Here is my messy solution, note that $$(2a+1)^2-4b=4a^2+4a+1-4b$$, doing the similiar things to others, adding them up: $$a^2+b^2+c^2=3$$note that $4a^2+4a-4b-4=0 \leftrightarrow b=a^2+a-1$, replacing b as this expression in $(2b+1)^2-4c$, we get $c=(a^2+a-1)^22+(a^2+a-1)-1$ , replacing these expressions to $a^2+b^2+c^2$, factoring, we get ( factoring can be look tricky, I just found a=-1, 1 works in the start of the problem and factored with the help of a-1 and a+1) $$a^2 + (-1 + a) a^3 (1 + a)^3 (2 + a)-1=0$$Now we will divide into cases: Case 1. $a \le -2$ Obviously does not work. Case 2. $ -1>a > -2$ Which does not satisfy hence $a^2$ is atleast 1 and $(-1+a)a^3(1+a)^3(2+a)\neq 0$ Case 3. $a=-1$. Works! Case 4. $1>a>-1$ the expression will be between $-2$ and $-3$, hence can't be 0 ( expressing factors as inequalities yielts the result) Case 5. $a=1$, works! Cse 6. $ a >1$, doesn't work, the same idea as Case 2. Hence the only answers are $a=1,-1$, and putting these values to the other expressions, we can find $b,c$ I should be more clear in the solution, I am sorry

Here is the official one, slightly tricky. We equivalently have $a^2 + a - b = b^2 + b - c = c^2 + c - a = 1$. If any of the variables equals $(-1)$, then we quickly get that all of them are equal to $(-1)$. Otherwise adding the equations gives $a^2 + b^2 + c^2 = 3$, while multiplying $a(a+1) = b+1$, $b(b+1) = c+1$, $c(c+1) = a+1$ and dividing by $(a+1)(b+1)(c+1) \neq 0$ yields $abc = 1$ and hence $a^2b^2c^2 = 1$. But $a^2 + b^2 + c^2 \geq 3\sqrt[3]{a^2b^2c^2} = 3$ by AM-GM and so equality must hold, forcing $a^2 = b^2 = c^2 = 1$, i.e. $a=b=c=1$, which satisfies the initial system.

Note that this problem is just the particular case $n=3$ of Mexico 2011 NMO P3 (https://artofproblemsolving.com/community/c6h594720p3528050).

@above lol I admit I did not know that The problem is indeed not really original, but came out from something different.

Here is an approach (inspred by that of lora, but cleaner), where the greedy idea of reducing the number of variables and equations leads to a nasty, though soluble, equation. This was the best idea for a lot of contestants, but unfortunately nobody among these succeeded and barely one of them gave useful arguments after performing the main factorization. (I do admit it took me some time as well, particularly the case $a\in (-2,-1)$.) So try not to be greedy when you do not succeed in being greedy! With $b = a^2 + a - 1$ and $c = b^2 + b - 1 = a^4 + 2a^3 - a - 1$, substituting $c^2 + c - a = 1$ and simplifying lead to $a^8 + 4 a^7 + 4 a^6 - 2 a^5 - 5 a^4 - 2 a^3 + a^2 - 1 = 0$, which (e.g. by Horner's method) can be factorized as $$ (a - 1)(a + 1)(a^6 + 4 a^5 + 5 a^4 + 2 a^3 + 1) = 0. $$For $a=1$ we get $a=b=c=1$, and for $a=-1$ we get $a=b=c=-1$. We will now show that $a^6 + 4 a^5 + 5 a^4 + 2 a^3 + 1 > 0$. If $a \geq 0$ this is clear, while for $a \leq -2$ we use $a^3(a+2)(a+1)^2 + 1$. For $a \in [-1,0)$ we use $a^4(a+2)^2 + a^4 + 2a^3 + 1 = a^4(a+2)^2 + a^3(a+1) + (a^3 + 1)$. For $a\in (-2,-1)$ write for convenience $a=-z-1$, $z \in (0,1)$, and the expression becomes $z^6 + 2z^5 - 2z^3 - z^2 + 1$. We have $2z^5 - 2z^3 - z^2 + 1 = (z-1)(z+1)(2z^3 - 1) \geq 0$ for $2z^3 - 1 \leq 0$ (and with $z^6 > 0$ we are done), while for $2z^3 - 1 > 0$ we rewrite the left-hand side as $(z^3-1)^2 + z^2(2z^3 - 1) > 0$. This completes the proof.