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  "body": "\n## Statement\n\nLet $p \\ge 10$ be a [[prime number|0SN0MAEKYX]] with decimal expansion $p = a_m 10^m + \\dots + a_1 10 + a_0$, where $m \\ge 1$, $a_i \\in \\set{0, 1, \\dots, 9}$ and $a_m \\ne 0$.\nThen the polynomial $f = a_m x^m + \\dots + a_1 x + a_0$ is [[irreducible|00X2A9G9W3]] in $\\Z[x]$.\n\n## Remarks\n\nProof idea: since the coefficients are digits, every complex root $\\alpha$ of $f$ satisfies $\\abs{10 - \\alpha} \u003e 1$.\nSo every factor $g \\in \\Z[x]$ of positive degree has $\\abs{g(10)} \u003e 1$.\nA factorisation $f = g h$ into factors of positive degree would then write $p = f(10) = g(10) h(10)$ as a product of two integers of absolute value greater than $1$.\nA constant factor is also excluded: it would divide all digits and $p$, but the digits are less than $p$.\nExample: $113$ is prime, so $x^2 + x + 3$ is irreducible; indeed its discriminant $1 - 12 = -11$ is negative.\nThe theorem also holds in every base $b \\ge 2$ in place of $10$.\nThe converse is false: $x^2 + 5$ is irreducible, but its value $105$ at $10$ is not prime.\n\n## Explanations\n\n### L1 human\n\nWrite a prime in decimal and read its digits as the coefficients of a polynomial.\nThat polynomial cannot be factored over the integers.\nFor example, the prime $1013$ gives $x^3 + x + 3$, which is irreducible.\n\n### L2 human\n\nThe criterion links primality of a value with irreducibility of the polynomial.\nThe reverse direction, that an irreducible polynomial with suitable conditions takes infinitely many prime values, is the Bunyakovsky conjecture, which is open in every degree at least $2$.\nThe proof is a good example of locating roots in the complex plane to prove an algebraic fact, like [[Perron's criterion|5BCRQYVYR8]].\n",
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