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I’m interested in pairs of rational numbers $a, b$ in the interval $(0,\frac12)$ such that $$\cos(a\pi) = \cos^2(b\pi)$$

Certainly $a=\frac13$, $b=\frac14$ is a solution. I suspect that this is the only solution – as a sanity check, I verified this numerically for denominators less than 200 – but I can’t currently see how to prove it.

I have the feeling there’s a simple proof that I’m not quite seeing, maybe involving expressing the cosines in terms of roots of unity? They’re all algebraic numbers, of course.

Blue
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Robin Houston
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  • The pair $(1/3,1/4)$ does not solve your equation since LHS is $\cos(π/3)=1/2$ and RHS is $\cos(π^2/16).$ As explained in my answer below, there is no solution. – Allawonder Aug 29 '19 at 17:17
  • @Allawonder I’m sorry for the confusion my notation has caused you. I hope the question is clearer now. – Robin Houston Aug 29 '19 at 17:21
  • Oh, in that case, I can only ask if you've noticed that your equation is equivalent to $$2\cos(aπ)-\cos(2πb)=1.$$ I checked, and there are no solutions for $a=2b.$ – Allawonder Aug 29 '19 at 17:41

2 Answers2

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Not quite simple, but this approach works here. Let $a=2u/w$ and $b=v/w$ with positive integers $u,v,w$.

Denoting $\zeta=e^{2\pi i/w}$, we have $2(\zeta^u+\zeta^{-u})=2+\zeta^v+\zeta^{-v}$. This is a polynomial equation w.r.t. $\zeta$, hence $$2(\zeta^{au}+\zeta^{-au})=2+\zeta^{av}+\zeta^{-av},\qquad 1\leqslant a\leqslant w,\ \gcd(a,w)=1$$ by the same "cyclotomic argument" as in the linked answer. Likewise, we sum over $a$ and use $$\frac{1}{\varphi(w)}\sum_{\substack{1\leqslant a\leqslant w\\\gcd(a,w)=1}}\zeta^{an}=\rho(d):=\frac{\mu(d)}{\varphi(d)},\qquad d=\frac{w}{\gcd(n,w)};$$ so, denoting $x=w/\gcd(u,w)$ and $y=w/\gcd(v,w)$, we get $\color{blue}{2\rho(x)=1+\rho(y)}$.

Examining the range of $\rho$, we see that this is possible only in the following cases:

  1. $\rho(x)=\rho(y)=1$. Thus, $x=y=1$ and $a/2,b\in\mathbb{N}$, out of range.
  2. $\rho(x)=1/2$ and $\rho(y)=0$. Thus, $x=6$ and $$a=\frac{2u}{w}=\frac{2}{x}\frac{u}{\gcd(u,w)}\in\frac{1}{3}\mathbb{N}\implies a=\frac{1}{3};$$ and we find the solution $(1/3,1/4)$ you know.
  3. $\rho(x)=1/4$ and $\rho(y)=-1/2$. This time $y=3$ and $b=1/3$, with no solution.
  4. $\rho(x)=0$ and $\rho(y)=-1$. Thus, $y=2$, leading to $a=b=1/2$.
metamorphy
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Your equation is equivalent to $$-2\sin\left(\frac{aπ+(bπ)^2}{2}\right)\sin\left(\frac{aπ-(bπ)^2}{2}\right)=0,$$ which gives the system (letting both factors vanish) $$aπ+(bπ)^2=2πj,\,\,aπ-(bπ)^2=2πk,$$ where $j,k$ are integers. You can now solve for $a,b$ and hope that your last condition $a,b\in(0,1/2)$ will indeed fix a unique pair $a,b.$ Adding the equations and simplifying gives that $a=\text{some integer}.$ But if this is the case, we must have that $b$ must be irrational. Indeed, $$b=\sqrt{\frac{2j-a}{π}},$$ which is clearly irrational. Something similar occurs if you decouple the equations. Thus, there are no such solutions as you seek. Indeed, we've proved that there are no rational solutions at all, not just in your interval.

Allawonder
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  • I’m sorry for my ambiguous notation! I intended the right-hand side to be read as $(\cos(bπ))^2$. I’ll amend the question to make that clear. – Robin Houston Aug 29 '19 at 17:18