<s>[INST] Solve the equation in integers $a,b$: $20a^3-b^3=1$ 
Solve the equation in integers $a,b$: $$20a^3-b^3=1.$$

Assume that $a \neq 0$. Then simplifying and rearranging the equation gives $$20a^3 = 2^2 \cdot 5 \cdot a^3 = b^3+1 = (b+1)(b^2-b+1).$$ Note that neither factors can be divisible by $3$, since both of them must be divisible by $3$ but $b^2-b+1 \not \equiv 0 \pmod{9}$ and $b^2-b+1 = 3$ gives a contradiction. Also since $b$ is odd, then $b+1$ is even and $b^2-b+1$ is odd. Also, $b^2-b+1 \not \equiv 0 \pmod{5}$, so $b+1 \equiv 0 \pmod{20}$.
Then since $$\gcd(b+1,b^2-b+1) = \gcd(b+1,-2b+1) = \gcd(b+1,3),$$ it follows that $(b+1)$ and $(b^2-b+1)$ are relatively prime. Thus $b+1 = 20n^3$ and $b^2-b+1 = m^3$ where $m,n$ are relatively prime. I then thought about bounding $b^2-b+1$ between two perfect cubes, but I didn't see how to do that.
Then since two solutions for $(a,b)$ are $(0,-1)$ and $(7,19)$ I wanted to show that $(b+1) \mid 20$, or equivalently that $a$ and $b+1$ are relatively prime. So suppose that for some prime $p$ we have $a \equiv 0 \pmod{p}$ and $b+1 \equiv 0 \pmod{p}$. It then follows that $b+1 \equiv 0 \pmod{p^3}$, so that $b = p^3k-1$ for some integer $k$. Then we have $$(p^3k-1)^2-(p^3k-1)+1 = p^3k(p^3k-3)+3 = m^3.$$ How can we get a contradiction from here?
 [/INST] From your work, after $b+1=20n^3$ and $b^2-b+1=m^3$, multiply second by $64$
$$64b^2-64b+64=(4m)^3 \ \ \ \Longrightarrow \ \ \ (8b-4)^2+48=(4m)^3$$
All solutions to $y^2+48=x^3$ in integers are known: $(x, y)= (4,\pm4), (28,\pm148)$.
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