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The Math Problem Sherlock Holmes Couldn’t Solve, But You Can – Mind Your Decisions
The new series Young Sherlock on Amazon Prime Video features an interesting math problem that baffles our hero. Solve for x in the equation:
x5 + x4 + x3 + x2 + x + 1 = 0
I share 2 different methods to solve this problem in a new video.
As usual, watch the video for a solution.
The Math Problem Young Sherlock Holmes Couldn’t Solve, But You Can
Or keep reading.
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Answer To The Math Problem Sherlock Holmes Couldn’t Solve, But You Can
(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).
In the episode Sherlock is confused by the problem. However, the same blackboard contains a derivation of the solution.
Method 1: polynomial division
Let f(x) = x5 + x4 + x3 + x2 + x + 1
By the Rational Root Theorem, the only candidates for rational roots are ±1/1 = ±1. Testing values we find f(1) = 6 and f(-1) = 0. Thus x = -1 is a root, and x + 1 is a factor of f.
Let’s divide f by x + 1.
Thus we have:
f(x) = (x + 1)(x4 + x2 + 1)
We can further factor the quartic by adding and subtracting x2.
x4 + x2 + 1
= x4 + x2 + x2 – x2 + 1
= x4 + 2x2 + 1 – x2
= (x2 + 1)2 – x2
= (x2 + 1 + x2)(x2 + 1 – x2)
Thus we have:
f(x) = (x + 1)(x2 + 1 + x2)(x2 + 1 – x2)
The roots of this equation correspond to the zeroes of the factored equations. The linear factor gives the root of x = -1, and the quadratic equations are easily solvable by the quadratic equation. Thus we get the 5 roots:
x = 1
x = -1/2 ± i√3/2
x = 1/2 ± i√3/2
Method 2: roots of unity
There is a well-known formula for n an integer greater than 2:
xn – 1 = (x – 1)(xn-1 + xn-2 + … + 1)
This is easily verifiable by expanding and comparing coefficients. In our problem we have:
x6 – 1 = (x – 1)(x5 + x4 + … + 1)
The roots of the equation correspond to the roots of the factors. The linear factor has a root of x = 1, which we already verified is not a factor of the 5th degree equation. So the 5 roots of this equation are the 6 roots of the equation x6 – 1 excluding x = 1.
x6 – 1 = 0
x6 = 1
We want to find the 6 roots of 1 (also called unity). This is easily done by putting 1 in the complex plane in polar form as e2πk for integers k.
1 = e2πk
11/6 = [e2πk]1/6
11/6 = e2πk/6
We will get 6 distinct values from k = 0, 1, 2, …, 5, giving:
e0 = 1
e2π/6 = 1/2 + i√3/2
e4π/6 = -1/2 + i√3/2
e6π/6 = -1
e8π/6 = -1/2 – i√3/2
e10π/6 = 1/2 – i√3/2
We know x = 1 is not a root of x5 + x4 + x3 + x2 + x + 1, so we have its 5 roots are:
e2π/6 = 1/2 + i√3/2
e4π/6 = -1/2 + i√3/2
e6π/6 = -1
e8π/6 = -1/2 – i√3/2
e10π/6 = 1/2 – i√3/2
What a fun problem! Credit to the math consultants on the show Young Sherlock.
References
The Mathematical Crimes of Young Sherlock
https://theconversation.com/the-mathematical-crimes-of-the-young-sherlock-holmes-series-278812
Christopher Scott Vaughen
https://www.youtube.com/watch?v=tRCkPqQTg8k
IGN clip
https://www.youtube.com/watch?v=WPC8mAcl0iw
Prime Video clip
https://www.youtube.com/watch?v=7DOsYvn3S7c
Young Sherlock Prime Video
https://www.amazon.com/gp/video/detail/B0G2BTDG5G/
WolframAlpha
https://www.wolframalpha.com/input?i=1%5E%281%2F6%29
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