Let the equation to be solved be
$$
Q: v^2=6u^4-2u^2+4u+1
$$
This is an Elliptic Curve so there is a map for almost all the points to a Weierstrass form Elliptic Curve $E$
$$
E: y^2+a_1 xy+a_3 y=x^3+a_2x^2+a_4x+a_6
$$
For lucky cases, $E$ has only finitely many rational points so we can compute the inverse to check which ones are integral in $Q$. However in this case $E$ will have rank $1 \implies$ infinite rational points so this does not work directly.
There is a way to find the finitely many integer points for Weierstrass form Elliptic Curves. Seems like there is an adaptation of those methods (Elliptic Logarithms) for the quartic curve case. Magma has an implementation of it so you can use that for solving your problem.
1. Magma solver
Magma has a quartic solver exactly for your case. You can go to the online calculator and use the command (integer inputs)
IntegralQuarticPoints([a,b,c,d,e],[u,v])
for solving
$$
V^2=aU^4+bU^3+cU^2+dU+e
$$
where $[u,v]$ is a known integer point. Putting in
IntegralQuarticPoints([6,0,-2,4,1],[0,1])
nets you the solutions $(u,v)$:
[ -2, -9 ],[ 1, 3 ],[ -1, 1 ],[ 0, -1 ],[ 5, -61 ],[ 4, -39 ]
So it says that you have found all of them.
The algorithm is based on this paper. This seems to be the same methods for finding integer points on Weierstrass form Elliptic Curves but adapted for the quartic curve case.
2. Birational Equivalence to an Weierstrass form Elliptic Curve
The equation $Q$ is an Elliptic Curve so there exists a Birational Transformation to a Weierstrass form Elliptic Curve. More concretely:
Every point $(u,v), u\neq 0$ on the curve
$$
Q: v^2 = 6u^4-2u^2+4u+1
$$
can be map onto
$$
E: y^2+4xy=x^3-6x^2-24x+144
$$
via
$$
\begin{align*}
x &= \frac{(4 u + 2 (1 + v))}{u^2}, & y &= \frac{(-8 u^2 + 2 (4 u - 2 u^2) +
4 (1 + v))}{u^3}
\end{align*}
$$
An inverse map for $(x,y),y\neq 0$ from $E$ to $Q$ is
$$
\begin{align*}
u &= \frac{2 (x - 6)}{y}, &v &= \frac{(72 x - 24 x^2 + 2 x^3 + 24 y - 4 x y - y^2)}{y^2}
\end{align*}
$$
The transformation can be found on page 105 of this handbook.
For $u=0$, the integer points are $(0,\pm 1)$ on $Q$. For every other integer point $(u,v),u\neq 0$ there exists a map onto a point $(x,y)$ on $E$, so we can attempt to take the inverse of all rational points on $E$ (excluding $y=0$) to see which ones map to integral $(u,v)$ on $Q$.
Edit 1: if $(u,v),u\neq 0$ maps to $y=0$ then solving the transform gives $v = -1 - 2 u + 3 u^2$. Then putting back to $Q$ we get $u=0,4$.
If $rank(E)=0$ then there are only finitely many points to check, but in this case $rank(E)=1$ so there are infinitely many rational points and I am stuck here.
The condition we need is
$$
u=\frac{2(x-6)}{y}
$$
is integral, where $(x,y),y\neq 0$ is a rational point on
$$
y^2+4xy = x^3-6x^2-24x+144
$$
but this does not seem to be sufficient to solve it directly.
Edit 2: Group Structure of $E$
The Elliptic Curve $E$ has Mordell-Weil group structure
$$
\mathbb Z \times \mathbb Z/2\mathbb Z
$$
Where torsion is $T = (-6,12)$ and generator $G=(12,12)$. This can be obtained from here via translation $x=X-1$ followed by $y=Y+2X$.
Therefore all the points $P$ on $E$ can be described as
$$
P = [k]G\oplus [\pm 1]T
$$
Some of the "small points" on $E$ and its reverse map to $(u,v)$ are
$$
\begin{align*}
G = (12 , 12) &\mapsto(1,3)\\
T\oplus G=(0, -12) &\mapsto(1,-3)\\
[2]G = (3, -3) &\mapsto(-2,9)\\
T\oplus [2]G = (6, -24) &\mapsto(0,-1)\\
[3]G = (-4, 20) &\mapsto(-1,-1)\\
[4]G = (-15/4,-39/8) &\mapsto(4,-39)\\
T\oplus [5]G = (144/25, -12/125) &\mapsto (5,61)
\end{align*}
$$
These generated the solutions we are looking for.
Notation: Let $ \{ x \} $ denote the remainder when $x$ is divided by $p$.
Let $ \lfloor \sqrt{ b} \rfloor = c$.
Claim:
One of the $c$ values $\{a\}, \{2a\}, \ldots , \{ca\}$ is either $< \frac{p}{c} $ or $ > \frac{(c-1)p}{c}.$
Proof:
By contradiction. Suppose that none of them satisfy the condition. Then by Pigeonhole Principle applied to these distinct $c$ values as pigeons and the $c-2$ ranges
$$ ( \frac{p}{c}, \frac{2p}{c}), ( \frac{2p}{c}, \frac{3p}{c}), \ldots , ( \frac{(c-2)p}{c}, \frac{(c-1)p}{c}) $$
as holes, then at least 2 of these values lie in the same hole.
Let $n_1 > n_2$ and $ \{ n_1 a \}, \{ n_2 a \}$ lie in the same hole.
Then $ | \{n_1a\} - \{n_2 a\} | < \frac{ p}{c}$, and so $\{ (n_1 - n_2) a \}$ is either $< \frac{p}{c}$ or $> \frac{(c-1)p}{c}$, as desired.
Note that we have to take the "larger index minus the smaller index" to ensure that $ n_ 1 - n_2 > 0 $ is one of the $c$ values.
Corollary:
First case: If $\{na\} < \frac{p}{c}$ with $0 < n \leq c$, then $\{na\}, \{2na\}, \{3na\}, \ldots \{cna\}$ is a sequence of $c$ terms
that form an increasing arithmetic progression (in that order).
Second case: if $\{na\} > \frac{(c-1)p}{c}$ with $0 < n \leq c$, then $\{na\}, \{2na\}, \{3na\}, \ldots \{cna\}$ is a sequence of $c$ terms that form a decreasing arithmetic progression (in that order).
Notice that $ cn \leq c^2 \leq b$, so we do indeed have $c$ terms.
Notes:
- Notice that we could force the indices to be of the form $n, 2n, \ldots cn$, and the remainders to be properly sorted. Though, stating this in the question would have given away too much for an Olympiad problem.
- The first case of "terms are of the form $na, 2na, 3na, \ldots$ and the
remainders are of the form $r, 2r, 3r, \ldots$" was easy to observe.
It was wishful thinking that this always existed.
- When playing with $p = 17, b = 9, a = 7$, I couldn't find any that fit the first case. I then realized that the sequence $\{14\}, \{28\}, \{42\} = 14, 11, 8$ was a good match for what I wanted, which gave the second case. This choice led to "the indices are of the form $n, 2n, \ldots cn$ and the remainders are arranged in order".
- For this set of $p,a,b$, there are other arithmetic progressions like $\{7\}, \{21\},\{35\} = 7, 4, 1$. This choice would have led to "the indices form an arithmetic progression and the remainders are arranged in order". There were no other arithmetic progressions, so it's safe to guess that one of these patterns must work.
- The condition that $p$ is prime is used to ensure that the remainders are distinct. In particular, we could have relaxed the condition to $ 1 = \gcd(a, p) = \gcd (b, p) $ instead.
Best Answer
This is a modification of Aryaaaaan's answer aimed at making the sequence of numbers of divisors a nonconstant arithmetic progression. I will use two facts: first, the Green-Tao theorem (there exist arbitrarily long arithmetic progressions of primes), and second, the elementary fact that the set of unit fractions $\{1, \frac{1}{2}, \frac{1}{3}, \dots\}$ contains arbitrarily long arithmetic progressions. The simplest example of the latter is the $n$-term arithmetic progression $\frac{1}{n!}, \frac{2}{n!}, \dots, \frac{n}{n!}$.
Fix a positive integer $n$, an arithmetic progression $p_1, \dots, p_n$ of primes, and a sequence $m_1, \dots, m_n$ whose reciprocals form an arithmetic progression. Set $N = \prod_{i=1}^n p_i^{m_i-1}$, which has exactly $m_1 m_2 \cdots m_n$ divisors. Now consider the arithmetic progression $(N p_1, \dots, N p_n)$. We can compute the number of divisors of each term from its prime factorization: namely, $N p_i$ has $m_1 m_2 \cdots (m_i+1) \cdots m_n$ divisors. Dividing out by the number of factors of $N$, we get: $$ \frac{\sigma_0(N p_i)}{\sigma_0(N)} = \frac{m_i + 1}{m_i} = 1 + \frac{1}{m_i}. $$ By assumption, these numbers form an arithmetic progression, so the sequence $\sigma_0(N p_i)$ does as well.
For example, if $n = 5$, you can choose $(p_i) = (5, 11, 17, 23, 29)$ and $(m_i) = (30, 20, 15, 12, 10)$, which leads to $N = 5^{29} \cdot 11^{19} \cdot 17^{14} \cdot 23^{11} \cdot 29^{9}$. Then $5N, 11N, 17N, 23N$, and $29N$ respectively have $1116000, 1134000, 1152000, 1170000$, and $1188000$ divisors.