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

  • $\begingroup$ NOTE: once an answer is accepted, the next one in the series will be posted shortly after. Side Note: to get $\bigstar$ use \bigstar in MathJax, and to get $\varepsilon$ use \varepsilon. This could come in handy when writing an answer to this $\endgroup$ – NODO55 Jan 17 '18 at 18:10
  • $\begingroup$ Additional note to proving only solution: you do not need to prove Eq.1's conclusions to get your answer accepted for this puzzle, as proving that is not easy at all. You are more than welcome to though. $\endgroup$ – NODO55 Jan 17 '18 at 18:26
  • $\begingroup$ Nice twist, the negative integers. I believe I've covered everything now. $\endgroup$ – Glorfindel Jan 17 '18 at 18:54

$\bigstar = 21$


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, $\bigstar = 3 \times 3 + 2 \times 6 = 21$.


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.

  • $\begingroup$ You sure about your conclusion for Eq.1 ? $\endgroup$ – NODO55 Jan 17 '18 at 18:30
  • $\begingroup$ @NODO55 yes, you can see that later on I use V. to determine if $\alpha = 2$ or $\beta = 2$. $\endgroup$ – Glorfindel Jan 17 '18 at 18:32
  • 4
    $\begingroup$ Why not $\alpha=-2,\beta=-4$? $\endgroup$ – Mike Earnest Jan 17 '18 at 18:36
  • $\begingroup$ @MikeEarnest thanks. I checked it out, and the value for $\bigstar$ is still 21, but this is definitely something which is part of the puzzle. $\endgroup$ – Glorfindel Jan 17 '18 at 18:54
  • $\begingroup$ $\alpha$ and $\beta$ cannot have opposite signs, as one would end up larger than 1 and the other smaller than 1. Other than that, nice job. Just edit that tiny bit. $\endgroup$ – NODO55 Jan 17 '18 at 18:58

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.

  • $\begingroup$ you only got half of the proof covered. You missed something with your first conclusion. $\endgroup$ – NODO55 Jan 17 '18 at 18:28
  • $\begingroup$ The solution is not the entire set of symbols, just the number that substitutes $\bigstar$, as I mentioned in section "What is a solution". Therefore, is there is only one possible value for $\bigstar$, it has one single solution. The definition on uniqueness of a solution, is not defined at all. $\endgroup$ – NODO55 Jan 17 '18 at 18:34
  • $\begingroup$ @NODO55: I've fixed the only problem I could see with my treatment of Equation I; is that what you were referring to? I've also removed the words about the solution being non-unique; I was assuming that a solution was a solution for all variables. $\endgroup$ – Michael Seifert Jan 17 '18 at 18:38
  • $\begingroup$ Note Mike Earnest 's comment on Glorfindel's answer $\endgroup$ – NODO55 Jan 17 '18 at 18:40
  • $\begingroup$ @NODO55: Ah, I misread your comment about negative integers. Let me see here... $\endgroup$ – Michael Seifert Jan 17 '18 at 18:42




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 $\eta = 1000$

  • $\begingroup$ I think you confused one of the operations. $\endgroup$ – NODO55 Jan 17 '18 at 18:37
  • $\begingroup$ @NODO55 yup, just fixed it. $\endgroup$ – Sebastián Mestre Jan 17 '18 at 18:39
  • $\begingroup$ Now try to find the conclusion you missed from Eq.1 $\endgroup$ – NODO55 Jan 17 '18 at 18:43
  • $\begingroup$ took me a while but i found that (-2)^(-4) = (-4)^(-2) $\endgroup$ – Sebastián Mestre Jan 17 '18 at 19:04

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