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  "body": "\n## Statement\n\nLet $n \\ge 1$ be an integer, and let $r_4(n)$ be the number of quadruples $(a, b, c, d)$ of [[integers|0C4JS9RF5M]] with $a^2 + b^2 + c^2 + d^2 = n$.\nThen\n$$r_4(n) = 8 \\sum_{\\substack{d \\mid n \\\\ 4 \\nmid d}} d.$$\nLet $\\sigma$ be the [[sum of divisors|06TK51263Q]], and let $m$ be the largest odd divisor of $n$.\nEquivalently, $r_4(n) = 8\\sigma(n)$ if $n$ is odd, and $r_4(n) = 24\\sigma(m)$ if $n$ is even.\n\n## Remarks\n\nOne proof uses the series $\\sum_{n \\in \\Z} q^{n^2}$ with $q = e^{2 \\pi i \\tau}$, which is $\\theta(2\\tau)$ for the theta constant $\\theta$ of the [[theta function|2F5CYQJVAF]].\nIts fourth power is $\\sum_{n \\ge 0} r_4(n) q^n$.\nThe function $\\theta(2\\tau)^4$ is a [[modular form|GWKN5VG78A]] of weight $2$ for the [[congruence subgroup|1H6R856P3C]] $\\Gamma_0(4)$.\nThat space has dimension $2$ and is spanned by Eisenstein series, and comparing two coefficients gives the formula.\nElementary proofs also exist.\nExample: $n = 4$ has divisors $1, 2$ not divisible by $4$, so $r_4(4) = 8 \\cdot 3 = 24$.\nCheck: the solutions are the $8$ quadruples with one entry $\\pm 2$ and three zeros, and the $16$ quadruples with all entries $\\pm 1$.\nSince the sum always contains $d = 1$, $r_4(n) \\ge 8 \u003e 0$, which gives [[Lagrange's four-square theorem|0KYYSMEVHH]].\n\n## Explanations\n\n### L1 human\n\nEvery positive integer is a sum of four squares, and this theorem says exactly in how many ways.\nThe count is $8$ times the sum of the divisors of $n$ that are not multiples of $4$.\nOrder and signs count: $1 = 1^2 + 0^2 + 0^2 + 0^2$ gives $8$ representations, one for each position and sign.\nIt turns an existence result into an exact formula.\n\n### L2 human\n\nThe theorem is a model case of how modular forms solve counting problems.\nA generating function of a counting problem is shown to be modular, and the finite dimension of the space of modular forms identifies it.\nFor two squares the same method gives $r_2(n) = 4 \\sum_{d \\mid n,\\, d \\text{ odd}} \\chi(d)$, where $\\chi(d) = (-1)^{(d-1)/2}$.\nFor larger numbers of squares, cusp forms can appear, and the exact formula then has a cusp form term besides the divisor sums.\n",
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