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While discussing about prime numbers with other users, I noticed that:

$(1)$ There are very few pairs of palindromic prime numbers that do not contain the digit $1$ and that have products which are palindromes.

Ex : $[2, 3], [2, 30203], [2, 30403]$

$(2)$ For the large range that was tested on PARI/GP, I noticed that all the palindromic primes that yielded palindromic products were composed only of the digits $0, 2, 3, 4$

$(3)$ The palindromic primes of these pairs are equal to $2$ or always exist in the range of $3 \times 10^k$ to $4 \times 10^k$ where $k \in \Bbb{+Z}, 0$

User Peter helped me get the following results for $a, b \lt 10^7$ on PARI/GP:

[2, 3] 
[2, 30203] 
[2, 30403] 
[2, 32323] 
[2, 32423] 
[2, 3002003] 
[2, 3222223] 
[2, 3223223] 
[2, 3233323] 
[2, 3304033] 
[2, 3343433] 
[2, 3400043] 
[2, 3424243] 
[2, 3443443] 
[2, 3444443] 
[3, 30203] 
[3, 32323] 
[3, 3002003] 
[3, 3222223] 
[3, 3223223] 
[3, 3233323] 
[30203, 3002003]

Questions:

(1) Are there a finite number of pairs of $a, b$, where $a, b$ are palindromic primes that do not contain the digit $1$ and $ab$ is a palindrome?

(2) Are all the palindromic primes that yielded palindromic products composed only of the digits $0, 2, 3, 4$?

(3) Are all the palindromic primes of these pairs equal to $2$ or always exist in the range of $3 \times 10^k$ to $4 \times 10^k$ where $k \in \Bbb{+Z}, 0$?

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  • 1
    $\begingroup$ I suspect the answer to (1) is no, but I don't have a proof. Note that (2) implies (3) because primes greater than $2$ cannot end in $2$. As for (2), I don't think this property is really particular to prime palindromes. In particular, it appears that if $a$ and $b$ are palindromes that do not contain $1$ and are not divisible by $11$, and $a\cdot b$ is a palindrome, then $a$ and $b$ consist of only the digits $0, 2, 3, 4$. Moreover, if one of them contains the digit $4$, then the other consists of only $0$ and $2$. This is based on Python code searching for such pairs. $\endgroup$ – kccu May 2 '19 at 22:05
  • $\begingroup$ @kccu I agree, and have reached similar conclusions without proof. If written as a long multiplication, any carry in the calculation seems to imply a palindrome product of two palindromes is impossible, which in turn implies only digits $0,2,3,4$ appear. It's not known whether there are infinite palindromic primes, or whether there are infinite primes containing only restricted digits, such as only $0,2,3,4$. Hence, if (2) is true, then proving (1) in the negative (i.e. that there are infinite such pairs) is certainly difficult. $\endgroup$ – nickgard May 3 '19 at 11:07

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