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?

Equation,  $(a-b)^3(a+b)^2 = (c^2 + 2(a-b) + 1)$
Thanks @Piquito, for reviewing my previous answer. 
By mistake I solved "OP' equation as a quartic rather than a quantic. 
If we put the condition, $(a=b+c)$ then we get:
$c^3(2b+c)^2=(c+1)^2$
Which has solution at, $(b,c)=[(1/2),(1)]$
And so we get, $(a,b,c)=[(3/2),(1/2),(1)]$
