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Joined: 17 Jul 2007
Posts: 1128
Location: JRZ
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...squares. Here's a roundabout proof using the facts about Gaussian integers I've developed so far. (There are easier and more elegant ways, but this hints at the role that geometry will play in our investigations in algebraic number theory.)
Suppose you can inscribe a regular polygon with vertices on the integer lattice in the plane. We may consider the integer lattice to be the Gaussian integers. Without loss of generality, we may assume one of the vertices is the origin by translation. Furthermore, let w be one of the vertices adjacent to the origin. Multiplying each vertex by w* scales and rotates the polygon around the origin and sends w to N(w), a positive integer. So we may assume that one of the vertices adjacent to the origin is a positive integer k.
Now let z be the other vertex adjacent to the origin. We must have |z| = k, since our polygon is regular. The angle between k and z in the counterclockwise direction must be π(n-2)/n (serious edit), and this is precisely the argument of z, and so we have z = keπi(n-2)/n. In particular, note that:
zn = kneπi(n-2) = Ā±kn
is a rational integer. I will show that this implies n = 4.
Now z factors as a product:
where u is a unit, all exponents are nonnegative integers, πi and πj are distinct irrational Gaussian primes if i ≠ j with norm greater than 2, qi are positive rational Gaussian primes. We may divide z by a positive integer without changing its argument, and so we may assume that vi = 0 for all i by dividing by qi as many times as possible. Similarly, since πiπi* is a positive integer, we may assume that min{si, ti} = 0 by dividing by πiπi* as many times as possible. And noting that (1+i)2 = 2i, we can assume r2 = 0 or 1 by dividing by 2 as many times as possible and absorbing any leftover factors of i into the unit u. Thus we may assume:
without changing the argument, where u is a unit, r2 = 0 or 1, si > 0 for all i, and πi and πj are distinct primes if i ≠ j.
Thus we still have zn is a rational integer. But if k > 0, i.e. if z is divisible by irrational Gaussian primes other than 1+i, then clearly so is zn. But then, since zn is rational, it must be divisible by the conjugate of any number that divides it (just express it as the product of two numbers and then take conjugates). This contradicts the prime factorization of z, as we have assumed if a rational prime of norm greater than 2 divides z, its conjugate doesn't. This z must be equal to u or u(1+i) for some unit u.
If z = u, then the argument of z is mπ/2 for some integer m; if z = u(1+i), then the argument of z is (2m+1)π/2 for some integer m. Clearly 0 < π(n-2)/n < π, and so we the only choices for the argument of z are π/4, π/2, and 3π/4. The first is impossible for any value of n, as (n-2)/n is minimal when n = 3. π/2 occurs when n = 4, and 3π/4 occurs when n = 8. So n = 4 or n = 8.
There's one last loose end to tie up. I claimed that n = 4 is the only possibility. Indeed, n = 8 is impossible. Instead of considering z to be a vertex adjacent to the origin, let z be the vertex opposite the origin. Then check that the argument of z is 3π/8, which is impossible if z is a Gaussian integer. So n = 4, and our polygon is a square. |
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