proof of Thue’s Lemma
Let be a prime congruent to 1 mod 4.
We prove the uniqueness first: Suppose
where without loss of generality, we can assume and even, and odd, , and thus that . Let and , and compute
whence . If , cancel the factor of to get a new equation with , so we can write
for some positive integer . Then
By Euler’s criterion (or by Gauss’s lemma), the congruence
(2) tells us
Write . We get
whence , as desired.
To prove Thue’s lemma in another way, we will imitate a part of the proof of Lagrange’s four-square theorem. From (1), we know that the equation
has a solution with, we may assume, . It is enough to show that if , then there exists such that and
If is even, then and are both even or both odd; therefore, in the identity
both summands are integers, and we can just take and conclude.
If is odd, write and with and . We get
for some . But consider the identity
On the left is , and on the right we see
Thus we can divide the equation
Remark: The solutions of the congruence (1) are explicitly
|Title||proof of Thue’s Lemma|
|Date of creation||2013-03-22 13:19:08|
|Last modified on||2013-03-22 13:19:08|
|Last modified by||mathcam (2727)|