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  "body": "\n## Definition\n\nLet $n \\ge 0$ be an integer.\nA *Littlewood polynomial* of degree $n$ is a polynomial\n$$p(z) = \\sum_{k=0}^{n} a_k z^k$$\nin one complex variable $z$ whose coefficients satisfy $a_k \\in \\set{-1, 1}$ for $k = 0, \\ldots, n$.\n\n## Remarks\n\nOn the unit circle, $\\theta \\mapsto p(e^{i\\theta})$ is a [[trigonometric polynomial|J3NJQ5FZJ5]] with coefficients $a_k$.\nBy orthogonality of the functions $e^{ik\\theta}$,\n$$\\frac{1}{2\\pi} \\int_{-\\pi}^{\\pi} \\abs{p(e^{i\\theta})}^2 \\, d\\theta = \\sum_{k=0}^{n} a_k^2 = n + 1.$$\nSo $\\max_{\\abs{z} = 1} \\abs{p(z)} \\ge \\sqrt{n+1}$ and $\\min_{\\abs{z} = 1} \\abs{p(z)} \\le \\sqrt{n+1}$.\nLittlewood asked whether there are constants $0 \u003c A \\le B$ and Littlewood polynomials of every degree with $A \\sqrt{n+1} \\le \\abs{p(z)} \\le B \\sqrt{n+1}$ on $\\abs{z} = 1$.\nBalister, Booth, Bollobás, Morris and Tiba answered this positively in 2019.\n\nExample: $p(z) = 1 + z - z^2$ has degree $2$ and coefficients $1, 1, -1$, all in $\\set{-1, 1}$.\nAt $z = 1$ its value is $1$, and at $z = -1$ it is $-1$.\n\nNon-example: $q(z) = 1 + z^2 = 1 + 0 \\cdot z + z^2$ is not a Littlewood polynomial, since its coefficient of $z$ is $0$.\nPolynomials with coefficients in $\\set{-1, 0, 1}$ form a different, larger class.\n",
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