<s>[INST] Solve in integers the equation $(a-b)^3(a+b)^2 = c^2+2(a-b)+1$ $$(a-b)^3(a+b)^2 = c^2+2(a-b)+1$$
So it is evident  that a solution to the equation $(a, b, c) = (0, 1, 0)$. The expression $c^2+1$ never divides $4$. Now let's assume that $c^2+1$ is even, then we can write it in the form $2k$ where $k$ is odd. Rewrite the equation as
$$ ((a-b)(a+b))^2 = \frac{c^2+1}{a-b}+2$$
Now if $a-b$ is even then we get a contradiction as the LHS would be even but the RHS would be odd. and if $a-b$ is odd, then LHS would be odd but the  RHS would be even another contradiction. Therefore $c^2+1$ is odd and $a-b$ is odd. In fact some further analysis in $mod$ $4$ would yield that $a-b$ is of the form $4k-1$.
That is how far I got and right now I am stuck. Any ideas?
 [/INST] If $p\mid a-b$ then $c^2\equiv -1\pmod p$, so $p=2$ or $p\equiv 1\pmod 4$. You already ruled out $p=2$, hence $a-b$ is $\pm$ a (possibly empty) product of primes $\equiv 1\pmod 4$. In case of $+$ sign, this give $a-b\equiv 1\pmod4$ overall, whereas you already know $a-b\equiv -1\pmod 4$.
It follows that $a-b$ is negative. Let $d=b-a>0$.
Then
$$1\le c^2+1=d(2-d^2(a+b)^2)$$
implies 
$$ d^2(a+b)^2\le 2.$$
As $a+b=0$ is impossible when $a-b$ is odd, we conclude $d=1$ and $a+b=\pm1$, so either $a=0$, $b=1$, and then $c=0$ as already found; or $a=-1$, $b=0$, and again $c=0$.
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