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): $$ \text{I. }a\times a=a \\ \space \\ \text{II. }a<b<b^a+c^a+c^a \\ \space \\ \text{III. }a-b<c<a \\ \space \\ \text{IV. }c\times d+d\times d=b\times b\times d - e\times e \\ \space \\ \text{V. }a-c-e<d \\ \space \\ \text{VI. }d-c-e\times b=\bigstar $$

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 $f:S_1\to\Bbb Z$).

A complete answer is a correct answer which also proves that the solution is the only solution. In other words, there is no other possible value for $\bigstar$.

How is an Answer Accepted?

After the puzzle is asked, a one day grace period will be given, in which no answer will be accepted. After that day passes, the complete answer which makes the least assumptions will be accepted. If no complete answer will appear within the grace period, the first complete answer that appears after the grace period will be accepted.

Good luck!

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

Previous puzzles in the series:

Initial pack: #1 #2 #3 #4 #5 #6 #7

Next Puzzle


1 Answer 1


We claim that:


First note that:

$a\times a=a\implies a=0\text{ or }1$


If $a=1$, from $\text{II}$ we have $b<b+c+c\implies c>0$


We also have $c<1$ from $\text{III}$, but $c$ is integral and therefore can't be between $0$ and $1$


$a\neq1\implies a=0$

From $\text{II}$:

$0<b<b^a+c^a+c^a=1+1+1=3$, so $b=1\text{ or }2$


$a-b<c<a$, so $b$ must be at least $2$ since all three values are integral


$b=2$ and we deduce that from $\text{III}$, $-2<c<0\implies c=-1$ since $c$ is integral

Now, substituting into $\text{IV}$ and $\text{V}$:

$-d+d^2=4d-e^2$ and $1-e<d$

We know that:

Squares are non-negative, and since $a=0$, $e\neq0$, so $e^2\geq1$

which allows us to deduce:

$-d+d^2<4d\implies d^2-5d<0$

But we can factorise this to:


Which implies:

$d\in[1,4]$, though since $b=2$ we know $d\neq2$

Case bashing:

$d=1\implies-1+1=4-e^2\implies e=\pm2$, but since $b=2$, $e=-2$, but this doesn't satisfy $1-e<d\iff3<1$

$d=3\implies-3+9=12-e^2\implies e=\pm\sqrt{6}$, which is not integral in either case

$d=4\implies-4+16=16-e^2\implies e=\pm2$, but since $b=2$, $e=-2$ (note that this satisfies $1-e<d\iff 3<4$



  • $\begingroup$ Why does $a\times a$ must be 1 or 0? $\endgroup$
    – NODO55
    Commented Jan 21, 2018 at 16:09
  • 1
    $\begingroup$ @NODO55 since $a\times a=a\implies a^2-a=0\implies a(a-1)=0\implies a=0\text{ or }a-1=0\implies a=0\text{ or }1$ $\endgroup$
    – boboquack
    Commented Jan 21, 2018 at 21:19

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