# Cyclic squaring

Find a complex number $$a$$, which satisfies the following conditions:

1. If you square $$a$$, you get a different complex number $$b$$.
2. If you square $$b$$, you get a complex number $$c$$, which is different from $$a$$ and $$b$$.
3. If you square $$c$$, you get $$a$$.
It turns out that there is more than one solution for $$a$$. What is the sum of all possible solutions?
There is a "puzzling"-like solution, which doesn't require a computer.
• This was asked as a Riddler question on 538. Jun 23 at 16:49
• @ParclyTaxel You're right – I wasn't aware of this question. But this question goes one step further by asking for the sum of the solutions, which gives a smoother solution than $a$ itself. Jun 23 at 16:58

For the source of this problem see the comment on the question. Anyway

we have $$a=c^2=b^4=a^8$$, so (since $$a=0$$ and $$a=1$$ can be quickly ruled out) $$a^7=1$$, i.e. $$a$$ is a non-trivial seventh root of unity. There are six possible solutions for $$a$$$$e^{2ik\pi/7}$$ for $$1\le k\le6$$ – and their sum is $$-1$$ as can be seen by noting that all seven roots form a regular heptagon in the complex plane (and must therefore sum to $$0$$), one of which is $$1$$.

• A simple way to reach the final solution without plotting is to realise that the roots form a regular heptagon centred on the origin. Jun 23 at 16:56
• @hexomino I know this in my head, as Maths Stack extraordinaire. Jun 23 at 16:59
• You can alternatively use their polynomial that they are roots of (so bypassing roots of unity idea) combined with Vieta's relations to find their sum. Jun 24 at 0:14
• @TheBestMagician Vieta is overkill here as we know the solutions are x,x^2,x^3,...,x^6 and the cyclotomic poly is 1+x+x^2+x^3+...+x^6 from which the desired sum can be read oiff directly. Jun 24 at 5:44

For anyone who's ever heard of roots of unity or cyclotomic polynomials this is very simple, so let's assume we haven't.

We will only use one not completely elementary tool and that is the fact that an n-th degree polynomial has n complex roots (counting multiplicities).

Substituting a for c for b in the given system of equations yields an order 8 polynomial in a, therefore there are 8 solutions (the exact number is not important; all we will use is the fact that there are more than zero and fewer than infinity solutions other than zero and one). Exactly two of those are ruled out because of a=b, viz. a=1 and a=0. We can therefore determine the sum of all nonzero solutions and then subtract 1.

Observe that if a and a' are solutions other than zero then so are 1/a and aa'. From this it follows that if we take one solution a and form the products aa1,aa2,... for all nonzero solutions ai then each solution is replaced with a different solution but the set as a whole is unchanged. It follows that the sum S must satisfy aS=S for all solutions a which is only possible if S=0; subtracting 1 we get the final answer -1.