Find the value of $\bigstar$: Puzzle 4 - In Between

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

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 Puzzle 3

Next Puzzle

Answer 1

Solution:

$\bigstar = 7$

Explanation:

Plug III. into II.: $\delta×\alpha+\beta=\gamma+\delta$
Subtract I.: $(\delta-\alpha)×\alpha=\delta$
This works if $\alpha=2$ and $\delta=4$ (assumption A)
By IV. $β^\alpha+16=\beta^\delta+16 \implies β$ = -1, 0 or 1 – no other numbers have the same value for different exponents.
So by I. $\gamma$ = 3, 4 or 5.
And by III. $\varepsilon$ = 7, 8 or 9. Of these numbers, 9 is the only square, so
by V. $\zeta = 3$, $\varepsilon = 9 \implies \gamma = 5$ and $β = 1$.
Now VI. says $\eta = 6$
So in VII, $\gamma = 5$ and $\alpha^\zeta = 8$ and 6 is already taken, so 7 is the only solution for $\bigstar$.

Proof of assumption A:

Because $\delta/\alpha=\delta-\alpha$ is an integer, $\alpha | \delta$. Let's say $\delta = n \times \alpha$, so the equation becomes $(n\alpha-\alpha)\times\alpha = n \times \alpha$; $\alpha$ cannot be 0 because it would imply $\beta=\gamma$, so $(n-1)\alpha = n \implies \alpha = \frac{n}{n-1}$. This is only an integer if $n=0$ (leading to $\alpha = 0$ which isn't possible) or $n=2$, which proves the assumption.

Answer 2

Combining equations I, II, and III, we have

$\delta \alpha - \alpha^2 = \epsilon - \gamma = \delta$, which means that $\alpha^2 - \delta \alpha - \delta = 0$. This means that $\alpha = \frac{1}{2} \left( \delta \pm \sqrt{ \delta^2 - 4 \delta} \right)$. The quantity in the square root must be a square number: $\delta^2 - 4 \delta = m^2$ for some $m$. But $\delta^2 - 4 \delta + 4 = (\delta - 2)^2$ is also a square number. Thus, $m^2 + 4 = n^2$. The only two square numbers that differ by 4 are 0 and 4, so $m^2 = \delta(\delta - 4) = 0$. $\delta$ cannot be 0 by III, so $\delta = 4$ and $\alpha = 2$.

Applying these results to Eq. IV, we have

$\beta^2 + 16 = \beta^4 + 16$, or $\beta^2 (\beta^2 - 1) = 0$. Thus, $\beta = 0, 1$, or $-1$. From Eq. II, we have $8 + \beta = \epsilon$, and from Eq. V we have $\epsilon$ is a square number. Thus, $\beta = 1$, $\epsilon = 9$, $\zeta = 3$, $\gamma = 5$, and $\eta = 6$.

Thus, we can conclude that $\bigstar$ is

7, since we have $5 < \bigstar < 2^3 = 8$, and we know that $\bigstar \neq \eta = 6$.