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- It is sufficient to prove the theorem for every odd prime number p. This immediately follows from Euler's four-square identity .
The residues of a2 modulo p are distinct for every a between 0 and .
To see this, take some a and define c as a2 mod p.
a is a root of the polynomial over the field .
So is .
In a field K, any polynomial of degree n has at most n distinct roots ,
so there are no other a with this property, in particular not among 0 to .
Similarly, for b taking integral values between 0 and , the are distinct.
By the pigeonhole principle, there are a and b in this range, for which a2 and are congruent modulo p, that is for which
Now let m be the smallest positive integer such that mp is the sum of four squares, . We show by contradiction that m equals 1: supposing it is not the case, we prove the existence of a positive integer r less than m, for which rp is also the sum of four squares .
For this purpose, we consider for each x'i the y'i which is in the same residue class modulo m and between and m/2 . It follows that , for some strictly positive integer r less than m.
Finally, another appeal to Euler's four-square identity shows that . But the fact that each x'i is congruent to its corresponding y'i implies that all of the zi are divisible by m. Indeed,
It follows that, for , , and this is in contradiction with the minimality of m.
In the descent above, we must rule out both the case , and also the case . For both of those cases, one can check that would be a multiple of m2, contradicting the fact that p is a prime greater than m. (en)
- The Hurwitz quaternions consist of all quaternions with integer components and all quaternions with half-integer components. These two sets can be combined into a single formula
where are integers. Thus, the quaternion components are either all integers or all half-integers, depending on whether is even or odd, respectively. The set of Hurwitz quaternions forms a ring; that is to say, the sum or product of any two Hurwitz quaternions is likewise a Hurwitz quaternion.
The (arithmetic, or field) norm of a rational quaternion is the nonnegative rational number
where is the conjugate of . Note that the norm of a Hurwitz quaternion is always an integer.
Since quaternion multiplication is associative, and real numbers commute with other quaternions, the norm of a product of quaternions equals the product of the norms:
For any , . It follows easily that is a unit in the ring of Hurwitz quaternions if and only if .
The proof of the main theorem begins by reduction to the case of prime numbers. Euler's four-square identity implies that if Lagrange's four-square theorem holds for two numbers, it holds for the product of the two numbers. Since any natural number can be factored into powers of primes, it suffices to prove the theorem for prime numbers. It is true for . To show this for an odd prime integer , represent it as a quaternion and assume for now that it is not a Hurwitz irreducible; that is, it can be factored into two non-unit Hurwitz quaternions
The norms of are integers such that
and . This shows that both and are equal to , and is the sum of four squares
If it happens that the chosen has half-integer coefficients, it can be replaced by another Hurwitz quaternion. Choose in such a way that has even integer coefficients. Then
Since has even integer coefficients, will have integer coefficients and can be used instead of the original to give a representation of as the sum of four squares.
As for showing that is not a Hurwitz irreducible, Lagrange proved that any odd prime divides at least one number of the form , where and are integers. This can be seen as follows: since is prime, can hold for integers , only when . Thus, the set of squares contains distinct residues modulo . Likewise, contains residues. Since there are only residues in total, and , the sets and must intersect.
The number can be factored in Hurwitz quaternions:
The norm on Hurwitz quaternions satisfies a form of the Euclidean property: for any quaternion with rational coefficients we can choose a Hurwitz quaternion so that by first choosing so that and then so that for . Then we obtain
It follows that for any Hurwitz quaternions with , there exists a Hurwitz quaternion such that
The ring of Hurwitz quaternions is not commutative, hence it is not an actual Euclidean domain, and it does not have unique factorization in the usual sense. Nevertheless, the property above implies that every right ideal is principal. Thus, there is a Hurwitz quaternion such that
In particular, for some Hurwitz quaternion . If were a unit, would be a multiple of , however this is impossible as is not a Hurwitz quaternion for . Similarly, if were a unit, we would have
so divides , which again contradicts the fact that is not a Hurwitz quaternion. Thus, is not Hurwitz irreducible, as claimed. (en)
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