Find the value of $\bigstar$: Puzzle 7 - Boss Battle

Question

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

Answer 1

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 $\bigstar = 6$.

Answer 2

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

$6$.