# The special machine

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

• Why downvote? There is a beautiful solution. Jun 26 '20 at 12:47

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.

• Yes, this is my solution! Jun 26 '20 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 $$-5+\root^{1024}\of{5}$$.

• Can you compose an elegant proof for the general formula? Jun 26 '20 at 12:52
• @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. Jun 26 '20 at 12:54