At the end of the proof of lemma 10, lemma 8 is cited. In order to use it and finish the contradiction, we need to show $n$ is not a multiple of $3.$ However, I don't see any contradiction in having $n \equiv \pm 2 \mod 8$ and $n \equiv 0 \mod 3.$ I also asked on MSE but not even a halfpenny of thoughts were given.


Proof: [Anglin - The square pyramid puzzle](https://doi.org/10.1080/00029890.1990.11995558).



Elementary proof of cannonball problem: why can't $n$ be a multiple of $3$?
Lemma 8 is used to conclude that the second factor on the left is $-1$. Note that this factor is
$$
\left( \frac{5}{u\_{2^s}}\right)
$$
meaning you apply Lemma 8 for $m=2^s$ and **not** for $n$. It is obvious that $m$ is not a multiple of $3$.


