David Ding: Riddler Express May 7, 2021

Trio of Squares

Can you find three distinct numbers such that the second is the square of the first, the third is the square of the second, and the first is the square of the third? Assuming you can, what are three such numbers?

The first of the three numbers is the all non-1 complex solutions to the equation

a^{7} = 1

Explanation:

Without the 'distinct' clause, we can easily think of two numbers, 0 and 1, who fit the description. But alas, is life ever that easy? Let us have a triplet,

(a, b, c)

, that represents the three numbers in the puzzle respectively. From the givens of the problem, we have:

\begin{aligned} b & = a^{2} \\ c & = b^{2} = a^{4} \\ a & = c^{2} = b^{4} = a^{8} \end{aligned}

Therefore, we are left to solve the following equation given that

a \neq 0

and

a \neq 1

\begin{aligned} a^{8} & = a \\ a^{7} & = 1 \end{aligned}

Since

a \neq 0

, we can cancel out an

a

from both sides of the equation.

Adding complexity to the situation, no pun intended, we have 7 roots for the above polynomial equation. Of course, we must ignore the

a = 1

case as that would not make the other two numbers distinct from

a

. This leaves us with the following, for

n = 1, 2, 3, 4, 5, 6

\begin{aligned} a^{7} & = 1 \\ a^{7} & = e^{2 π n i} \\ a & = e^{\frac{2 π n i}{7}} \end{aligned}

When

n = 7

, we get back the real number 1 for

e^{\frac{2 π n i}{7}}

, and the cycle repeats beyond that every multiple of 7 for

n

. Also remember

e^{(2 k) π n i} = 1

for all integer values of

k

So we have six triplets for

(a, b, c)

that satisfy our answer:

$n$	$a$	$b = a^{2}$	$c = b^{2}$	$c^{2}$
1	$e^{\frac{2 π i}{7}}$	$e^{\frac{4 π i}{7}}$	$e^{\frac{8 π i}{7}}$	$e^{\frac{16 π i}{7}}$ = $e^{\frac{(14 + 2) π i}{7}}$ = $e^{2 π i} e^{\frac{2 π i}{7}} = e^{\frac{2 π i}{7}} = a$
2	$e^{\frac{4 π i}{7}}$	$e^{\frac{8 π i}{7}}$	$e^{\frac{16 π i}{7}}$	$e^{\frac{32 π i}{7}}$ = $e^{\frac{(28 + 4) π i}{7}}$ = $e^{4 π i} e^{\frac{4 π i}{7}} = e^{\frac{4 π i}{7}} = a$
3	$e^{\frac{6 π i}{7}}$	$e^{\frac{12 π i}{7}}$	$e^{\frac{24 π i}{7}}$	$e^{\frac{48 π i}{7}}$ = $e^{\frac{(42 + 6) π i}{7}}$ = $e^{6 π i} e^{\frac{6 π i}{7}} = e^{\frac{6 π i}{7}} = a$
4	$e^{\frac{8 π i}{7}}$	$e^{\frac{16 π i}{7}}$	$e^{\frac{32 π i}{7}}$	$e^{\frac{64 π i}{7}}$ = $e^{\frac{(56 + 8) π i}{7}}$ = $e^{8 π i} e^{\frac{8 π i}{7}} = e^{\frac{8 π i}{7}} = a$
5	$e^{\frac{10 π i}{7}}$	$e^{\frac{20 π i}{7}}$	$e^{\frac{40 π i}{7}}$	$e^{\frac{80 π i}{7}}$ = $e^{\frac{(70 + 10) π i}{7}}$ = $e^{10 π i} e^{\frac{10 π i}{7}} = e^{\frac{10 π i}{7}} = a$
6	$e^{\frac{12 π i}{7}}$	$e^{\frac{24 π i}{7}}$	$e^{\frac{48 π i}{7}}$	$e^{\frac{96 π i}{7}}$ = $e^{\frac{(84 + 12) π i}{7}}$ = $e^{12 π i} e^{\frac{12 π i}{7}} = e^{\frac{12 π i}{7}} = a$

n

a

b = a^{2}

c = b^{2}

c^{2}

e^{\frac{2 π i}{7}}

e^{\frac{4 π i}{7}}

e^{\frac{8 π i}{7}}

e^{\frac{16 π i}{7}}

e^{\frac{(14 + 2) π i}{7}}

e^{2 π i} e^{\frac{2 π i}{7}} = e^{\frac{2 π i}{7}} = a

e^{\frac{4 π i}{7}}

e^{\frac{8 π i}{7}}

e^{\frac{16 π i}{7}}

e^{\frac{32 π i}{7}}

e^{\frac{(28 + 4) π i}{7}}

e^{4 π i} e^{\frac{4 π i}{7}} = e^{\frac{4 π i}{7}} = a

e^{\frac{6 π i}{7}}

e^{\frac{12 π i}{7}}

e^{\frac{24 π i}{7}}

e^{\frac{48 π i}{7}}

e^{\frac{(42 + 6) π i}{7}}

e^{6 π i} e^{\frac{6 π i}{7}} = e^{\frac{6 π i}{7}} = a

e^{\frac{8 π i}{7}}

e^{\frac{16 π i}{7}}

e^{\frac{32 π i}{7}}

e^{\frac{64 π i}{7}}

e^{\frac{(56 + 8) π i}{7}}

e^{8 π i} e^{\frac{8 π i}{7}} = e^{\frac{8 π i}{7}} = a

e^{\frac{10 π i}{7}}

e^{\frac{20 π i}{7}}

e^{\frac{40 π i}{7}}

e^{\frac{80 π i}{7}}

e^{\frac{(70 + 10) π i}{7}}

e^{10 π i} e^{\frac{10 π i}{7}} = e^{\frac{10 π i}{7}} = a

e^{\frac{12 π i}{7}}

e^{\frac{24 π i}{7}}

e^{\frac{48 π i}{7}}

e^{\frac{96 π i}{7}}

e^{\frac{(84 + 12) π i}{7}}

e^{12 π i} e^{\frac{12 π i}{7}} = e^{\frac{12 π i}{7}} = a

This puzzle was "complex" after all!

Riddler Express: May 7, 2021

Trio of Squares