Problem 3

Let

P=A_1A_2\cdots A_k

be a convex polygon in the plane. The vertices

A_1,A_2,\dots,A_k

have integer coordinates and lie on a circle. Let

S

be the area of

P

. An odd positive integer

n

is given such that the square of the length of each side of

P

is an integer divisible by

n

. Prove that

2S

is an integer divisible by

n

Step 5 of 5: Conclusion by minimal counterexample and combining prime powers

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In plain words

The contradiction rules out any minimal counterexample beyond the settled triangle case, so induction closes the gap up to the original polygon.

\text{no counterexample exists}\ \Rightarrow\ n=p^e\mid 2S\ \Rightarrow\ n\mid2S\ \text{for general odd } n

Detailed analysis

The contradiction of the previous step shows no minimal counterexample with $k\ge3$ vertices beyond the triangle case can exist: either a diagonal itself has $p^e$ dividing its squared length (in which case induct on the two smaller cyclic sub-polygons it creates, each with strictly fewer vertices, both already known to satisfy the divisibility, and $2S$ for the whole polygon is the sum of $2S$ for the two pieces), or no such diagonal exists and the valuation argument of the previous step gives a contradiction. By strong induction starting from the triangle base case (Step 1), every convex cyclic lattice polygon with all squared side lengths divisible by $p^e$ has $p^e\mid2S$ . Since $n$ (odd) is a product of such prime powers $p^e$ and the corresponding divisibility of $2S$ by each prime power combines (as $2S$ is a single fixed integer) into divisibility by their product $n$ , we conclude $n\mid2S$ .