There is a machine, which when you put a number card which has written $x$ on it, it will output a number card with $$x^2+10x+20$$ on it. John has a number card on his hand, then repeated the following procedure ten times:

Put the card on his hand in the machine, and get the output.

John found out that he had the number card $0$ on his hand after repeating the procedures.

What number could have been written on John’s card at first?

There is a clever solution with an 'aha' moment.

Problem by me

  • 3
    $\begingroup$ Why downvote? There is a beautiful solution. $\endgroup$ Commented Jun 26, 2020 at 12:47

2 Answers 2


This is probably similar to msh210's idea but if we write

$$p(x) = (x+5)^2 - 5 $$ Then, we notice that $$ p(p(x)) = (p(x) + 5)^2 - 5 =((x+5)^2 - 5 + 5)^2 - 5 = (x+5)^4 - 5 $$ and for $k$ iterations $$p^k(x) = p(p(\ldots p(x)\ldots)) = (x+5)^{2^k} - 5$$ so we just need to solve $$p^{10}(x) = (x+5)^{1024} - 5 = 0 $$ which has solutions $$x = -5 + 5^{1/1024}$$ where the $1024$th root can be any such root. This is what others have found, just expressed in a different way.

  • $\begingroup$ Yes, this is my solution! $\endgroup$ Commented Jun 26, 2020 at 14:13

First, note that the minimum of $x^2+10x+20$ is equal to $-5$, so equations like $x^2+10x+20=y$ are only solvable for $y \ge -5$

Now solve the general case equation: $x^2+10x+20=y$. The solutions are $-5-\sqrt{y+5}$ and $-5+\sqrt{y+5}$. Now we have to repeat the process 10 times, starting with $y = 0$. If we choose the new $y$ to be the first solution (except for the last iteration), the next equation won't be solvable.

The first solution is $-5+\sqrt{5}$, the second solution is $-5+\root^4\of{5}$, the third solution is $-5+\root^8\of{5}$ and so on. In general, the $n$th solution is $-5+5^{(1/2)^n}$. The solution on the ninth iteration is thus $-5+\root^{512}\of{5}$.

Therefore, the two possible (real) starting numbers are the solutions of $x^2+10x+20 = -5+\root^{512}\of{5}$. These solutions are (using the general case formula above) $-5-\root^{1024}\of{5}$ and $-5+\root^{1024}\of{5}$.

  • $\begingroup$ Can you compose an elegant proof for the general formula? $\endgroup$ Commented Jun 26, 2020 at 12:52
  • $\begingroup$ @CulverKwan The -5+5^(1/2)^n one? It can be seen that to get the next iteration of the formula, when calculating the next solution, we add 5, take the square root and subtract 5 again, so every time after the first iteration, the first $-5$ is unchanged, while the second $5$'s power halves every iteration. $\endgroup$ Commented Jun 26, 2020 at 12:54

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