Number Theory Solutions: Siegel’s Ellipse

William F. Hammond

The following problem was on the last assignment:

Can you find the smallest integer m ≥ 2 with the property that there are no integers x, y for which

5x^{2} + 11y^{2} ≡ 1 (mod m) ?

The answer to the question is “No” since there is no smallest such m. In other words, even though there is obviously no integer point on the ellipse

5x^{2} + 11y^{2} = 1

there are, for each modulus m, points (x, y) with integer coordinates for which the congruence mod m is satisfied. In fact, the mathematician C. L. Siegel (1896 – 1981) produced this as an example of an ellipse that for each modulus m is “equivalent” to the ellipse

x^{2} + 55y^{2} = 1 .

Proof. To show the existence of solutions (x, y) mod m for each m it is enough by the Chinese Remainder Theorem to treat the case where m is the power of a prime.

If m is the power of an odd prime, or is any positive odd number, then, letting u_{m} be a multiplicative inverse of 4 mod m one has

5u_{m}^{2} + 11u_{m}^{2} ≡ 16u_{m}^{2} ≡ 1 (mod m) .

One then proceeds to treat the case m = 2^{n}, n ≥ 1 recursively. Bear in mind that the value of an integer t mod 2^{n} determines the value of t^{2} mod 2^{n+1}.

It is obvious that (x, y) is a solution mod 4 if and only if

x ≡ 1 and y ≡ 0 (mod 2)

since all odd squares are 1 mod 4.

However, (1, 0) is not a solution mod 8. One sees that every solution mod 8 must be congruent to (±1, 2) mod 4, while (±1, ±2) and (±3, ±2) are all of the distinct solutions mod 8. Of the distinct solutions mod 8 only (±1, ±2) are solutions mod 16, while the distinct solutions mod 16 are (±1, ±2), (±7, ±2), (±1, ±6), and (±7, ±6). These observations lead one to guess that there might be 2^{n} solutions mod 2^{n} for all n ≥ 1.

Suppose that (x_{n}, y_{n}) is a solution mod 2^{n} for n ≥ 3. Then there is an integer u_{n} such that

5x_{n}^{2} + 11y_{n}^{2} = 1 + 2^{n}u_{n} ,

and the validity of this relation depends only on (x_{n}, y_{n}) mod 2^{n−1}. If (x_{n+1}, y_{n+1}) is to be a solution mod 2^{n+1} that reduces mod 2^{n−1} to (x_{n}, y_{n}), then one must have

\{ \begin{matrix}x_{n+1} = x_{n} + 2^{n−1}s \\ y_{n+1} = y_{n} + 2^{n−1}t \end{matrix}

for some integers s, t. Then let

u_{n+1} = \frac{5x_{n+1}^{2} + 11y_{n+1}^{2} − 1}{2^{n+1}} .

u_{n+1} is certainly a rational number, and it is an integer if and only if 2^{n+1}u_{n+1} ≡ 0 (mod 2^{n+1}) .

After some computation one sees that the last condition becomes

2^{n}u_{n} + 2^{n}(5x_{n}s + 11y_{n}t) + 2^{2n−2}(…) ≡ 0 (mod 2^{n+1})

which is equivalent to

u_{n} + 5x_{n}s + 11y_{n}t ≡ u_{n} + s ≡ 0 (mod 2)

Thus, one has a solution (x_{n+1}, y_{n+1}) mod 2^{n+1} by choosing s ≡ u_{n} mod 2 and t arbitarily. In fact, the free choice mod 2 for t is the reason why the number of solutions (x_{n}, y_{n}) mod 2^{n} is 2^{n}.