Sums of Four Squares

This is a direct algebraic verification (technique: brute-force expansion, made transparent via quaternion multiplication). Consider the quaternions

in the quaternion algebra $\mathbb{H}$ with $i^2=j^2=k^2=-1$, $ij=k=-ji$, $jk=i=-kj$, $ki=j=-ik$. For a quaternion $\alpha=a_1+a_2i+a_3j+a_4k$ define its conjugate $\bar\alpha = a_1-a_2i-a_3j-a_4k$ and its norm $N(\alpha)=\alpha\bar\alpha = a_1^2+a_2^2+a_3^2+a_4^2$, which is a nonnegative real scalar (the cross terms cancel because $i,j,k$ anticommute pairwise and each squares to $-1$; this is a direct computation from the multiplication rules above). Quaternion multiplication is associative (it is a specific instance of an associative algebra defined by bilinear extension of the rules on basis elements $1,i,j,k$; associativity is checked once on the 4×4×4 table of basis products and then follows for all quaternions by bilinearity). Conjugation is an anti-automorphism: $\overline{\alpha\beta} = \bar\beta\,\bar\alpha$, verified directly from the defining rules by checking it on $1,i,j,k$ and extending bilinearly. Hence for any quaternions $\alpha,\beta$,

using that $N(\beta)$ is a real scalar and hence commutes with $\alpha$. This proves $N(\alpha)N(\beta) = N(\alpha\beta)$. Now compute $\alpha\beta$ explicitly using the multiplication table

and collect the coefficient of $1, i, j, k$ in

Expanding term by term (16 products) and grouping by basis element gives exactly

with $c_1,c_2,c_3,c_4$ as stated. (For instance the coefficient of $1$ collects $a_1b_1$ from $1\cdot 1$, $-a_2b_2$ from $i\cdot i = -1$ times $a_2b_2$, and similarly $-a_3b_3$, $-a_4b_4$ from $j^2,k^2$, giving $c_1=a_1b_1+a_2b_2+a_3b_3+a_4b_4$; the other coefficients are computed the same way using the table.) Then

which is precisely the claimed identity. $\blacksquare$

Immediate from Euler's Four-Square Identity: write $m=a_1^2+a_2^2+a_3^2+a_4^2$ and $n=b_1^2+b_2^2+b_3^2+b_4^2$, then $mn = c_1^2+c_2^2+c_3^2+c_4^2$ for the explicit integers $c_1,\dots,c_4$ (integer because each $c_i$ is an integer linear combination of integer $a_i,b_j$). $\blacksquare$

Immediate by inspection. $\blacksquare$

This is a pigeonhole argument (technique: counting residues, due essentially to Lagrange/Euler and streamlined by later authors such as Ireland–Rosen). Consider the two sets

Each set has exactly $\frac{p+1}{2}$ elements. Indeed, for $0\le a_1 < a_2 \le \frac{p-1}{2}$, if $a_1^2\equiv a_2^2 \pmod p$ then $p \mid (a_2-a_1)(a_2+a_1)$; since $0 < a_2+a_1 \le p-1 < p$, $p$ cannot divide $a_2+a_1$, so $p \mid (a_2-a_1)$, forcing $a_1=a_2$ because $0\le a_2-a_1 < p$. Hence the map $a \mapsto a^2 \bmod p$ is injective on $\{0,1,\dots,\frac{p-1}{2}\}$, giving $|S| = \frac{p-1}{2}+1 = \frac{p+1}{2}$ distinct residues. The identical injectivity argument applied to $b\mapsto -1-b^2 \bmod p$ (composing an injective squaring map with the bijection $x \mapsto -1-x \bmod p$) shows $|T| = \frac{p+1}{2}$ as well. Both $S$ and $T$ are subsets of the set of residues $\mathbb{Z}/p\mathbb{Z}$, which has exactly $p$ elements. Since

the pigeonhole principle forces $S \cap T \neq \emptyset$. So there exist $a,b$ with $0\le a,b\le \frac{p-1}{2}$ such that

Thus $a^2+b^2+1^2+0^2 = mp$ for some non-negative integer $m$. We bound $m$: since $0\le a,b \le \frac{p-1}{2}$,

for $p\ge 2$ (in fact $\frac{(p-1)^2}{2}+1 < p^2$ certainly holds for $p \ge 2$). Hence

Also $m\ge 1$ since $a^2+b^2+1 \ge 1 > 0$ and $mp$ must be a positive multiple of $p$ (it cannot be $0$ since $a^2+b^2+1\ge1$). Thus $1 \le m < p$, as claimed. $\blacksquare$

This is Lagrange's infinite descent (technique: minimal counterexample / descent, sharpened using Euler's identity — the modern streamlined form is due to later expositors, e.g. as in Ireland–Rosen). By Lemma: Some small multiple of $p$ is a sum of four squares, the set of positive integers $m$ such that $mp$ is a sum of four squares is non-empty and contains an element less than $p$; by well-ordering it has a least element $m$, and $1\le m < p$ (in particular $m\ge 1$ exists and $m\le$ the bound already proved, so a minimum with $1\le m<p$ exists). Suppose, for contradiction, that $m>1$. Write

for some integers $a,b,c,d$.

Step 1: $m$ is odd. Suppose $m$ is even. Then $a^2+b^2+c^2+d^2 \equiv 0 \pmod 2$. Among any four integers whose squares sum to an even number, we can always pair $a,b,c,d$ into two pairs of equal parity: reduce mod 2, each of $a,b,c,d$ is $0$ or $1$, and $a^2+b^2+c^2+d^2$ even means the number of odd terms among $a,b,c,d$ is even (0, 2, or 4). In every such case, we may reorder $a,b,c,d$ so that $a\equiv b\pmod 2$ and $c\equiv d \pmod 2$ (if $0$ or $4$ are odd, any pairing works since all agree in a block of parity; if exactly $2$ are odd, pair the two odd ones together and the two even ones together). Then $a+b$, $a-b$, $c+d$, $c-d$ are all even, so

This exhibits $\frac{m}{2}p$ as a sum of four squares (of integers, since the four halved quantities are integers), and $\frac{m}{2} < m$ is a smaller positive integer with $\frac{m}{2}p$ representable — contradicting minimality of $m$. Hence $m$ must be odd.

Step 2: reduce $a,b,c,d$ modulo $m$. Since $m$ is odd and, by hypothesis, $m>1$, we have $m \ge 3$. Choose integers $a_1,b_1,c_1,d_1$ with

(possible since $m$ is odd: every residue class mod $m$ has a unique representative strictly between $-m/2$ and $m/2$). Then

so $a_1^2+b_1^2+c_1^2+d_1^2 = mn$ for some non-negative integer $n$. Moreover, since each of $a_1,b_1,c_1,d_1$ satisfies $|a_1| < m/2$ etc.,

so $n < m$.

Step 3: rule out $n=0$. If $n=0$, then $a_1=b_1=c_1=d_1=0$, meaning $m \mid a$, $m\mid b$, $m\mid c$, $m\mid d$. Then $m^2 \mid a^2+b^2+c^2+d^2 = mp$, so $m \mid p$. Since $1 < m < p$ and $p$ is prime, this is impossible (the only positive divisors of $p$ are $1$ and $p$). Hence $n \neq 0$, so $n \ge 1$.

Step 4: apply Euler's identity to $(mp)(mn) = m^2 pn$. We have two representations

so by Euler's Four-Square Identity,

where, using the identity with $(a_1,a_2,a_3,a_4) \to (a,b,c,d)$ and $(b_1,b_2,b_3,b_4)\to(a_1,b_1,c_1,d_1)$,

Step 5: show each $e_i$ is divisible by $m$, so that $e_i/m$ is an integer. Since $a_1\equiv a, b_1 \equiv b, c_1\equiv c, d_1\equiv d \pmod m$, we compute modulo $m$:

So $m \mid e_1,e_2,e_3,e_4$, and we may set $f_i = e_i/m \in \mathbb{Z}$ for $i=1,2,3,4$. Dividing the identity $m^2pn = e_1^2+e_2^2+e_3^2+e_4^2$ by $m^2$ gives

Step 6: contradiction. This exhibits $pn$ as a sum of four squares, with $1 \le n < m$ (we showed $n\ge1$ in Step 3 and $n<m$ in Step 2). This contradicts the minimality of $m$ as the smallest positive integer with $mp$ a sum of four squares. Therefore the assumption $m>1$ is false, and $m=1$. $\blacksquare$

(Proof of Lagrange's Four-Square Theorem.) We assemble the pieces (technique: multiplicative reduction to primes, then descent on primes — this is the full, non-circular proof). Base cases. By Lemma: Base cases, $0,1,2$ are each sums of four squares. Odd primes. Let $p$ be an odd prime. By Lemma: Some small multiple of $p$ is a sum of four squares, the set $\{m \ge 1 : mp \text{ is a sum of four squares}\}$ is non-empty. By Lemma: Descent step, its least element is $1$, i.e., $p$ itself is a sum of four squares. General $n$. Every integer $n\ge 2$ factors as a product of primes,

where each $q_i \in \{2\} \cup \{\text{odd primes}\}$ (with repetition allowed). Each $q_i$ is a sum of four squares, by the base case $q_i=2$ or by the odd-prime case just proved. By repeated application of Corollary: Multiplicative closure (an induction on $r$: the product of the first $k$ factors is a sum of four squares, and multiplying by the $(k+1)$-st factor — itself a sum of four squares — preserves this by the corollary), $n = q_1\cdots q_r$ is a sum of four squares. Together with $n=0,1$ from the base case, every non-negative integer $n$ is a sum of four squares. $\blacksquare$