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 1 of 5: Reduce to prime powers, and settle the triangle case

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In plain words

It suffices to check each odd prime power dividing $n$ separately, and for a triangle the symmetric form of Heron's formula makes the divisibility visible term by term.

S=\tfrac14\sqrt{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}

Detailed analysis

It suffices to prove the statement for $n=p^e$ , $p$ an odd prime, $e\ge1$ (the general odd $n$ then follows since the $2S$ -divisibility by each prime power factor of $n$ combines to divisibility by $n$ ). For $k=3$ (a triangle with sides $a,b,c$ ), Heron's formula in symmetric form gives $S=\tfrac14\sqrt{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4}$ , so $16S^2=2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4$ . If $p^e\mid\gcd(a^2,b^2,c^2)$ then $p^e$ divides each $a^2,b^2,c^2$ , so $p^{2e}$ divides every term on the right, hence $p^{2e}\mid16S^2$ , and since $p$ is odd this gives $p^{2e}\mid S^2$ up to the factor of $16=2^4$ which contributes no odd prime, so $p^e\mid 2S$ .