Find the value of $\bigstar$: Puzzle 2 - Switch-a-roo

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

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

Next Puzzle

Answer 1

$\bigstar = 21$

Explanation:

Substituting VI. and VII. in IX. gives $$\bigstar = \delta\times\delta+\alpha\times\gamma$$ The only solution to I. where $\alpha$ and $\beta$ are (non-equal) positive integers is $4^2 = 2^4$ (but if $\alpha = 2$ or $\beta = 2$ is unclear yet).
That means that by II., $\gamma = 6$, by III., $\delta = 3$, by IV. $\varepsilon = 9$.
From V. it follows that $\alpha = 2$ and $\beta = 4$ (and not the other way around).
Therefore, $\bigstar = 3 \times 3 + 2 \times 6 = 21$.

Addendum:

We must still prove that values of $\zeta$, $\eta$ and $\theta$ exists (and are different from the other symbols. An example is $\eta = 5$, which implies $\theta = 7$ and $\zeta = 14$.

Mike Earnest mentions another possibility; because the symbols must be integers, but the 'intermediate' results not,

$\alpha$ and $\beta$ can be $-2$ and $-4$ (or vice versa).
That means that by II., $\gamma = -6$, by III., $\delta = -3$, by IV. $\varepsilon = -9$.
From V. it follows that $\alpha = -2$ and $\beta = -4$ (and not the other way around).
Therefore, $\bigstar = -3 \times -3 + -2 \times -6 = 21$ (again).

The addendum applies here as well.

Answer 2

Equation I implies

$\alpha = 2$ and $\beta = 4$ (or vice versa), or $\alpha = -2$ and $\beta = -4$ (or vice versa).

We can break this down into a few cases:

Neither $\alpha$ nor $\beta$ can be zero. Suppose $\alpha$ and $\beta$ are both positive. Then Equation I is equivalent to $(\ln \alpha)/\alpha = (\ln \beta)/\beta$. The function $f(x) = (\ln x)/x$ has a single maximum at $x = e$, as can be seen by taking the derivative of $f$. This implies that either $\alpha < e < \beta$ or $\beta < e < \alpha$. Suppose that $\alpha < \beta$. The only positive integers smaller than $e$ are 1 and 2, so $\alpha = 1$ or $\alpha = 2$. But if $\alpha = 1$, $\alpha^\beta = 1$ and $\beta^\alpha \neq 1$. So we must have $\alpha = 2$ and $\beta = 4$. By a similar logic, if $\beta < \alpha$ then $\beta = 2$ and $\alpha = 4$.

Second case:

If $\alpha$ and $\beta$ are both negative, then we have $(-1)^\beta \frac{1}{(-\alpha)^{-\beta}} = (-1)^\alpha \frac{1}{(-\beta)^{-\alpha}}.$ If both sides are to have the same sign, $\alpha$ and $\beta$ must both be even or both odd, and we have to have $(-\alpha)^{-\beta} = (-\beta)^{-\alpha}$. By the logic of the previous part, $-\alpha = 2$ and $-\beta = 4$ or vice versa.

Third case:

Finally, if $\alpha$ and $\beta$ have opposite sign, then WLOG let $\alpha < 0 < \beta$. Then we have $(-1)^\beta (-\alpha)^\beta = 1/\beta^{-\alpha}$, with $- \alpha > 0$. The right-hand side is obviously positive, and if the left-hand side is to be positive, we must have $\beta$ even. But then we have the left-hand side necessary greater than or equal to 1, while the right-hand side is strictly less than 1. So no such solution exists.

Regardless of the choice made here, we have from equations II–IV

$\gamma = \pm 6$, $\delta = \pm 3$, and $\varepsilon = \pm 9$.

Equation V then implies that

$\beta = \pm 4$ and $\alpha = \pm 2$, since the equation is not satisfied if $\alpha = \pm 4$ and $\beta = \pm 2$.

We can then find $\bigstar$ by noting that

adding equations VI & VII yields $\zeta + \theta = \delta \times \delta + \alpha \times \gamma = 9 + 12 = {\bf 21}.$ Note that $\alpha \times \gamma$ and $\delta \times \delta$ are positive regardless of the signs of $\alpha$, $\beta$, $\gamma$, and $\delta$ (they are all either all positive or all negative.)

Note that

$\eta$, $\zeta$, and $\theta$ are not uniquely determined by this system of equations; in fact, there are infinitely many possible values for each one.

Answer 3

Answer:

$$\bigstar=21$$

Proof:

The only pair of integers for which the conditions $a^b=b^a$ and $a\neq b$ are 4 and 2. initially it seems arbitrary which one is which. therefore: $$ \alpha=2, \beta=4 \lor \alpha=4,\beta=2 \lor \alpha=-4, \beta=-2 \lor \alpha=-2, \beta=-4$$ depending on which is correct, we get $$\gamma=\alpha+\beta\rightarrow \gamma=6 \lor \gamma = -6$$ $$\delta+\delta=\gamma \rightarrow \delta = 3 \lor \delta=-3$$ $$\varepsilon=\delta+\gamma\rightarrow \varepsilon = 9 \lor \varepsilon = -9$$ using the following expression we eliminate some posibilities for $\alpha$ and $\beta$ $$\delta^\beta=\varepsilon^\alpha$$ we have four options but the ones that give a true equialities are: $$3^4=9^2 \land 3^{-4}=9^{-2}$$ therefore $\alpha=2$, $\beta=4$ or $\alpha=-2$, $\beta=-4$. continuing to solve we see that- regardless of sign-: $$\zeta=\delta\times\delta+\eta=9+\eta$$ $$\theta=\alpha\times\gamma-\eta=12-\eta$$ FINALLY: $$\bigstar=\zeta+\theta=9+\eta+12-\eta=21$$ $\eta$ ended up not getting a value but for completeness sake i wil let $\eta = 1000$