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My main question is: Is there a criteria for solving a quartic Diophantine equation of the form

$$x^4 + ax^3 + bx^2 + ax + d = 0 \tag 1$$

We have the restriction $d \ne 1$.

Here's my effort in obtaining a criteria and I would like the solution to be verified.

EDIT: As @mathlove points out in comments, this approach is incorrect. I am leaving the question as-is for other readers.

Eqn (1) is almost palindromic (or what Wikipedia calls as [Quasi-palindromic][1]). The suggested variable transformation for quartics of the form

$$a_0 x^4 + a_1 x^3 + a_2 x^2 + a_1mx + a_0 m^2 \tag 2$$

is to divide by $x^2$ and apply the change of variable $$z = x + \frac{m} {x} \tag 3$$ and obtain

$$ a_0 z^2 + a_1 z + a_2 - 2ma_0 = 0 \tag 4 $$

Using Eqns. (2),(3) and (4) and with $m=1$, our Eqn. (1) becomes

$$z^2 + az + b - 2d = 0 \tag 5$$

Using the quadratic formula

$$z = \frac {-a \pm \sqrt {a^2 - 4(b-2d)}} 2 \tag 6$$

This has real solutions iff the discriminant $$D = a^2 - 4b + 8d \gt 0. \tag 7$$

This is a necessary and sufficient condition for a real solution.

From the Integral Root Theorem, $x$, the root of Eqn. (1) must divide $d$.

So, the best I could come up with as a solvability criteria for Eqn. (1) is

$$ \begin{align} a^2 - 4b + 8d \gt 0 \land x \in \{d_i : \text{the set of divisors of $d$}\} \tag 8 \end{align} $$

Can this criteria be refined or is this the best we can get?

vvg
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    In the first place, I think that (1) is not of the form (2) unless $d=1$. The second (7) (there are two (7)s) is wrong when $(a,b,d)=(4,1,1)$ for which (1) has no integer solutions. – mathlove Dec 07 '22 at 09:10
  • @mathlove: No, $d \ne 1$. Eqn. (2) and (1) are both almost palindromic. We have $a = a_1 m$ and $d = a_0 m^2 = m^2$ since $a_0 = 1$. Since my question wasn't explicit about $d \ne 1$, I have edited it. – vvg Dec 07 '22 at 09:39
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    "(1) becomes $z^2 + az + b - 2d = 0\ \ $ (5)" I don't think so because we have $(x+\frac 1x)^2+a(x+\frac 1x)+b-2d=0$ which is $x^2+2+\frac{1}{x^2}+ax+\frac ax+b-2d=0$. Multiplying the both sides by $x^2$, we get $x^4+ax^3+(2+b-2d)x^2+ax+1=0$ which is not (1). – mathlove Dec 07 '22 at 09:51
  • I think I see the issue. In mapping the coeffiients between (1) and (2) we are saying $a = a_1$ and $a = a_1 m \implies m = 1$ and $a_0 = 1$ and $d = a_0 m^2 = m^2 \implies d = 1$. – vvg Dec 07 '22 at 10:45

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