# Pairs of palindromic primes without $1$ and have a palindromic product

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 $$k \in \Bbb{+Z}, 0$$?

• 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. – kccu May 2 '19 at 22:05
• @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. – nickgard May 3 '19 at 11:07