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  "body": "\n## Statement\n\nLet $R$ be a commutative ring, let $E$ be a linear recurrence with constant coefficients over $R$, and let $q \\in R$.\nThen the sequence $(q^n)_{n \\ge 0}$ is a [[solution|0ZV2YKKZ58]] of $E$ if and only if $q$ is a root of the [[characteristic polynomial|0BWDJGQYST]] $\\chi_E$.\n\n## Remarks\n\nProof idea: for the sequence $q^n$ the recurrence at index $n$ reads $q^n \\chi_E(q) = 0$. At $n = 0$ this is $\\chi_E(q) = 0$.\nOver a field, if $\\chi_E$ has $d$ distinct roots $q_1, \\dots, q_d$, every solution is $u_n = a_1 q_1^n + \\dots + a_d q_d^n$ with constants $a_i$.\n[[Binet's formula|0404D8TN7Y]] is the case of the Fibonacci recurrence.\n\n## Explanations\n\n### L1 human\n\nTry a solution of the form $u_n = q^n$. Put it into the recurrence and divide by $q^n$.\nWhat is left is the equation $\\chi_E(q) = 0$.\n",
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  "name": "Geometric solutions come from roots of the characteristic polynomial",
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