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 ${\displaystyle u=1+l^{2}+m^{2}}$, where l and m are integers.^[8] This can be seen as follows: since p is prime, ${\displaystyle a^{2}\equiv b^{2}{\pmod {p}}}$ can hold for integers ${\displaystyle a,b}$, only when ${\displaystyle a\equiv \pm b{\pmod {p}}}$. Thus, the set ${\displaystyle X=\{0^{2},1^{2},\dots ,((p-1)/2)^{2}\}}$ of squares contains ${\displaystyle (p+1)/2}$ distinct residues modulo p. Likewise, ${\displaystyle Y=\{-(1+x):x\in X\}}$ contains ${\displaystyle (p+1)/2}$ residues. Since there are only p residues in total, and ${\displaystyle |X|+|Y|=p+1>p}$, the sets X and Y must intersect.

The number u can be factored in Hurwitz quaternions: $${\displaystyle 1+l^{2}+m^{2}=(1+l\;\mathbf {i} +m\;\mathbf {j} )(1-l\;\mathbf {i} -m\;\mathbf {j} ).}$$

The norm on Hurwitz quaternions satisfies a form of the Euclidean property: for any quaternion ${\displaystyle \alpha =a_{0}+a_{1}\mathbf {i} +a_{2}\mathbf {j} +a_{3}\mathbf {k} }$ with rational coefficients we can choose a Hurwitz quaternion ${\displaystyle \beta =b_{0}+b_{1}\mathbf {i} +b_{2}\mathbf {j} +b_{3}\mathbf {k} }$ so that ${\displaystyle \mathrm {N} (\alpha -\beta )<1}$ by first choosing ${\displaystyle b_{0}}$ so that ${\displaystyle |a_{0}-b_{0}|\leq 1/4}$ and then ${\displaystyle b_{1},b_{2},b_{3}}$ so that ${\displaystyle |a_{i}-b_{i}|\leq 1/2}$ for ${\displaystyle i=1,2,3}$. Then we obtain $${\displaystyle {\begin{aligned}\mathrm {N} (\alpha -\beta )&=(a_{0}-b_{0})^{2}+(a_{1}-b_{1})^{2}+(a_{2}-b_{2})^{2}+(a_{3}-b_{3})^{2}\\&\leq \left({\frac {1}{4}}\right)^{2}+\left({\frac {1}{2}}\right)^{2}+\left({\frac {1}{2}}\right)^{2}+\left({\frac {1}{2}}\right)^{2}={\frac {13}{16}}<1.\end{aligned}}}$$

It follows that for any Hurwitz quaternions ${\displaystyle \alpha ,\beta }$ with ${\displaystyle \alpha \neq 0}$, there exists a Hurwitz quaternion ${\displaystyle \gamma }$ such that $${\displaystyle \mathrm {N} (\beta -\alpha \gamma )<\mathrm {N} (\alpha ).}$$

The ring H 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 ${\displaystyle \alpha }$ such that $${\displaystyle \alpha H=pH+(1-l\;\mathbf {i} -m\;\mathbf {j} )H.}$$

In particular, ${\displaystyle p=\alpha \beta }$ for some Hurwitz quaternion ${\displaystyle \beta }$. If ${\displaystyle \beta }$ were a unit, ${\displaystyle 1-l\;\mathbf {i} -m\;\mathbf {j} }$ would be a multiple of p, however this is impossible as ${\displaystyle 1/p-l/p\;\mathbf {i} -m/p\;\mathbf {j} }$ is not a Hurwitz quaternion for ${\displaystyle p>2}$. Similarly, if ${\displaystyle \alpha }$ were a unit, we would have $${\displaystyle (1+l\;\mathbf {i} +m\;\mathbf {j} )H=(1+l\;\mathbf {i} +m\;\mathbf {j} )pH+(1+l\;\mathbf {i} +m\;\mathbf {j} )(1-l\;\mathbf {i} -m\;\mathbf {j} )H\subseteq pH}$$ so p divides ${\displaystyle 1+l\;\mathbf {i} +m\;\mathbf {j} }$, which again contradicts the fact that ${\displaystyle 1/p-l/p\;\mathbf {i} -m/p\;\mathbf {j} }$ is not a Hurwitz quaternion. Thus, p is not Hurwitz irreducible, as claimed.