Consider a right square pyramid whose base has side length $2r$ and whose height is $h$. Let the dihedral angle between the base and each triangular side be $\theta$, and the dihedral angle between adjacent triangular sides be $\phi$. We have $$\cos(\theta)=\frac{r}{\sqrt{r^2 + h^2}}$$ and $$\cos(\phi)=\frac{-r^2}{r^2+h^2}$$ so in particular $$\cos^2(\theta)=-\cos(\phi).$$


This square pyramid is scissors congruent to a cube iff its Dehn invariant is zero, for which it is necessary that $\pi$ be expressible as a rational linear combination of $\theta$ and $\phi$.


If $r=h$ then this condition is satisfied, and indeed such a square pyramid is scissors congruent to a cube. There is no other nontrivial way to satisfy this condition with both $\theta$ and $\phi$ being rational multiples of $\pi$, as shown by [this answer on MSE](https://math.stackexchange.com/a/3338539/49965).


It seems unlikely that there are any other solutions at all, but I have no idea how to even begin to prove such a thing. To ask a precise



> 
> **Question**: Are there any pairs $\theta$, $\phi$ such that $\cos^2(\theta)=\cos(\phi)$ and $\pi$ is expressible as a rational linear combination of $\theta$ and $\phi$, except where $\cos(\phi)\in\{0,\frac12,1\}$?
> 
> 
> Equivalently, are there any pairs $\theta$, $\phi$ such that $\cos^2(\theta)=\cos(\phi)$, and $\pi$ is expressible as a rational linear combination of $\theta$ and $\phi$, but $\theta$ and $\phi$ are not rational multiples of $\pi$.
> 
> 
> 




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**Added later**: I asked the question above in the expectation that the answer would be no, which would then imply the nonexistence of other right square pyramids that are scissors congruent to a cube. Unfortunately it is quite easy to see that the answer is actually yes, as Daniil Rudenko points out below.


This shows that I asked the wrong precise question. The right question is more complicated, and is described – and answered – by user145307.



Which right square pyramids are scissors congruent to a cube?
This is not the question you want to ask. (The actual question asked is easy by a continuity argument.) If the sides of the pyramid have length $2r$ and the height is $h$, then the other side lengths of the pyramid have length $\sqrt{2r^2 + h^2}$. You want to ask whether the element 


$$\xi = (2r \otimes \theta) + (\sqrt{h^2 + 2 r^2} \otimes \phi)$$


is trivial in the Dehn group. For this to be true, either:


1. $\theta$ and $\phi$ are both rational multiples of $\pi$ (case already done).
2. $r$ is a rational multiple $0 < v < 1/\sqrt{2}$ of $\sqrt{h^2 + 2 r^2}$, and so


$$\xi = \sqrt{h^2 + 2 r^2} \otimes (2 v \theta + \phi);$$


then one moreover demands $2 v \theta + \phi$ is a rational multiple of $\pi$. This is a much more stringent requirement.
In fact, it never happens, by the following elementary but tedious argument.


By scaling, we can assume that $h = 1$. Hence it follows that $r^2$ is a rational multiple of $1 + 2 r^2$, which certainly implies that $r^2$ is rational. So let $r^2 = t$. Thus we require that 


$$\frac{r}{\sqrt{1 + 2 r^2}} = \sqrt{\frac{t}{1 + 2 t}} = v$$


is rational. It follows that we have
$$r^2 = t = \frac{v^2}{1 - 2 v^2}$$
for some rational $v$. Thus we can rephrase the problem as follows:



> 
> Find all rational $0 < v < 1/\sqrt{2}$ with
>  $$\cos(\theta) = \frac{r}{\sqrt{r^2 + 1}} = \frac{v}{\sqrt{1 - v^2}},$$
> $$\cos(\phi) = \frac{-r^2}{r^2 + 1} = \frac{-v^2}{1 - v^2},$$
>  and such that
>  $$(2 v \theta + \phi) \in {\mathbf{Q}} \pi.$$
> 
> 
> 


Let us do an example to explain how one can eliminate any specific $v$.
Take the case of $v = 1/2$. We find that
 $$\cos(\theta) = 1/\sqrt{3}, \quad \cos(\phi) = -1/3,$$
 from which we deduce (for example) that
 $$\cos(2 v \theta + \phi) = \cos(\theta + \phi) = - \frac{5}{3 \sqrt{3}}.$$
 If $\alpha = 2v \theta + \phi$ is a multiple of $\pi$, then
 $$\cos(\alpha) = \frac{\zeta + \zeta^{-1}}{2},$$
 where $\zeta = e^{i \alpha}$ is a root of unity. But knowing $\cos(\alpha)$ one can
 solve for $\zeta$ and then determine if it is a root of unity or not from its minimal polynomial.
 In the example above, we win immediately
 because $2 \cos(\alpha)$ should be an algebraic integer and it is not.
 The cases $v = 1/3$ and $v = 2/5$ can be handled in a very similar way: if $v = 1/3$, then $\cos(\theta) = 1/(2 \sqrt{2})$ and $\cos(\phi) = -1/8$, and,
 with $\alpha = 3 (2v \theta + \phi)$,
 $$\cos(\alpha) = \cos(2 \theta + 3 \phi) = \frac{87}{256},$$
 and if $v = 3/5$, then $\cos(\theta) = 3/5$ and $\cos(\phi) =9/25$, and
 with $\alpha = 5 (2v \theta + \phi)$,
 $$\cos(\alpha) = \cos(6 \theta + 5 \phi) = \frac{3617721}{4194304}.$$
 In both cases, the corresponding $\zeta$ is manifestly not a root of unity because $2 \cos(\alpha)$
 is not an algebraic integer.


Returning to the original problem,
the first thing we will do is prove that ${\mathbf{Q}}(e^{2 v i \theta})$ is an abelian extension.
We may write
$$e^{2 v i \theta} = e^{(2 v \theta + \phi) i} \cdot e^{- \phi i}.$$
Since $2 v \theta + \phi \in {\mathbf{Q}} \pi$, the first factor is a root of unity and so lies in an
abelian extension. On the other hand, the second term is simply
$$\cos(\phi) - i \cdot \sin(\phi).$$
Since $\cos(\phi) \in {\mathbf{Q}}$, it follows that $i \cdot \sin(\phi) = \sqrt{\cos^2(\phi) - 1}$ lives 
in an imaginary quadratic extension of ${\mathbf{Q}}$. In particular,
$e^{- i \phi}$ clearly lies inside an abelian extension. Taken together, we deduce:



> 
> The extension ${\mathbf{Q}}(e^{2 v i \theta})$ is abelian.
> 
> 
> 


We also have the explicit formulae
$$e^{i \theta} = \cos(\theta) + i \sin(\theta) = \frac{v + i \sqrt{1 - 2 v^2}}{\sqrt{1 - v^2}},$$
and then squaring:
$$e^{2 i \theta} = \frac{ 3 v^2 - 1 + 2 v \sqrt{2 v^2 - 1}}{1 - v^2}.$$


Let $v = a/b$ with $(a,b) = 1$, this becomes
$$e^{2 i \theta} = \frac{ 3 a^2 - b^2 + 2 a \sqrt{2 a^2 - b^2 }}{(b^2 - a^2)}.$$
Let $E = {\mathbf{Q}}(\sqrt{2 a^2 - b^2 })$, which is an imaginary
quadratic extension of ${\mathbf{Q}}$. (The condition that $a/b = v < 1/\sqrt{2}$
implies that $b^2 > 2 a^2$.)
Let us write
$$x = 3 a^2 - b^2 + 2 a \sqrt{2 a^2 - b^2 } \in \mathcal{O}\_E.$$
Note that $N(x) = (b^2 - a^2)^2$.
Secondly, note that
$$e^{2 v i \theta} = (e^{2 i a\theta})^{1/b}$$


By Galois theory, $E(\alpha^{1/b})$
can only be an abelian extension of ${\mathbf{Q}}$ (or even of $E$) under the following conditions:


1. $\alpha$ is a perfect $b$th power in $E$.
2. $b$ is even and $\alpha$ is a perfect $b/2$th power in $E$.


(Added explanation: You can work prime by prime on the divisors $p$ of $b$. Suppose that $\alpha$ is not a perfect $p$th power for a prime $p > 3$. Then the Galois closure of $E(\alpha^{1/p})$ is $E(\alpha^{1/p},\zeta\_p)$ and contains the automorphism $\tau: \alpha^{1/p} \rightarrow \zeta\_p \alpha^{1/p}$. But then Galois group of $E(\zeta\_p)$ over $E$ contains an element $\sigma$ which fixes $\alpha$ and sends $\zeta\_p$ to $\zeta^i\_p \ne \zeta\_p$. Then $\tau$ and $\sigma$ do not commute. The same argument works if $p = 3$ as long as $E(\zeta\_p) \ne E$, which can only happen if $E = \mathbf{Q}(\sqrt{-3})$. But $2 a^2 - b^2 \ne - 3 c^2$ by $3$-adic considerations. The same argument works for "$p = 4$" as well, as long as $E(\zeta\_4) \ne E$, which can only happen if $E = \mathbf{Q}(\sqrt{-1})$. But $2 a^2 - b^2 = - c^2$ can't happen when $b$ is even (which is the only relevant case) because then $a$ is odd and $2 a^2 - b^2$ is $2 \mod 4$.)


Moreover, since$(a,b) = 1$, if $e^{2 i a \theta}$ is a perfect $b$ or $b/2$th power, then so is $e^{2 i \theta}$.
In particular, $e^{4 i \theta}$ is a $b$th power in $E$. Note also that
$$e^{4 i \theta} = \frac{x^2}{N(x)} = \frac{x^2}{x \overline{x}} = \frac{x}{\overline{x}},$$
so $x/\overline{x}$
is a perfect $b$th power in $E$.
Our goal is now to prove that the ideal $(x)$ is (almost) a $b$th power,
and deduce that $N(x) = (b^2 - a^2)^2$ is (almost) a $b$th power.


Suppose that $\mathfrak{p}$ is a prime ideal of $\mathcal{O}\_E$ which divides both $x$ and $\overline{x}$.
I claim that $\mathfrak{p}$ divides $2$. Note that $N(x) = (b^2 - a^2)^2$, so 
so $a^2 \equiv b^2 \mod \mathfrak{p}$.
But then
$$x + \overline{x} = 2(3 a^2 - b^2) \equiv 4 a^2 \mod \mathfrak{p}.$$
If $a \in \mathfrak{p}$ then also $b \in \mathfrak{p}$ contradicting that $(a,b) = 1$. Hence $(x,\overline{x})$
is only divisible by primes above $2$. 
Since $x/\overline{x}$ is a $b$th power, and 
 $(x,\overline{x})$ is supported at primes above $2$, we deduce that $(x) = I^b \cdot J$
where $J$ is supported at primes above $2$. We deduce that



> 
> $$N(x) = (b^2 - a^2)^2 = n^b \cdot 2^k.$$
> 
> 
> 


where $n$ is an odd integer. Let us first consider the case when $a$ and $b$ are not both odd.
In this case $k$ is trivial, and $(b^2 - a^2)^2 = n^b$. Since $b^2 - a^2 = 1$ has no solutions
in positive integers, it follows that $b^2 - a^2 > 1$, but then
$$b^4 > (b^2 - a^2)^2 \ge 2^b,$$
from which we deduce that $b < 16$.
Now suppose that $a$ and $b$ are both odd. Since $(b-a,a+b) = 2$ in this case,
it must be the case that
$$b-a = r^b 2^u, b+a = s^b 2^v,$$
where one of $u$ and $v$ must be equal to $1$.


1. Case 1: $u = 1$. If $b-a=2$, then from the inequality
$v < 1/\sqrt{2}$ and $a < b/\sqrt{2}$, we deduce that 
$$2 = b - a > b(1 - 1/\sqrt{2}),$$
and thus $b < 7$. If $b-a=2 \cdot r^b$ with $r > 1$, then
$$b \ge b - a \ge 2^{b+1},$$
which is impossible.
2. Case 2: $v = 1$. We have $a+b = 2 \cdot s^b$, and now
$$2b > a + b \ge 2^{b+1},$$
which once again is impossible.


Putting this together, we deduce that $b < 16$.
Checking all the cases with $b < 16$
(noting that $a/b < 1/\sqrt{2}$ and $(a,b) = 1$) we find that the only possible pairs
are 
$$(a,b) = (1,2), (1,3), (3,5), \ \text{or} \ v =1/2, 1/3, 3/5.$$
But these cases have already been considered previously.


