Problem 3

Assume now that every diagonal has squared length with $p$-adic valuation $<e$. Let $\nu_p$ be the valuation on nonzero rationals, and put $u_i=\nu_p(OA_i^2)$, $t_i=\nu_p(A_iA_{i+1}^2)$, and $d=\nu_p(A_1A_k^2)$. The two segments $OA_1,OA_k$ are sides, so $u_1,u_k\ge e$; the other $OA_i$ are diagonals, so $u_i<e$ for $2\le i\le k-1$. Also $t_i\ge e$ because $A_iA_{i+1}$ is a side, whereas $d<e$ because $A_1A_k$ is a diagonal. From $q_i=\dfrac{A_iA_{i+1}^2}{OA_i^2\cdot OA_{i+1}^2}$ and $q=\dfrac{A_1A_k^2}{OA_1^2\cdot OA_k^2}$ we obtain $\nu_p(q_i)=t_i-u_i-u_{i+1}$ and $\nu_p(q)=d-u_1-u_k$. Hence, at the two ends, $\nu_p(q_1)-\nu_p(q)=(t_1-d)+(u_k-u_2)>0$ and $\nu_p(q_{k-1})-\nu_p(q)=(t_{k-1}-d)+(u_1-u_{k-1})>0$: each parenthesis is at least $1$. For $2\le i\le k-2$, $\nu_p(q_i)-\nu_p(q)=(t_i-d)+(u_1+u_k-u_i-u_{i+1})>0$, since the first parenthesis is at least $1$ and the second is at least $2$. Thus $\nu_p(q_i)>\nu_p(q)$ for every $i$, equivalently $\nu_p(q_i^2)>\nu_p(q^2)$. We now use the square-root lemma: if positive rationals satisfy $\sum_i\sqrt{x_i}=\sqrt{x}$, then there is a positive rational $b$ such that all $\sqrt{x_i/b}$ and $\sqrt{x/b}$ are rational. To see why, write each radicand as a rational square times a squarefree positive integer; the square roots of distinct squarefree integers are linearly independent over $\mathbb Q$, and positivity prevents cancellation between different square classes. Applying the lemma to the Ptolemy relation gives positive rationals $r_i=\sqrt{q_i/b}$ and $r=\sqrt{q/b}$ with $\sum_i r_i=r$. Since $\nu_p(r_i^2)=\nu_p(q_i)-\nu_p(b)$ and $\nu_p(r^2)=\nu_p(q)-\nu_p(b)$, the strict inequality above yields $\nu_p(r_i)>\nu_p(r)$ for every $i$. The ultrametric inequality gives $\nu_p(\sum_i r_i)\ge\min_i\nu_p(r_i)>\nu_p(r)$, contradicting $\sum_i r_i=r$. Therefore the assumption that all diagonals fail is impossible.