Diophantine equation from the Latvian Baltic Way team selection competition 2019 So here is the problem statement:

Find all integer triples $(a, b, c)$ such that

$(a-b)^3(a+b)^2 = c^2 + 2(a-b) + 1$

The only things I have so far figured out is that (-1, 0, 0) and (0, 1, 0) are solution, gcd((a-b), c) = 1 and that c must be even.

Any ideas? Maybe even a general solution to the problem?
Consider the equation already found:-
$$(a-b)\left((a^2-b^2)^2-2\right)=c^2+1,$$
where $c^2+1$ is odd or singly even. 

If $a-b$ is even then the LHS would be divisible by 4. Therefore $a-b$ is odd and $c$ is even. 

Then $(a^2-b^2)^2-2$ is of the form $4k-1$ and it is therefore either $-1$ or has a prime divisor of the form $4k-1$. However, $c^2+1$ cannot have a prime factor of the form $4k-1$ and so $$(a^2-b^2)^2-2=-1.$$

Therefore $a^2-b^2$ is $-1$ or $1$ and then one of $a$ and $b$ is $0$. We also have $b-a=c^2+1$ and so the solutions for $(a,b,c)$ are $(0,1,0)$ or $(-1,0,0).$
Above equation shown below:

$(a-b)^3(a+b)^2 = (c^2 + 2(a-b) + 1)$ ----$(1)$

Equation $(1)$ has parameteric solution given below:

$(a,b,c)=[(2m^2+2),(2m^2+1),(2m)]$

For, $m=7$ we get:

$(a,b,c)=(100,99,14)$