This puzzle replaces all numbers with other symbols.


All symbols abide to the following rules:

  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?


What is a Correct Answer?


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

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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 $1-e<d\iff 3<4$



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  • $\begingroup$ Why does $a\times a$ must be 1 or 0? $\endgroup$ – NODO55 Jan 21 '18 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 Jan 21 '18 at 21:19

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