mathematics - Find the value of $bigstar$: Puzzle 9 - Options

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, I recommend you solve Puzzle 1 first.

All symbols follow these 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 $\beta$ in the puzzle, $\alpha\neq\beta$.


3. The following equations are satisfied (this is the heart of the puzzle): $$ \text{I. }a^a=a \\ \space \\ \text{II. }a+a+a=b \\ \space \\ \text{III. }c

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.

Can you prove that there is only one possible value for $\bigstar$, and find that value?

Side Note: to get $\bigstar$ use $\bigstar$, and to get $\text^$ use $\text^$

---

Previous puzzles:

Introduction: #1 #2 #3 #4 #5 #6 #7

Inequalities: #8

Next Puzzle

Answer

We must have

$\bigstar=1$.

Proof:

V is true when f=g, of course, but that won't do; also when {f,g}={2,4} or when {f,g} = {-2,-4}; and those are all the integer solutions to V. So f,g are 2,4 in some order or -2,-4 in some order. The latter is impossible because VI says positive < (f+g) . positive, which is impossible if f+g is negative.

Now

I is true only when a is -1 or +1. (You might argue for 0, but II forbids that since then we'd have b=a.) By II, b is then -3 or +3.

Let's look first at the case where

a=1, b=3. Then III says c < d < c^3(3-c). Can we have c>0? No! The LHS is positive so the RHS had better be too, so c<3. But 1,2,3 are already taken, contradiction. Can we have c=0? No! We get 0 < d < 0. What if c<0? Then the bounds on d are the wrong way around, because c^3(3-c) is always more negative than c. (Because it's c times c^2(3-c); the latter is the product of two positive factors, at least one of which must be >1.)

So

a=-1, b=-3. Now III says c < d < c^-3.(-3-c). If c>0 then that RHS is negative, so we have positive < something < negative, contradiction; if c=0 then it says 0 < d < 0, contradiction, so c<0. Then either c=-2 or c<-3. In the former case we have -2 < d < -1/8.(-3 - -2) = 1/8 which (since -1 is already taken) requires d=0. Then IV says -1 . (-2+0) = e^2 which is impossible. So c<-3. Then c < d < 1/c^3 . (-3-c) and the RHS always lies between -1 and 0. So d is negative (and must in fact be <= -2, since -1 is already taken) but less negative than c. IV then says that -(c+d) = e^2.

Note now that

f+g=6 so VI says c^2+d^2 < 6e^2 = -6(c+d). That's quite constraining. In particular it implies -(c+d) < 12. (Because c^2+d^2 >= (c+d)^2/2, so (c+d)^2/2 < -6(c+d); dividing by the positive quantity -(c+d)/2 we get -(c+d) < 12.) This thing has to be a square. Which square? Well, c <= -4 and d <= -2, so 6 = 2+4 <= -(c+d) <= 12. The only square in this range is 9.

But

the only ways to make 9 with our given constraints so far are (-7,-2) and (-5,-4). We can readily check that the former violates VI. So c=-5 and d=-4. And we have e^2=9 so e must be 3. (-3 is already taken.)

Our remaining conditions now look like

-2 < h < 2^f and h < $\bigstar$ < 3-h. The latter implies 3-h >= h+2 or h <= 1/2, so in fact we must have h=0 and then $\bigstar$ is either 1 or 2 -- but 2 is already taken, so $\bigstar$ is 1.

We have nailed down

specific values for all variables except f,g, which could be either 2,4 or 4,2.