mathematics - Find the value of $bigstar$: Puzzle 1 - Evaluation

This puzzle replaces all numbers with other symbols.

Your job, as the title suggests, is to find what number fits in the place of $\bigstar$.

All symbols abide to the following rules:

1. Each symbol represents integers and only integers. This means fractions and irrational numbers like $\sqrt2$ are not allowed. However, negative numbers and zero are allowed.


2. Each symbol represents a unique number. This means that for any two symbols $\alpha$ and $\beta$ which are in the same puzzle, $\alpha\neq\beta$.


3. The following equations are satisfied (this is the heart of the puzzle): $$ \text{I. }\alpha\times\alpha=\alpha \\ \space \\ \text{II. }\alpha+\alpha=\beta \\ \space \\ \text{III. }\beta+\alpha=\gamma \\ \space \\ \text{IV. }\gamma\times\beta=\delta \\ \space \\ \text{V. }\delta\times\gamma=\varepsilon \\ \space \\ \text{VI. }\varepsilon-\beta=\bigstar $$

What is a Solution?

A solution is an integer value for $\bigstar$, such that, for the group of symbols in the puzzle $S_1$ there exists a one-to-one function $f:S_1\to\Bbb Z$ which, after replacing all provided symbols using this function, satisfies all given equations.

What is a Correct Answer?

An answer is considered correct if you can prove that a certain value for $\bigstar$ is a solution. This can be done easily by getting a function from every symbol in the puzzle to the correct integers (that is, find an example for $f:S_1\to\Bbb Z$).

An answer will be accepted if it is the first correct answer to also prove that the solution is the only solution. In other words, there is no other possible value for $\bigstar$.

Good luck!

Next Puzzle

Answer

I. α×α=α

The only values that satisfy this are 0 and 1

II. α+α=β

β=2α, but since 2.Each symbol represents a unique number, α and β both cannot be 0, so α=1 and β=2

III. β+α=γ

Plugging in our known values, γ=2+1=3

IV. γ×β=δ

Plugging in more values, δ=3*2=6

V. δ×γ=ε

Plugging in more values, ε=6*3=18

VI. ε−β=★

Plugging in more values, ★=18-2=16

Proof

This all relies on α=1
$$α*α=α$$ $$α*α-α=0$$ $$α^2-α=0$$ Using the quadratic formula... $$aα^2+bα+c=0$$ $$a=1, b=-1, c=0$$ $$α=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ The two values for α... $$α=\frac{-(-1)+\sqrt{(-1)^2-4*1*0}}{2*1}=\frac{1+1}{2}=1$$ $$α=\frac{-(-1)-\sqrt{(-1)^2-4*1*0}}{2*1}=\frac{1-1}{2}=0$$ But since $β=2α$, and symbols are unique, α and β cannot be 0