Make many "Morrie's Laws"

Geometry Level 5

Richard Feynman was fascinated by the equation $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)=\dfrac1{2^3}$ , which he named "Morrie's Law" after a childhood friend, Morrie Jacobs, who told him the rule. More generally, how many positive integers $n<90$ are there such that $\displaystyle \prod_{k=0}^{m-1}\cos(2^kn^\circ)=\dfrac{1}{2^m}$ for some positive integer $m$ ?

Inspiration: here and here .

The answer is 14.

1 solution

Kartik Sharma
Sep 8, 2015

Great problem! No wonder why it could fascinate even Mr. Feynman!

$\displaystyle \prod_{k=0}^{m-1}{cos(2^kn^{\circ})} = \frac{1}{2^m}$

LHS becomes

$\displaystyle \frac{1}{2^m sin(n^{\circ})} 2sin(n^{\circ})cos(n^{\circ}) \prod_{k=1}^{m-1}{2 \ cos(2^kn^{\circ})}$

Repeatedly using $2 sin(x) cos(x) = sin(2x)$ ,

$\displaystyle = \frac{sin(2^m n^{\circ})}{2^m sin(n^{\circ})}$

implies that

$\displaystyle \frac{sin(2^m n^{\circ})}{2^m sin(n^{\circ})} = \frac{1}{2^m} \Rightarrow sin(2^m n^{\circ}) = sin(n^{\circ})$

$\displaystyle \Rightarrow 2^m n = 180k + (-1)^k n$ for some positive integer $k$ .

Now, it becomes a number theory problem. We'll take up the 2 possible cases

Case 1 - $k$ is even

$\displaystyle 2^m n = 360 k + n \Rightarrow 2^m n \equiv n\pmod{360}$

If $\text{gcd}(n,360) = 1$ then our equation becomes $2^m \equiv 1 \pmod{360}$ which has no solutions since $\text{gcd}(m,360) \neq 1$ .

Hence, $\text{gcd}(n,360) \neq 1$ , it is such that $\text{gcd}\left(m,\frac{360}{\text{gcd}(n,360)}\right) = 1$ . One can see very easily that $\text{gcd}(n,360)$ is at least $8$ . Or $n$ is a multiple of $8$ and it is given that $n<90$ . So, $n$ can take the following values - $8,16, 24, 32, 40, 48, 56, 64, 72, 80, 88$

Case 2 - $k$ is odd

$\displaystyle 2^m n = 360k - 180 - n \Rightarrow 2^m n \equiv 180-n \pmod{360}$

Let us define $n' = 180 - n$ , then $2^m (180 - n') \equiv n' \pmod{360}$

$m$ is a positive integer, hence, $-2^m n' \equiv n' \pmod{360}$

And if $\text{gcd}(n,360) = 1$ then our equation becomes $2^m \equiv -1 \pmod{360}$ which has no solutions since $\text{gcd}(m,360) \neq 1$ .

Hence, $\text{gcd}(n',360) \neq 1$ , it is such that $\text{gcd}\left(m,\frac{360}{\text{gcd}(n',360)}\right) = 1$ . One can see very easily that $\text{gcd}(n',360)$ is at least $8$ . But now, not all multiples give the answer and experimentation(or probably programming) tells us that only $n' = 8*15, 8*18, 8*20$ gives us the answer. So, $n = 180 - n'$ takes up the values $20, 36, 60$ . I don't know how to find this thing. Anyone can help?

As a result, we have $14$ solutions namely $8,16, 20, 24, 32, 36, 40, 48, 56, 60, 64, 72, 80, 88$ .

@Otto Bretscher Can you give me any hints on how to find solutions of Case 2 in a better way?

Kartik Sharma - 5 years, 9 months ago

Very nice! Thanks! I did it in essentially the same way; maybe somebody else has a more elegant solution.

I wrote the second case as $n(2^m+1)=(2k-1)180$ for some positive integer $k$ , or, after division by 4, we have $\frac{n}{4}(2^m+1)=(2k-1)45$ . The LHS must be divisible by 45. Since $2^m$ is always congruent to 1, 2, 4, or 8 $\pmod{15}$ , the factor $2^m+1$ cannot be divisible by 45 or even 15. If $2^m+1$ is divisible by 9 (when $m=3$ , for example), then $\frac{n}{4}$ must be an odd multiple of 5 , so $n=20$ or $n=60$ . If $2^m+1$ is divisible by 5 (when $m=2$ , for example), then $\frac{n}{4}$ must be an odd multiple of 9, so $n=36$ since $n<90$ .

I wrote your first case as $n(2^m-1)=360k$ . Now $n$ must be divisible by 8 and we can write $\frac{n}{8}(2^m-1)=45k$ . We can let $m=\lambda(45)=12$ [so that $2^{12}\equiv{1}\pmod{45}$ ] to see that $n$ can be any multiple of 8, so $n=8,16,...,80,88$ . See this .

Otto Bretscher - 5 years, 9 months ago

Oh thanks! You eased it a lot! Now, it looks so obvious.

Kartik Sharma - 5 years, 9 months ago

Make many "Morrie's Laws"

Inspiration: here and here .

The answer is 14.

1 solution

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