Find the value of $\bigstar$: Puzzle 3 - Substitution

Question

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 by the following rules:

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

$$ \text{I. }\alpha+\beta=\gamma\times\delta \\ \space \\ \text{II. }\gamma+\varepsilon=\delta \\ \space \\ \text{III. }\gamma-\varepsilon\times\varepsilon=\beta \\ \space \\ \text{IV. }\varepsilon\times\delta+\gamma=\alpha \\ \space \\ \text{V. }\zeta\times\zeta=\zeta \\ \space \\ \text{VI. }\zeta+\gamma=\varepsilon \\ \space \\ \text{VII. }\alpha-\delta-\beta=\bigstar $$

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!

Previous puzzles in the series:

Puzzle 1 Puzzle 2

Next Puzzle

Answer 1

From V and VI, we have that

$\zeta = 1$. Eq. V implies that $\zeta^2 - \zeta = \zeta(\zeta - 1) = 0$, which means $\zeta = 0$ or $\zeta = 1$. But if $\zeta = 0$, then $\gamma = \epsilon$, which is forbidden.

These then imply, from II, III, and VI, that

$\varepsilon = \gamma + 1$, $\delta = 2 \gamma + 1$, and $\beta = \gamma - (\gamma + 1)^2$.

From I and IV we have that

$\alpha = \gamma + (2 \gamma + 1)(\gamma + 1) = 2 \gamma^2 + 4 \gamma + 1$, and $\alpha = \gamma (2 \gamma + 1) - \gamma + (\gamma + 1)^2 = 3 \gamma^2 + 2 \gamma + 1$. Equating these two, we have $\gamma^2 - 2 \gamma = 0$, implying that $\gamma = 2$ or $\gamma = 0$. But if $\gamma = 0$, then by II $\epsilon = \delta$. Thus, $\gamma = 2$.

From here, we can plug things in:

$ \alpha = 2 + (5\times 3) = 17$ from above; $\beta = 2 - 3^2 = -7$; $\delta = 5$; and thus $\bigstar = 19$.

Answer 2

$\bigstar = 19$

Explanation ('brute-force' substitution, there's probably a more elegant solution):

V. $\zeta = 1$ (the other solution, 0 would violate VI. because the symbols are unique numbers)
VI. $\varepsilon = \gamma + 1$
II. $2\gamma + 1 = \delta$
IV. $(\gamma + 1)(2\gamma + 1) + \gamma = \alpha$
=> $2\gamma^2 + 4\gamma + 1 = \alpha$ (!)
III. $\gamma - (\gamma + 1)(\gamma + 1) = \beta$
=> -$\gamma^2 - \gamma - 1 = \beta$ (!!)
I. $\gamma^2 + 3\gamma = \gamma\times\delta$
=> $\delta = \gamma + 3 (!!!)$ ($\gamma = 0$ would violate II. because of uniqueness)
II. $2\gamma + 1 = \gamma + 3$
=> $\gamma = 2$
By (!), $\alpha = 17$
By (!!), $\beta = -7$
By (!!!), $\delta = 5$
So $\alpha−\delta−\beta = \bigstar = 19$.

Answer 3

$\bigstar=19$

Explanation:

V implies that $\zeta=1$ (because VI implies that $\zeta\not=0$. Then, VI implies $\varepsilon=\gamma+1$ and II implies $\delta=2\gamma+1$. From there we can deduce using III and IV that $\alpha=2\gamma^2+4\gamma+1$ and $\beta=-\gamma^2-\gamma-1$, which turns 1 into $$\gamma^2+3\gamma=2\gamma^2+\gamma.$$ This quadratic equation has two solutions, $\gamma\in\{0,2\}$, but $\gamma=0$ implies $\varepsilon=\zeta=1$ which violates the rules. Therefore, $$\gamma=2\\\varepsilon=3\\\delta=5\\\alpha=17\\\beta=-7\\\bigstar=19$$

Answer 4

We know from (v) and (2): $$\zeta=1$$
From (iii) and (iv) we get:
$$\alpha+\beta=2\gamma+\epsilon(\delta-\epsilon)$$
From (ii) we therefore have:
$$\alpha+\beta=2\gamma+\epsilon\gamma$$
From (i) we deduce:
$$\gamma\delta=2\gamma+\epsilon\gamma$$
so that:
$$\delta=2+\epsilon$$
(ii) now tells us:
$$\gamma=2$$
and (vi) and (ii) give:
$$\epsilon=3, \delta=5$$
(iv) - (iii) gives: $$\alpha-\beta=\epsilon(\delta+\epsilon)=3\times8=24$$
and so:
$$\bigstar=24-5=19$$