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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

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  • $\begingroup$ 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 19:11
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    $\begingroup$ (You maybe should consider making these with letters, not symbols, so answering them doesn't require tedious mucking around with MathJax in spoiler-tags. The effort in answering should be in coming up with the solution, more than in figuring out how to make its formatting work.) $\endgroup$ – Rubio Jan 17 '18 at 19:22
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    $\begingroup$ Noted. I will start doing do in Puzzle 5 $\endgroup$ – NODO55 Jan 17 '18 at 20:06
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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 $\bigstar = 19$.

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  • $\begingroup$ Nice, just make sure to also include proof as to why $\zeta$ must be either 1 or 0. $\endgroup$ – NODO55 Jan 17 '18 at 20:13
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$\bigstar = 19$

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 $\alpha−\delta−\beta = \bigstar = 19$.

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  • $\begingroup$ Pretty sure you've got an error going from the fourth to the fifth line of your second spoiler block. $\endgroup$ – Michael Seifert Jan 17 '18 at 19:26
  • $\begingroup$ @MichaelSeifert eh ... yes. Thanks. $\endgroup$ – Glorfindel Jan 17 '18 at 19:31
  • $\begingroup$ Wait... is your answer the one in block 1 or block 2? $\endgroup$ – NODO55 Jan 17 '18 at 20:08
  • $\begingroup$ o_O ... solved a mistake in the derivation, but forgot to update the solution itself ... $\endgroup$ – Glorfindel Jan 17 '18 at 20:10
  • $\begingroup$ Nice, just make sure to also include proof as to why $\zeta$ must be either 1 or 0. $\endgroup$ – NODO55 Jan 17 '18 at 20:13
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$\bigstar=19$

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, $$\gamma=2\\\varepsilon=3\\\delta=5\\\alpha=17\\\beta=-7\\\bigstar=19$$

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  • $\begingroup$ the solutions for that quadratic equation is not {2,0} $\endgroup$ – NODO55 Jan 17 '18 at 20:11
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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:
$$\bigstar=24-5=19$$

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