Blackboard problem with polynomial

Question

The blackboard contains the two numbers 0 and 120. Each minute you are allowed to write an additional number $ x $ on the blackboard if $ x$ has not been written before on the blackboard and if $ x $ is the integer root of some polynomial $ P(x) $ in one variable whose coefficients are numbers on the blackboard. The game ends when you can not write further numbers on the blackboard.

Which numbers will be written on the blackboard when the game ends?

Answer 1

By the rational root theorem, any integer root of the polynomial $$a_nx^n + \dots + a_1x + a_0$$ is a divisor of its constant term $a_0$, if $a_0 \neq 0$.

We will never let $a_0 = 0$, since that means we can divide the whole polynomial by $x$ (repeatedly, if necessary; yielding a root $0$ each time, which is initially already on the board) and get a polynomial (with the same non-zero roots) for which $a_0 \neq 0$.

Thus initially, $a_0 = 120$, our only choice. From this, suppose we get a new root $p$, a divisor of $a_0 = 120$. Then the next minute, we can choose from $\{p, 120\}$, both divisors of 120, etc. Clearly, the integer roots we generate by this process will always be divisors of 120, positive or negative.

We show how to achieve all divisors of 120:

• First, $120x+120$ gives us $x=-1$.

• Then, $(-1)x^{120}+(-1)x^{119}+\dots+(-1)x^1+120$ gives us $x=1$.

• Then, $x+120$ gives us $x=-120$.

We can get $2$, $3$ and $5$ using only the coefficients in $\{\pm 120, \pm 1\}$:

• $-x^7 + x^3 \pm 120$ gives us $x = \pm 2$.
• $-x^4 -x^3 -x^2 -x \pm 120$ gives us $x = \pm 3$.
• $-x^3 +x \pm 120$ gives us $x = \pm 5$.

Aravind found a nice trick -- now that we have $2$, $3$, and $5$ on the blackboard, the other roots follow by dividing from $120 = 2^3 \cdot 3 \cdot 5$. (If we have $p$ and $q$ on the board, we can use $px-q$ to get $x=q/p$, so first take $q=120$ and then repeat the process with $q \to q/p$ for a prime $p$.)

We can conlude that the solution set is precisely the set of positive and negative divisors of 120.

Answer 2

Every integer can be written on the board.

First polynomial: $$P(x) = 120x - 120$$

The root of this polynomial is 1.

Any integer:

$$P(x) = x - x$$

In this solution, it is important to note that constants are not coefficients. Also note that $-$ is an operator, not part of the coefficient. Had it been intended as the coefficient, I would have written $x + -x$ (or even more explicitly $1x + -1x$).

Edit: As an added shortcut (thanks to f''), you could simply use this polynomial from the initial board to retrieve any value for $x$:

$$P(x) = 0x$$

Answer 3

I believe the OP's intended answer is:

Exactly the divisors of 120. This is true of the non-zero numbers on the board at any stage. If $P(x)=0$, then we can write it as $xg(x)+d=0$ and $d$ is a divisor of 120, this implies that so is $x$.

StillLearning's answer shows that all of them can be obtained.

Once $2,3,5$ are obtained, other divisors of 120 can be obtained by repeated division, eg: $2x-120=0$ to get $60$, then $2x-60-0$ to get $30$ etc.

Answer 4

I started like this

$$\begin{align} 120x + 120 = 0 &\rightarrow& x &= &-1\\ - x^{120} - x^{119} - ... - x^2 - x^1 +120 = 0 &\rightarrow&x &= &1\\ x + 120 = 0 &\rightarrow& x& = &-120\\ x^7 - x^3 + 120 = 0 &\rightarrow& x &= &-2\\ x - 2 = 0 &\rightarrow& x &= &2\\ \end{align} $$

Now we have $-120, -2, -1, 0, 1, 2, 120$.

Having both $-1$ and $1$ means that I can always get the negated value of any new number.

But I can't crack this puzzle as I keep finding new numbers.

Wonder if I got it all wrong...

Edit:

Some more (thanks to Aravind for the first 3)...

$$ \begin{align} x^2 + 2x - 120 = 0 &\rightarrow& x &= &10\\ 2x - 10 = 0 &\rightarrow& x &= &5\\ 10x - 120 = 0 &\rightarrow& x &= &12\\ x^4 + x^3 + x^2 + x - 120 = 0 &\rightarrow& x &= &3\\ \end{align} $$

And

$$ \begin{align} 2x - 120 = 0 &\rightarrow& x &= &60\\ 3x - 120 = 0 &\rightarrow& x &= &40\\ 5x - 120 = 0 &\rightarrow& x &= &24\\ 5x^4 + 10x - 120 &\rightarrow& x &= &4\\ 4x - 120 = 0 &\rightarrow& x &= &30\\ 4x^2 + 4x -120 &\rightarrow& x &= &5\\ 5x -120 = 0 &\rightarrow& x &= &24\\ 3x^2 + 2x - 120 = 0 &\rightarrow& x &= &6\\ 6x - 120 = 0 &\rightarrow& x &= &20\\\\ -2x^2 - x + 120 = 0 &\rightarrow& x &= &-8\\\\ -8x + 120 = 0 &\rightarrow& x &= &15\\ \end{align} $$

So far I have

$\{0, 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120\}$ (and all the negated numbers as well).

As far as I understand from other answers/comments this should be the complete list. I still wonder how that is proved.