ which makes $n$ $3$ digit number starting with $1$ in its base. Therefore, $n$ should also end with $1$ in these bases and it follows that $n-1$ is common multiple of $\lfloor\sqrt{n}\rfloor$, $\lfloor\sqrt{n}\rfloor-1$ and $\lfloor\sqrt{n}\rfloor-2$. However, it can be easily proved that the lcm of $3$ consecutive numbers is at least the half of their product. Then one can see that $$(\lfloor\sqrt{n}\rfloor)(\lfloor\sqrt{n}\rfloor-1)(\lfloor\sqrt{n}\rfloor-2)>(\sqrt{n}-1)(\sqrt{n}-2)(\sqrt{n}-3)>n^{3/2}-6n>2(n-1)$$so this is contradiction.
Now, we only need to check the case $n\le 105$. Filtering by base $2$ gives possible candidates$$1, 2, 3, 5, 7, 9, 15, 17, 21, 27, 31, 33, 45, 51, 63, 65, 73, 85, 93, 99$$ and filtering them by base $3$ gives candidates$$1, 2, 3, 5, 7$$And these numbers are only massively palindromic number. They have only $0$ or $1$ bases which makes them more than $3$ digit number, but they are all palindromes in such bases.
Also, it is known that even palindromic in both bases $2$ and $3$ is extremely rare condition. See this question.