Number Theory Solutions: Siegel's Ellipse

The fol lowing problem was on the last as signment :

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



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



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



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 be a multiplicative
inverse of 4 mod m one has



One then proceeds to treat the case m = 2ⁿ, n ≥ 1 recursive ly. Bear in mind that the value of an integer
t mod 2ⁿ de termines the value of t² mod .

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 observatio ns lead one to guess that there might be 2ⁿ solutions mod 2ⁿ for all n ≥1.

Suppose that is a solution mod 2ⁿ for n ≥ 3. Then there is an integer such that



and the validity of this relation depends only on . If is to be a solution mod
that reduces mod to , then one must have



for some integers s, t. Then let



is certainly a rational number , and it is an integer if and only if .

After some computation one sees that the last condition becomes



which is equivalent to



Thus, one has a solution by choosing mod 2 and t arbitarily. In fact, the
free choice mod 2 for t is the reason why the number of solutions is 2ⁿ.