There's a straightforward elementary number theory proof that if four odd squares sum to a power of four, then the squares are $1$ and the power is $4$. Is there a geometric proof?
I am ultimately interested in representing finite groups as subgroups of groups of unit quaternions with rational coefficients with fixed denominators.
I'm not sure whether it's standard or not, but in the quaternion $\frac{1}{2} + \frac{1}{3}j$, I am referring to $\frac{1}{2}$ and $\frac{1}{3}$ as coefficients, by analogy with the coefficients of a polynomial.
For example, there are 8 unit quaternions with integral coefficients: $\pm1, \pm i, \pm j, \pm k$.
It's a little bit harder to see, but there are 24 unit quaternions with coefficients of the form $\frac{a}{2}$. 8 solutions corresponding to $4+0+0+0$ and 16 corresponding to $1+1+1+1$.
There are 104 unit quaternions with coefficients of the form $\frac{a}{3}$, 8 corresponding to $9+0+0+0$ and 96 corresponding to $4+4+1+0$.
However, for unit quaternions with coefficients of the form $\frac{a}{4}$, we're back down to 24.
I checked OEIS and found this sequence, and saw that 24 repeated itself a lot.
I guessed that the 24 entries corresponded to powers of 2.
Then I tried to prove it. Here's what I came up with after struggling to prove this for a little while.
Proof that if $a, b, c, d$ are odd squares summing to $4^k$, then $a=b=c=d=k=1$.
Suppose we have $a = 2e+1, b=2f+1, c=2g+1, d=2h+1$.
$$ (2e+1)^2 + (2f+1)^2 + (2g+1)^2 + (2h+1)^2 = 4^k $$
Rearranging terms we get:
$$ 4(1+e+e^2+f+f^2+g+g^2+h+h^2) = 4^k $$
$1+e+e^2+f+f^2+g+g^2+h+h^2$ is odd.
However, the only odd divisor of $4^k$ is $1$.
Thus, $1+e+e^2+f+f^2+g+g^2+h+h^2$ is $1$.
Thus $e=f=g=h=0$.
Thus $a=b=c=d=k=1$.
However, this proof doesn't seem very geometric.
Quaternion multiplication has a geometric interpretation and, beyond that, this result seems like it would imply that $S^3$ intersects the dyadic rational lattice $\mathbb{D}^4$ in finitely many places, which is also weird.
Is there a geometric proof?