While working on a proof, I ended up with the following Diophantine inequality of degree-6:
$$
a_1^6-2a_1^5+a_1^4 (1-4n)+8a_1^3 n+a_1^2 (4n^2+8n)-8a_1 n^2+16n^2 \ge 0 \tag 1
$$
The variable is $a_1$ and $n \in \mathbb{Z}$ is a known constant, $n \gt 0$. We want $a_1 \in \mathbb{Z}$ satisfying the inequality (1).
Using Descartes rule of signs
Positive roots. Number of sign changes in $f(a_1): (+,-,-,+,+,-,+) \implies 4$ sign changes. So, the number of positive roots is either $4, 2$ or $0$.
Negative roots. Number of sign changes in $f(-a_1): (+,+,+,-,+,+,+) \implies 2$ sign changes. So, the number of negative roots of $f$ is either $2$ or $0$.
Real roots: Therefore, there are either $6, 4, 2$ or $0$ real roots (positive or negative).
Numerical roots calculation
Assuming there are real roots, I can then use a numerical method like Newton-Raphson to calculate the roots and determine the search intervals (where the polynomial takes positive values). Once the intervals are found, I can check the integers in these ranges.
What are the bounds on the sizes of the intervals to be searched?
Is the approach right? Is there anything else I can do to find the roots algebraically?
Best Answer
All integer values of $a_1$ solve the inequality $(1)$.
Seeing $$ a_1^6-2a_1^5+a_1^4 (1-4n)+8a_1^3 n+a_1^2 (4n^2+8n)-8a_1 n^2+16n^2 \ge 0 \tag 1 $$ as a quadratic inequality on $n$, we get $$\underbrace{4(a_1^2-2a_1+4)}_{\text{positive}}n^2+(-4a_1^4+8a_1^3)n+a_1^6-2a_1^5+a_1^4\ge 0$$ The discriminant of LHS is $$D=-16(5a_1^2 - 10a_1 + 4)a_1^4$$ So, we see that $D\gt 0\iff a_1=1$.
For $a_1=1$, we get $4n(3n+1)\ge 0$ which is true.
In conclusion, all integer values of $a_1$ solve $(1)$.