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Main question: Which are the primes that can be represented as monic palindromic polynomials with integer coefficients of degree 4: $$f(x) = x^4+ ax^3+ bx^2+ ax + 1.$$

If we take $x=2$,

$$p = 16 + 8a + 4b + 2a + 1 = 10a + 4b + 17$$

If $p$ is an odd prime then, the equation $$10a + 4b + 17 = 2m + 1 = p$$ may be solved for integers $a, b, c$ by solving $$5a + 2b - m = -8.$$

This equation has an infinite number of solutions in $a,b,c$ and since there are an infinte number of primes, how does one characterize the intersection of these two sets and prove that it is non-empty?

Another line of thinking I have explored is that $f(x)$ is a degree 4 polynomial. If the prime $p$ is one of the values taken by the polynomial at $x = u$, i.e., $f(u) = p$ then surely we must be able to take another point $(v, f(v))$ on the polynomial that satisfies the palindrome condition with $a,b \in \mathbb{Z}$. Does this mean that every sufficiently large prime is representable as a palindrome polynomial form of degree 4?

Secondary question: There is this result in [Banks2004] that states that almost all palindromes in any base are composite. Does this result also extend to bases where digits are $[-2r, +2r]$ and radix is $2r+1, r \in \mathbb{Z}$? What if the coefficients are allowed to be any integer (absolute value of coefficient not necessarily less than the radix)?

I am sensing an apparent conflict between these two lines of reasoning in the main question versus the secondary question and I could be wrong. Could someone help clarify?

References

[Banks2004]: W. D. Banks, D. N. Hart & M. Sakata, "Almost all palindromes are composite", Math. Res. Lett. 11 (2004), 853-868. DOI: 10.4310/MRL.2004.v11.n6.a10. url (accessed Nov 6, 2022): https://dx.doi.org/10.4310/MRL.2004.v11.n6.a10

vvg
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    If we take $x=1$, we get $p = 2a+b+2$. Setting $b=1$ gives $p=2a+3$ which is the set of all odd primes.

    Still with $x=1$, $a=b=0$ gives us $p=2$.

    Are you looking instead for "Which primes can be represented for particular values of $a,b$?"

     – Eric Snyder Nov 07 '22 at 02:09

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COMMENT.- I think a way to get solutions can be asked for a particular prime $p$ finding integers solution of the equation $f(x)=p$. For example $$f(x)=3$$ have integer solutions $$(a,b,x)=(2n+1,5n-1,-2).$$ $$f(x)=5$$ have integer solutions for $$(a,b,x)=(2n,-5n-3,2)$$ and $$f(x)=7$$ have integer solutions $$(a,b,x)=(3n+1,10n-5),-3)$$ This gives an infinite set of palindromic $f(x)$ giving the primes $3,5$ and $7$ respectively.(Which of course does not prevent the appearance of other primes in each of the three considered polynomials).

Ataulfo
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  • OP here. @EricSnyder's comment below the question and this answer has clarified my question. Thanks. – vvg Nov 07 '22 at 03:51