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Latest revision as of 10:02, 11 March 2025

Problem

Let $S$ be the set of complex numbers of the form $c+di$ such that $c+di = (a+bi)^{12}$ for some integers $a$ and $b$. Find the largest integer that must divide $d$ for all numbers in $S$.

Solution

$d=\Im (z^{12})=\dfrac{z^{12}-\overline{z}^{12}}{2}=4\cdot\dfrac{z^6+\overline{z}^6}{2}\cdot\dfrac{z^3+\overline{z}^3}{2}\cdot\dfrac{z^3-\overline{z}^3}{2}=4\Re (z^6)\Re (z^3)\Im(z^3)$ Plug in $z=a+bi$ and factor to $d=f(a,b)=4ab(a+b)(a-b)(a^2 -3b^2)(3a^2-b^2)(a^4-14a^2b^2+b^4)$

Let $g$ be the desired $\gcd$ of all $d=f(a,b)$. Since $f(2,1)=-2^3\times 3^2\times 11\times 13$ and $f(3,2)=2^3\times 3^2\times 5\times 11\times 23\times 37$, our $g$ is at most $2^3\times 3^2\times 11=\boxed{792}$. We now prove that this is indeed the case:


(1) $2^3\mid g$. This is easy: $4ab(a-b)$ is always divisible by $8$ because one of $a,b,a-b$ is always even.

(2) $3^2\mid g$. First, $3\mid ab(a+b)(a-b)$ because always either $b\equiv 0\pmod{3}$ or $a\in\{ 0,\pm b\}\pmod{3}$. Second, $3\mid (a^2 -3b^2)(3a^2-b^2)(a^4-14a^2b^2+b^4)$ because either $3$ divides one of $a,b$ or $a^4-14a^2b^2+b^4\equiv 1-14+1\equiv 0\pmod{3}$ by FLT.

(3) $11\mid g$. For the sake of contradiction assume that $11$ does not divide any of $a,b,a+b,a-b,a^2-3b^2,3a^2-b^2$. This gives $b\not\equiv 0\pmod{11}$ and $a\not\in\{ 0,\pm b,\pm 2b,\pm 5b\}\pmod{11}\implies a\in\{ \pm 3b,\pm 4b\}\pmod{11}$. If $a\equiv \pm 3b\pmod{11}$, then $a^4-14a^2b^2+b^4\equiv (3b)^4-3(3b)^2b^2+b^4\equiv 55b^4\equiv 0\pmod{11}$. If $a\equiv \pm 4b\pmod{11}$, then $b\equiv \pm 3a\mod{11}$, and this case is symmetric to $a\equiv \pm 3b\pmod{11}$. So $11\nmid ab(a+b)(a-b)(a^2-3b^2)(3a^2-b^2)\implies 11\mid (a^4-14a^2b^2+b^4)$, and we're done.

See Also

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