# Find the value of $\bigstar$: Puzzle 10 - Uncertainty

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.

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\times a=a \\ \space \\ \text{II. }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?

# Good luck!

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 #9

1. Let's rewrite the equations to the more readable form: $$1.\ a^2=a$$ $$2.\ c $$3.\ b^c $$4.\ (d+1)b=c$$ $$5.\ d^2+2c $$6.\ e(b^2+1)=d$$ $$7.\ a+b^2+c+2d+e=\bigstar$$ 2. Now prove that all numbers (except possibly $$a$$) are nonzero. Firstly, from (1) we have $$a\leqslant 1$$ (because either $$a=0$$ or $$a=1$$), so $$b\leqslant 0$$ and $$c\leqslant-1$$, and $$c\ne0$$. Next, from (4) it's obvious that $$b\ne0$$, Now, since $$b^2>0$$, then $$2d$$ must be also $$>0$$ from (5), so $$d\ne0$$. Finally, from (6) we get $$e\ne0$$.
3. If we find a solution with $$a=1$$, we can safely replace it with $$a=0$$ (and other numbers the same), and it will affect neither the equations (since $$a$$ is only present in conditions (1) and (2), and it's easy to see that they will still hold when we replace $$a=1$$ with $$a=0$$, since we've already proven that $$b<0$$ (strictly negative)) nor the uniqueness of numbers (since we've proven that $$bcde\ne0$$), but will change the value of $$\bigstar$$ (as can be easily seen from (7)). So, to prove the uniqueness of $$\bigstar$$, we need to prove that $$a$$ is necessarily $$0$$ (but this is not sufficient).
4. We already know that $$b$$ and $$c$$ are both negative (from (2) and the fact that $$b\ne0$$), and $$d$$ and $$e$$ must be positive from (5) and (6). Now let's introduce $$x=-b=|b|$$ and $$y=-c=|c|$$, so all numbers in equations will be positive (of course, except $$a$$).
5. Now, we get $$d^2+2c=d^2-2y=e^2(x^2+1)^2-2e(x^2+1)x-2x=(e^2x^4-2ex^3+2ex^2-(2e+2)x+e^2) < x^2,$$ or $$e^2x^4-2ex^3+(2e-1)x^2-(2e+2)x+e^2<0$$ ($$e$$ and $$x$$ being positive integers). This polynomial (let's designate it $$P(x)$$) does monotonically increase on $$[2;+\infty)$$ (here we assume that $$x\geqslant 2$$, but see below for $$x=1$$) because its derivative $$P'(x)=4e^2x^3-6ex^2+(4e-2)x-(2e+2)$$ is positive for $$x=2$$ and any $$e\geqslant1$$ (being $$P'(2)=32e^2-18e-2$$), and $$P''(x)=12e^2x^2-12ex+(4e-2)$$ is positive for any $$x\geqslant2$$ and $$e\geqslant1$$ (so $$P'(x)$$ keeps its sign, also being positive everywhere for these values of $$x$$). So, that means that minimum value of $$P(x)$$ (for $$x\geqslant2$$) is $$P(2)=16e^2-16e+(8e-4)-(4e+4)+e^2=17e^2-12e-8$$. It can be less than zero only when $$e=1$$ (and therefore $$a=0$$, due to the uniqueness requirement).
6. When $$x=1$$, we get $$e^2-2e+(2e-1)-(2e+2)+e^2<0$$, or $$2e^2-2e-3<0$$, which is true again only for $$e=1$$.
7. So, $$e=1$$, $$a=0$$, $$d=x^2+1$$, $$x\geqslant1$$ and $$x^4-2x^3+x^2-4x+1<0$$ (so either $$x=1$$ or $$x=2$$, i.e. $$b=-1$$ or $$b=-2$$). If $$b=-1$$, we get $$d=2$$ and $$c=-3$$ (Jens' solution); if $$b=-2$$, we get $$d=5$$ and $$c=-12$$ (Oray's solution). In both cases, $$\bigstar=3$$.
8. Since we have proven that no other solutions exist, the proof is complete. Q.E.D.

If

$$a=0$$, $$b=-1$$, $$c=-3$$, $$d=2$$ and $$e=1$$ then the equations are fulfilled and $$\bigstar = 3$$

• is that the only solution though? – NODO55 Jan 3 '20 at 18:43
• This answer is exact copy of Oray... can we not just copy others answer and try to solve it ourself – Ms Designer Jan 4 '20 at 8:51
• @Ms Designer : Uh...this answer is not the same as Oray's answer and I posted first. – Jens Jan 4 '20 at 10:07

It seems a,b,c,d,e values are not unique; here is another answer;

a=0,b=-2;c=-12,d=5;e=1 but it still makes $$\bigstar$$=3.

• @nahmid d^2+2c<b2 – Oray Jan 3 '20 at 19:29
• could there be another solution? – NODO55 Jan 4 '20 at 0:35