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This puzzle replaces all numbers with other symbols.

Your job, as the title suggests, is to find what value fits in the place of $\bigstar$. To get the basic idea down, I recommend you solve Puzzle 1 first.

All symbols abide to the following rules:

  1. Each numerical 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): $$ \text{I. }a\times a=a \\ \space \\ \text{II. }b=a+a+a+a \\ \space \\ \text{III. }b^c+d^c=e^c \\ \space \\ \text{IV. }b-d=c\times d+f \\ \space \\ \text{V. }f=(e+a+a)\times g \\ \space \\ \text{VI. }b\times c=\bigstar+g \\ \space \\ \text{VII. }b\times\bigstar+c=a+f $$

What is a Solution?

A solution is a 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 these functions, 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 values (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:

#1 #2 #3 #4 #5 #6

Next Puzzle

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  • 1
    $\begingroup$ NOTE: This is the last puzzle in this series (that is already made). If you enjoyed this, tell me so that I know working on another 7 puzzles would be worth it. Side Note: to get $\bigstar$ use \bigstar in MathJax. This could come in handy when writing an answer to this. BONUS: Since this is the last puzzle in the series, which has never been solved by anyone other than me, the first person to solve this puzzle may grant himself the title of Mercury. $\endgroup$ – NODO55 Jan 18 '18 at 18:58
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    $\begingroup$ I have discovered a marvelous proof of the solutions for Eq. III, which this comment box is too small to contain. $\endgroup$ – Michael Seifert Jan 18 '18 at 19:40
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    $\begingroup$ @MichaelSeifert when writing an answer, make sure to use proper Fermatting, it helps out with seeing the solution more clearly. $\endgroup$ – NODO55 Jan 18 '18 at 20:53
  • $\begingroup$ To clarify: does your condition 2 mean that all the $a,b,c,d,e,f,g,\bigstar$ here must be distinct, or only that any two which appear in the same equation must be distinct? $\endgroup$ – Rand al'Thor Jan 18 '18 at 22:06
  • $\begingroup$ if they appear in the same puzzle (and yes, they do) they must be distinct. $\endgroup$ – NODO55 Jan 19 '18 at 7:30
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As far as I can tell, the solution is

$\bigstar = 6$.

Proof:

The first equation implies that $a^2 - a = a (a - 1) = 0$, which implies that $a = 1$ or $a = 0$. Since $4a = b$ and $a \neq b$, this means that $a = 1$ and $b = 4$.

Examining Eq. IV, we note that

We cannot have $c = -1$ or $d = 0$, as in either case we have $b = f$.

This fact will help us in examining Eq. III:

Eq. III is Fermat's Last Theorem in disguise. If $c = 0$, there is no solution. Assume temporarily that $e$ is non-zero. If either $d$ or $e$ are negative, the overall equation is equivalent to an equation of the form $b^c \pm |d|^c = \pm |e|^c$, which can be rearranged into the form $x^c + y^c = z^c$ for $\{x, y, z\}$ being some permutation of $\{b, |d|, |e|\}$–i.e., all positive integers. If $c > 0$, this will be the standard version of Fermat's Last Theorem, and so we know that $c \leq 2$. If $c < 0$, then we can rewrite this equation as $(yz)^{-c} + (xz)^{-c} = (xy)^{-c}$, and by the same logic, there cannot be a solution if $-c > 2$. We have already ruled out $c = +1 = a$, and $c = -1$ (in the block above). Thus, we have $c = -2$ or $c = +2$.

Taking these two cases in turn:

If $c = +2$, then $(b, \pm d, \pm e)$ form a Pythagorean triple. (Note that we already established $d \neq 0$.) The only Pythagorean triple containing $b = 4$ is $(4, 3, 5)$. (This can be proven by using standard results concerning the generation of Pythagorean triples.) Thus, $d = \pm 3$ and $e = \pm 5$. If $d = +3$, then Eq. IV implies that $f = -5$, and Eq. V implies that either $-5 = 7g$ or $-5 = (-3)g$, which means $g$ is not an integer. If $d = -3$, then Eq. IV implies that $f = 1 = a$, which is not allowed.

For the other case,

When $c = -2$, we cannot have $d = 0$ or $e = 0$. Under this assumption, Eq. III becomes $(de/4)^2 + e^2 = d^2$. Rearranging yields $e^2 = d^2 (1 -e^2/4)$. This implies that $|e| < 2$, as otherwise the right-hand side would be negative (or zero) while the left-hand side is positive. Thus, we must have $e = \pm 1$, which implies that $1 = d^2 * (3/4)$, and thus $d$ cannot be an integer.

All of this taken together then implies that

the bolded assumption above is false: $e = 0$, which implies that $d = -4$ and $c$ is odd.

From here, it's a straight run to the finish:

The remaining equations are equivalent to $8 = -4c + f$, $f = 2g$, $-4c = \bigstar + g$, and $4\bigstar + c = 1 + f$. This is a linear system of four equations and four unknowns, and it has a unique solution. Solving this system via a large amount of tedious algebra yields $c = 5$, $f = 28$, $g = 14$, and $\bigstar = 6$.

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  • $\begingroup$ I've probably missed a loophole somewhere in this proof. Any hints as to where it is would be appreciated. $\endgroup$ – Michael Seifert Jan 18 '18 at 21:27
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    $\begingroup$ Fermat's last theorem is only valid for natural numbers $b$, $d$, $e$. But you are allowed to use any integer numbers. $\endgroup$ – A. P. Jan 18 '18 at 22:04
  • $\begingroup$ ... And there's the loophole. I had a mental argument against negative values of $d$ and $e$ (now articulated), but I had missed the case where $e = 0$. Thanks, @A.P. $\endgroup$ – Michael Seifert Jan 18 '18 at 22:28
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  1. $I$ has two possible solutions:

    $a=0$ or $a=1$. These are the only solutions in $\mathbb{C}$ (let alone $\mathbb{Z}$) to the equation $a^2-a=0$.

  2. Given the above, we can tell from $II$ that:

    $a=1$ and $b=4$, since the alternative would be $a=b=0$ and we know $a\neq b$.

  3. Considering $III$ in light of Fermat's last theorem, we know that

    if we assume $d,e\geq0$, then $c\not>2$ and $c\not<-2$. Clearly $c$ cannot be $0$, since then the equation would be $1+1=1$ and invalid. Nor can it be $1$, since then $a=c$, contradiction. Nor can it be $-1$, since then equation $IV$ would yield $b=f$, contradiction. Nor can it be $-2$, since then $$4^{-2}+d^{-2}=e^{-2}\Rightarrow d^2+16=\Big(\frac{4d}{e}\Big)^2\Rightarrow d=3,\frac{4d}{e}=5.$$ So we must have $c=2,d=3,e=5$. Then $IV$ gives $f=-5$ and $V$ gives $-5=7g$, contradiction.

    Thus we must have

    at least one of $d$ and $e$ being non-positive. Clearly $d=0$ is impossible, since it would give $e=b$. Negative $d$ and positive $e$ would rearrange to an equation of the same FLT form with all bases positive, so that's out too. ($c=2,d=5,e=3$ is just as much of a contradiction from the next two equations as $c=2,d=3,e=5$ was.) Ditto for negative $d$ and negative $e$, while positive $d$ and negative $e$ is just impossible. So we must have $e=0$.

    I went with this to get

    $d=-4$ and $c$ odd.

  4. Now $IV,V,VI,VII$ become:

    • $f=8+4c$.

    • $f=2g$, so $g=4+2c$.

    • $4c=\bigstar+g$, so $\bigstar=2c-4$.

    • $4\bigstar+c=1+f$, so $9c-16=9+4c$, giving $c=5$.

So the solution is:

$a=1,b=4,c=5,d=-4,e=0,f=28,g=14$

with the variable $\bigstar$ being

$6$.

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  • $\begingroup$ You may now grant yourself the title of Mercury and brag about it. Work ongoing to find loopholes in the proofs $\endgroup$ – NODO55 Jan 19 '18 at 7:47

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