Gauss: Seventeenth Root of Unity Expressed As Radicals

Carl Friedrich Gauss was 18 when he was able to solve the seventeenth root of unity in terms of radicals. He not only made this discovery but also realized that this implied that it would be possible to construct a seventeen-sided polygon using only compass and ruler. Gauss later generalized these results on two levels:

(1) He proved that all roots of unity are expressible as radicals.
(2) A regular polygon is constructible by ruler and compass if and only if its number of sides is a product of distinct Fermat primes F_n (where F_n = 2²ⁿ + 1) and a power of 2.

In today's blog, I will show Gauss's solution for the seventeenth root of unity. In the next blog, I will prove that it is possible to construct a seventeen sided polygon known as the heptadecagon using only compass and ruler.

Gauss was so fascinated by the construction of the heptadecagon that he decided that he would dedicate his life to mathematics instead of philology.

Theorem 1: The seventeen root of unity is expressible as radicals

Proof:

(1) Let ζ be a primitive seventeen-root of unity.

(2) Then, the sixteen roots of unity not equal to 1 are:

ζ¹, ζ², ζ³, ζ⁴, ζ⁵, ζ⁶, ζ⁷, ζ⁸, ζ⁹, ζ¹⁰, ζ¹¹, ζ¹², ζ¹³, ζ¹⁴, ζ¹⁵, ζ¹⁶

(3) We can divide the above roots into the following sums:

x₁ = ζ¹ + ζ⁹ + ζ¹³ + ζ¹⁵ + ζ¹⁶ + ζ⁸ + ζ⁴ + ζ²

x₂ = ζ³ + ζ¹⁰ + ζ⁵ + ζ¹¹ + ζ¹⁴ + ζ⁷ + ζ¹² + ζ⁶

(4) Now, since (see Lemma 1, here):

1 + ζ + ζ² + ... + ζ¹⁶ = 0

We have:

x₁ + x₂ + 1 = 0

So that:

x₁ + x₂ = -1

(5) x¹ * x² = ( ζ¹ + ζ⁹ + ζ¹³ + ζ¹⁵ + ζ¹⁶ + ζ⁸ + ζ⁴ + ζ²)(ζ³ + ζ¹⁰ + ζ⁵ + ζ¹¹ + ζ¹⁴ + ζ⁷ + ζ¹² + ζ⁶) =

= ζ¹(ζ³ + ζ¹⁰ + ζ⁵ + ζ¹¹ + ζ¹⁴ + ζ⁷ + ζ¹² + ζ⁶) + ζ⁹(ζ³ + ζ¹⁰ + ζ⁵ + ζ¹¹ + ζ¹⁴ + ζ⁷ + ζ¹² + ζ⁶) + ζ¹³(ζ³ + ζ¹⁰ + ζ⁵ + ζ¹¹ + ζ¹⁴ + ζ⁷ + ζ¹² + ζ⁶) + ζ¹⁵(ζ³ + ζ¹⁰ + ζ⁵ + ζ¹¹ + ζ¹⁴ + ζ⁷ + ζ¹² + ζ⁶) + ζ¹⁶(ζ³ + ζ¹⁰ + ζ⁵ + ζ¹¹ + ζ¹⁴ + ζ⁷ + ζ¹² + ζ⁶) + ζ⁸(ζ³ + ζ¹⁰ + ζ⁵ + ζ¹¹ + ζ¹⁴ + ζ⁷ + ζ¹² + ζ⁶) + ζ⁴(ζ³ + ζ¹⁰ + ζ⁵ + ζ¹¹ + ζ¹⁴ + ζ⁷ + ζ¹² + ζ⁶) + ζ²(ζ³ + ζ¹⁰ + ζ⁵ + ζ¹¹ + ζ¹⁴ + ζ⁷ + ζ¹² + ζ⁶) =

= (ζ⁴ + ζ¹¹ + ζ⁶ + ζ¹² + ζ¹⁵ + ζ⁸ + ζ¹³ + ζ⁷) + (ζ¹² + ζ² + ζ¹⁴ + ζ³ + ζ⁶ + ζ¹⁶ + ζ⁴ + ζ¹⁵) +
(ζ¹⁶ + ζ⁶ + ζ + ζ⁷ + ζ¹⁰ + ζ³ + ζ⁸ + ζ²) + (ζ + ζ⁸ + ζ³ + ζ⁹ + ζ¹² + ζ⁵ + ζ¹⁰ + ζ⁴) +
(ζ² + ζ⁹ + ζ⁴ + ζ¹⁰ + ζ¹³ + ζ⁶ + ζ¹¹ + ζ⁵) + (ζ¹¹ + ζ + ζ¹³ + ζ² + ζ⁵ + ζ¹⁵ + ζ³ + ζ¹⁴) +
(ζ⁷ + ζ¹⁴ + ζ⁹ + ζ¹⁵ + ζ + ζ¹¹ + ζ¹⁶ + ζ¹⁰) + (ζ⁵ + ζ¹² + ζ⁷ + ζ¹³ + ζ¹⁶ + ζ⁹ + ζ¹⁴ + ζ⁸) =

= 4[ζ¹ + ζ² + ζ³ + ζ⁴ + ζ⁵ + ζ⁶ + ζ⁷ + ζ⁸ + ζ⁹ + ζ¹⁰ + ζ¹¹ + ζ¹² + ζ¹³ + ζ¹⁴ + ζ¹⁵ + ζ¹⁶] = 4[x₁ + x₂] = -4

(6) We can now use the two equations to form a quadratic equation:

x₂ = -1 -x₁

x₁*x₂ = -4

x₁(-1 -x₁) = -4

which gives us:

-x₁² - x₁ + 4 = 0

or if we multiply each side by -1

x₁² + x₁ - 4 = 0

(7) Now, it should also be clear that since each equation is symmetric, we could easily have said that:

x₁ = -1 - x₂

so it is clear that we also have:

x₂² + x₂ - 4 = 0

(8) This means that x₁, x₂ are found using the quadratic equation (see Theorem, here) to get:

x = (-1 ±√1 - (-16))/2 =

= (-1 ±√17)/2

So, we that we have:

x₁ = (-1 +√17)/2

x₂ = (-1 -√17)/2

(9) Now, let's define the following values:

y₁ = ζ¹ + ζ¹³ + ζ¹⁶ + ζ⁴

y₂ = ζ⁹ + ζ¹⁵ + ζ⁸ + ζ²

y₃ = ζ³ + ζ⁵ + ζ¹⁴ + ζ¹²

y₄ = ζ¹⁰ + ζ¹¹ + ζ⁷ + ζ⁶

(10) Now, it is clear that:

y₁ + y₂ = x₁

and

y₃ + y₄ = x₂

(11) Further, we have:

y₁y₂ = (ζ¹ + ζ¹³ + ζ¹⁶ + ζ⁴)(ζ⁹ + ζ¹⁵ + ζ⁸ + ζ²) =

= ζ¹(ζ⁹ + ζ¹⁵ + ζ⁸ + ζ²) + ζ¹³(ζ⁹ + ζ¹⁵ + ζ⁸ + ζ²) +
ζ¹⁶ (ζ⁹ + ζ¹⁵ + ζ⁸ + ζ²) + ζ⁴(ζ⁹ + ζ¹⁵ + ζ⁸ + ζ²) =

= (ζ¹⁰ + ζ¹⁶ + ζ⁹ + ζ³) + (ζ⁵ + ζ¹¹ + ζ⁴ + ζ¹⁵) +
(ζ⁸ + ζ¹⁴ + ζ⁷ + ζ) + (ζ¹³ + ζ² + ζ¹² + ζ⁶) =

= ζ + ζ² + ζ³ + ζ⁴ + ζ⁵ +ζ⁶ + ζ⁷ + ζ⁸ + ζ⁹ + ζ¹⁰ + ζ¹¹ + ζ¹² + ζ¹³ + ζ¹⁴ + ζ¹⁵ + ζ¹⁶ = x₁ + x₂ = -1

y₃y₄ = (ζ³ + ζ⁵ + ζ¹⁴ + ζ¹²)(ζ¹⁰ + ζ¹¹ + ζ⁷ + ζ⁶) =

= ζ³(ζ¹⁰ + ζ¹¹ + ζ⁷ + ζ⁶) + ζ⁵(ζ¹⁰ + ζ¹¹ + ζ⁷ + ζ⁶) +
ζ¹⁴(ζ¹⁰ + ζ¹¹ + ζ⁷ + ζ⁶) + ζ¹²(ζ¹⁰ + ζ¹¹ + ζ⁷ + ζ⁶) =

= (ζ¹³ + ζ¹⁴ + ζ¹⁰ + ζ⁹) + (ζ¹⁵ + ζ¹⁶ + ζ¹² + ζ¹¹) +
(ζ⁷ + ζ⁸ + ζ⁴ + ζ³) + (ζ⁵ + ζ⁶ + ζ² + ζ) =

= ζ +ζ² +ζ³ + ζ⁴ + ζ⁵ +ζ⁶ + ζ⁷ + ζ⁸ + ζ⁹ + ζ¹⁰ + ζ¹¹ + ζ¹² + ζ¹³ + ζ¹⁴ + ζ¹⁵ + ζ¹⁶ = -1

(12) So, this gives us:

y₂ = x₁ - y₁
y₄ = x₂ - y₃

y₁(x₁ - y₁) = -1
y₃(x₂ - y₃) =-1

So that:

x₁y₁ - y₁² + 1 = 0
x₂y₃ - y₃² + 1 = 0

(13) So that we can find the value for y₁, y₂ by solving:

y² - (x₁)y - 1 = 0

So that y = [(x₁) ± √(x₁² + 4)]/2

So we assign:

y₁ = [(x₁) + √(x₁² + 4)]/2

y₂ = [(x₁) - √(x₁² + 4)]/2

(14) We can find the value for y₃, y₄ by solving:

Y² - (x₂)Y - 1 = 0

So that Y = [(x₂) ± √(x₂² + 4)]/2

So we assign:

y₃ = [(x₂) + √(x₂² + 4)]/2

y₄ = [(x₂) - √(x₂² + 4)]/2

(15) Now, let's define the following values:

z₁ = ζ + ζ¹⁶

z₂ = ζ¹³ + ζ⁴

z₃ = ζ⁸ + ζ⁹

z₄ = ζ² + ζ¹⁵

z₅ = ζ³ + ζ¹⁴

z₆ = ζ⁵ + ζ¹²

z₇ = ζ⁷ + ζ¹⁰

z₈ = ζ⁶ + ζ¹¹

(16) We note that:

z₁ + z₂ = y₁

z₃ + z₄ = y₂

z₅ + z₆ = y₃

z₇ + z₈ = y₄

(17) If we multiply combinations together, we get:

z₁*z₂ = (ζ + ζ¹⁶)(ζ¹³ + ζ⁴) = ζ¹⁴ + ζ⁵ + ζ¹² + ζ³ = y₃

z₃*z₄ = (ζ⁸ + ζ⁹)( ζ² + ζ¹⁵) = ζ¹⁰ + ζ⁶ + ζ¹¹ + ζ⁷ = y₄

z₅*z₆ = (ζ³ + ζ¹⁴)(ζ⁵ + ζ¹²) = ζ⁸ + ζ¹⁵ + ζ² + ζ⁹ = y₂

z₇*z₈ = (ζ⁷ + ζ¹⁰)(ζ⁶ + ζ¹¹) = ζ¹³ + ζ + ζ¹⁶ + ζ⁴ = y₁

(18) So, this gives us:

z₂ = y₁ - z₁

z₄ = y₂ - z₃

z₆ = y₃ - z₅

z₈ = y₄ - z₇

which means:

z₁(y₁ - z₁) = y₃

z₃(y₂ - z₃) = y₄

z₅(y₃ - z₅) = y₂

z₇( y₄ - z₇ ) = y₁

(19) So, we find z₁, z₂ by solving for:

z₁² - z₁y₁ + y₃ = 0

so that:

z₁ = [y₁ + √(y₁)² - 4y₃)]/2

z₂ = [y₁ - √(y₁)² - 4y₃)]/2

(20) Finally, we can now solve for the seventeen root of unity since:

ζ¹⁶ = z₁ - ζ [From step #15 above]

ζ(z₁ - ζ) = 1 [This follows since ζ¹⁷ = 1]

z₁ζ - ζ² - 1 = 0

ζ² - z₁ζ + 1 = 0 [Multiplying both sides by -1]

ζ = (z₁ + √(z₁)² - 4)/2 [Using the solution to the quadratic equation, see here]

Working all this out gives us:

ζ =



References

• Tom Rike, "Fermat Numbers and the Heptadecagon", San Jose Math Circle.Org
• Heinrich Dorrie, 100 Great Problems of Elementary Mathematics
• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001