ラグランジュの四平方定理

It is sufficient to prove the theorem for every odd prime number p. This immediately follows from Euler's four-square identity (and from the fact that the theorem is true for the numbers 1 and 2).

The residues of a² modulo p are distinct for every a between 0 and (p − 1)/2 (inclusive). To see this, take some a and define c as a² mod p. a is a root of the polynomial x² − c over the field Z/pZ. So is p − a (which is different from a). In a field K, any polynomial of degree n has at most n distinct roots (Lagrange's theorem (number theory)), so there are no other a with this property, in particular not among 0 to (p − 1)/2.

Similarly, for b taking integral values between 0 and (p − 1)/2 (inclusive), the −b² − 1 are distinct. By the pigeonhole principle, there are a and b in this range, for which a² and −b² − 1 are congruent modulo p, that is for which $${\displaystyle a^{2}+b^{2}+1^{2}+0^{2}=np.}$$

Now let m be the smallest positive integer such that mp is the sum of four squares, x₁² + x₂² + x₃² + x₄² (we have just shown that there is some m (namely n) with this property, so there is a least one m, and it is smaller than p). 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 (this is in the spirit of the infinite descent^[7] method of Fermat).

For this purpose, we consider for each x_i the y_i which is in the same residue class modulo m and between (–m + 1)/2 and m/2 (possibly included). It follows that y₁² + y₂² + y₃² + y₄² = mr, for some strictly positive integer r less than m.

Finally, another appeal to Euler's four-square identity shows that mpmr = z₁² + z₂² + z₃² + z₄². But the fact that each x_i is congruent to its corresponding y_i implies that all of the z_i are divisible by m. Indeed, $${\displaystyle {\begin{cases}z_{1}&=x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}+x_{4}y_{4}&\equiv x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}&=mp\equiv 0&{\pmod {m}},\\z_{2}&=x_{1}y_{2}-x_{2}y_{1}+x_{3}y_{4}-x_{4}y_{3}&\equiv x_{1}x_{2}-x_{2}x_{1}+x_{3}x_{4}-x_{4}x_{3}&=0&{\pmod {m}},\\z_{3}&=x_{1}y_{3}-x_{2}y_{4}-x_{3}y_{1}+x_{4}y_{2}&\equiv x_{1}x_{3}-x_{2}x_{4}-x_{3}x_{1}+x_{4}x_{2}&=0&{\pmod {m}},\\z_{4}&=x_{1}y_{4}+x_{2}y_{3}-x_{3}y_{2}-x_{4}y_{1}&\equiv x_{1}x_{4}+x_{2}x_{3}-x_{3}x_{2}-x_{4}x_{1}&=0&{\pmod {m}}.\end{cases}}}$$

It follows that, for w_i = z_i/m, w₁² + w₂² + w₃² + w₄² = rp, and this is in contradiction with the minimality of m.

In the descent above, we must rule out both the case y₁ = y₂ = y₃ = y₄ = m/2 (which would give r = m and no descent), and also the case y₁ = y₂ = y₃ = y₄ = 0 (which would give r = 0 rather than strictly positive). For both of those cases, one can check that mp = x₁² + x₂² + x₃² + x₄² would be a multiple of m², contradicting the fact that p is a prime greater than m.

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 $${\displaystyle \alpha ={\frac {1}{2}}E_{0}(1+\mathbf {i} +\mathbf {j} +\mathbf {k} )+E_{1}\mathbf {i} +E_{2}\mathbf {j} +E_{3}\mathbf {k} =a_{0}+a_{1}\mathbf {i} +a_{2}\mathbf {j} +a_{3}\mathbf {k} }$$ where ${\displaystyle E_{0},E_{1},E_{2},E_{3}}$ are integers. Thus, the quaternion components ${\displaystyle a_{0},a_{1},a_{2},a_{3}}$ are either all integers or all half-integers, depending on whether ${\displaystyle E_{0}}$ 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${\displaystyle \mathrm {N} (\alpha )}$ of a rational quaternion ${\displaystyle \alpha }$ is the nonnegative rational number$${\displaystyle \mathrm {N} (\alpha )=\alpha {\bar {\alpha }}=a_{0}^{2}+a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}$$ where ${\displaystyle {\bar {\alpha }}=a_{0}-a_{1}\mathbf {i} -a_{2}\mathbf {j} -a_{3}\mathbf {k} }$ is the conjugate of ${\displaystyle \alpha }$. Note that the norm of a Hurwitz quaternion is always an integer. (If the coefficients are half-integers, then their squares are of the form ${\displaystyle {\tfrac {1}{4}}+n:n\in \mathbb {Z} }$, and the sum of four such numbers is 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: $${\displaystyle \mathrm {N} (\alpha \beta )=\alpha \beta ({\overline {\alpha \beta }})=\alpha \beta {\bar {\beta }}{\bar {\alpha }}=\alpha \mathrm {N} (\beta ){\bar {\alpha }}=\alpha {\bar {\alpha }}\mathrm {N} (\beta )=\mathrm {N} (\alpha )\mathrm {N} (\beta ).}$$

For any ${\displaystyle \alpha \neq 0}$, ${\displaystyle \alpha ^{-1}={\bar {\alpha }}\mathrm {N} (\alpha )^{-1}}$. It follows easily that ${\displaystyle \alpha }$ is a unit in the ring of Hurwitz quaternions if and only if ${\displaystyle \mathrm {N} (\alpha )=1}$.

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 ${\displaystyle 2=1^{2}+1^{2}+0^{2}+0^{2}}$. To show this for an odd prime integer p, represent it as a quaternion ${\displaystyle (p,0,0,0)}$ and assume for now (as we shall show later) that it is not a Hurwitz irreducible; that is, it can be factored into two non-unit Hurwitz quaternions $${\displaystyle p=\alpha \beta .}$$

The norms of ${\displaystyle p,\alpha ,\beta }$ are integers such that $${\displaystyle \mathrm {N} (p)=p^{2}=\mathrm {N} (\alpha \beta )=\mathrm {N} (\alpha )\mathrm {N} (\beta )}$$ and ${\displaystyle \mathrm {N} (\alpha ),\mathrm {N} (\beta )>1}$. This shows that both ${\displaystyle \mathrm {N} (\alpha )}$ and ${\displaystyle \mathrm {N} (\beta )}$ are equal to p (since they are integers), and p is the sum of four squares $${\displaystyle p=\mathrm {N} (\alpha )=a_{0}^{2}+a_{1}^{2}+a_{2}^{2}+a_{3}^{2}.}$$

If it happens that the ${\displaystyle \alpha }$ chosen has half-integer coefficients, it can be replaced by another Hurwitz quaternion. Choose ${\displaystyle \omega =(\pm 1\pm \mathbf {i} \pm \mathbf {j} \pm \mathbf {k} )/2}$ in such a way that ${\displaystyle \gamma \equiv \omega +\alpha }$ has even integer coefficients. Then $${\displaystyle p=({\bar {\gamma }}-{\bar {\omega }})\omega {\bar {\omega }}(\gamma -\omega )=({\bar {\gamma }}\omega -1)({\bar {\omega }}\gamma -1).}$$

Since ${\displaystyle \gamma }$ has even integer coefficients, ${\displaystyle ({\bar {\omega }}\gamma -1)}$ will have integer coefficients and can be used instead of the original ${\displaystyle \alpha }$ to give a representation of p as the sum of four squares.

As for showing that p is not a Hurwitz irreducible, Lagrange proved that any odd prime p divides at least one number of the form ${\displaystyle u=1+l^{2}+m^{2}}$ここで、lとmは整数である。^{[ 8 ]}これは次のように見ることができる。pは素数であるため、${\displaystyle a^{2}\equiv b^{2}{\pmod {p}}}$整数に対して成り立つ${\displaystyle a,b}$のみ${\displaystyle a\equiv \pm b{\pmod {p}}}$したがって、集合は${\displaystyle X=\{0^{2},1^{2},\dots ,((p-1)/2)^{2}\}}$正方形には${\displaystyle (p+1)/2}$pを法とする異なる残基。同様に、${\displaystyle Y=\{-(1+x):x\in X\}}$を含む${\displaystyle (p+1)/2}$残基。合計でp個の残基しかないため、${\displaystyle |X|+|Y|=p+1>p}$集合XとYは交差しなければならない。

数u は、フルヴィッツ四元数に因数分解できます。 $${\displaystyle 1+l^{2}+m^{2}=(1+l\;\mathbf {i} +m\;\mathbf {j} )(1-l\;\mathbf {i} -m\;\mathbf {j} ).}$$

フルヴィッツ四元数のノルムは、ユークリッドの性質の一形態を満たす。任意の四元数に対して${\displaystyle \alpha =a_{0}+a_{1}\mathbf {i} +a_{2}\mathbf {j} +a_{3}\mathbf {k} }$有理係数を用いると、フルヴィッツ四元数を選択できる。${\displaystyle \beta =b_{0}+b_{1}\mathbf {i} +b_{2}\mathbf {j} +b_{3}\mathbf {k} }$となることによって${\displaystyle \mathrm {N} (\alpha -\beta )<1}$まず最初に選択することで${\displaystyle b_{0}}$となることによって${\displaystyle |a_{0}-b_{0}|\leq 1/4}$その後${\displaystyle b_{1},b_{2},b_{3}}$となることによって${\displaystyle |a_{i}-b_{i}|\leq 1/2}$のために${\displaystyle i=1,2,3}$すると、次のようになります。 $${\displaystyle {\begin{aligned}\mathrm {N} (\alpha -\beta )&=(a_{0}-b_{0})^{2}+(a_{1}-b_{1})^{2}+(a_{2}-b_{2})^{2}+(a_{3}-b_{3})^{2}\\&\leq \left({\frac {1}{4}}\right)^{2}+\left({\frac {1}{2}}\right)^{2}+\left({\frac {1}{2}}\right)^{2}+\left({\frac {1}{2}}\right)^{2}={\frac {13}{16}}<1.\end{aligned}}}$$

したがって、任意のフルヴィッツ四元数に対して${\displaystyle \alpha ,\beta }$と${\displaystyle \alpha \neq 0}$フルヴィッツ四元数が存在する${\displaystyle \gamma }$そのため $${\displaystyle \mathrm {N} (\beta -\alpha \gamma )<\mathrm {N} (\alpha ).}$$

フルヴィッツ四元数の環Hは可換ではないため、実際のユークリッド整域ではなく、通常の意味での一意分解を持ちません。しかしながら、上記の性質は、すべての右イデアルが主イデアルであることを意味します。したがって、フルヴィッツ四元数 が存在します。${\displaystyle \alpha }$そのため $${\displaystyle \alpha H=pH+(1-l\;\mathbf {i} -m\;\mathbf {j} )H.}$$

特に、${\displaystyle p=\alpha \beta }$あるフルヴィッツ四元数に対して${\displaystyle \beta }$。 もし${\displaystyle \beta }$単位であった、${\displaystyle 1-l\;\mathbf {i} -m\;\mathbf {j} }$pの倍数になるはずですが、これは不可能です。${\displaystyle 1/p-l/p\;\mathbf {i} -m/p\;\mathbf {j} }$はフルヴィッツ四元数ではない${\displaystyle p>2}$同様に、${\displaystyle \alpha }$もしユニットだったら、 $${\displaystyle (1+l\;\mathbf {i} +m\;\mathbf {j} )H=(1+l\;\mathbf {i} +m\;\mathbf {j} )pH+(1+l\;\mathbf {i} +m\;\mathbf {j} )(1-l\;\mathbf {i} -m\;\mathbf {j} )H\subseteq pH}$$ したがって、p は割り切れる${\displaystyle 1+l\;\mathbf {i} +m\;\mathbf {j} }$これはまた、${\displaystyle 1/p-l/p\;\mathbf {i} -m/p\;\mathbf {j} }$はフルヴィッツ四元数ではない。したがって、主張されているように、 pはフルヴィッツ既約ではない。