(From a math competition) Question: Find all integers $a,b,c$ that satisfy:
$$a^3 - 3a^2b - 3c+2b^2 = c^3 -3ab^2 + 3c^2 +1 $$
What I have tried/attempted basically I've been looking for expansions such as $(a+b)^3$ etc. and I could find one $(c+1)^3$
so
$$a^3 - 3a^2b - 3c+2b^2 = c^3 -3ab^2 + 3c^2 +1 $$
$$ \Leftrightarrow a^3 -3a^2b+3ab^2 +2b^2 = c^3+3c^2+3c+1 $$
$$ \Leftrightarrow a^3 -3a^2b+3ab^2 +2b^2 =(c+1)^3$$
But other than that I'm not sure how to continue. . . .
(Not yet allowed to comment so this has to be provided as an answer)
Continuing - but not necessarily solving…
$(a-b)^3+b^3+2b^2=(c+1)^3$
With $b=0$ this gives an infinite set of answers with $a=c+1$.
For the rest, rewriting as
$(c+1)^3-(a-b)^3={b^2}(b+2)$ (or let's say $x^3-y^3={b^2}(b+2)$)
gives a series of "difference of two cubes" problems. My guess is that these can be taken in stages and handled with modular arguments. But I have to admit that I haven't actually tried.