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

  • $\begingroup$ 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
    Commented Jan 17, 2018 at 20:25

2 Answers 2



$\bigstar = 7$


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.

  • 1
    $\begingroup$ The upper bound in the last equation is $\alpha^{\zeta}$, not $\alpha^{\gamma}$, so there is a unique solution. $\endgroup$
    – A. P.
    Commented Jan 17, 2018 at 21:43
  • 1
    $\begingroup$ The mental typo strikes again ... $\endgroup$
    – Glorfindel
    Commented Jan 17, 2018 at 21:47
  • $\begingroup$ in your proof of ass. A, why does the equation $\alpha={n\over n-1}$ mean that $n$ is either 0 or 2? $\endgroup$
    – NODO55
    Commented Jan 18, 2018 at 5:14
  • $\begingroup$ @NODO55 e.g. 4/3, 11/10 or -2/-3 can never be an integer. That division only 'works' if $n-1$ is $\pm1$. $\endgroup$
    – Glorfindel
    Commented Jan 18, 2018 at 6:32
  • $\begingroup$ Again, why? Hint: try to get the equation of $\alpha$ to a form of $a+{b\over n-1}$ where $a,b$ are numbers, and see what you can derive from that. $\endgroup$
    – NODO55
    Commented Jan 18, 2018 at 6:37

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 $\bigstar \neq \eta = 6$.

  • $\begingroup$ "The only two square numbers that differ by 4 are 0 and 4". Why? $\endgroup$
    – NODO55
    Commented Jan 18, 2018 at 5:16
  • 1
    $\begingroup$ @NODO55: For any square number $n^2$, the next square number differs from it by $2n+1$. This means that for $n > 2$, the difference between consecutive square numbers is greater than 4. (And the difference between non-consecutive square numbers is even larger.) $\endgroup$ Commented Jan 18, 2018 at 12:14
  • $\begingroup$ Nice. Your proof is now complete $\endgroup$
    – NODO55
    Commented Jan 18, 2018 at 12:18

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