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Frictionless Jellyfish deleted his useful (but snide) answer, so here is some elaboration. Define

$r[q] = f2[q]^{24} + 2^{12}/f2[q]^{24} = q^{-1} - 24 + 4372q + O(q^{2})$,

where f2[.] is a Weber function. Weber functions satisfy a quadratic polynomials for Class 2 numbers such as 232:

$r[e^{-\pi \sqrt{58}}] = 64(((5 + \sqrt{29})/2)^{12} + 2^{12}/(5 + \sqrt{29})^{12}) = 24591257728 ~ = e^{\pi \sqrt{232}} - 24 + ...$.

Note that r[q] is specifically engineered to cancel square-roots and be an exact rational integer.

Alternatively we could work directly with $2^{12}/f2[q]^{24} = q^{-1} - 24 + 276q + O(q^{2})$:

$2^{12}/f2[e^{-\pi \sqrt{58}}]^{24} = 64(((5 + \sqrt{29})/2)^{12}$.

The number (5 $(5 + \sqrt{29})/2) sqrt{29})/2)$ is a Pisot number because its conjugate is less than unity, so the twelfth power is approximately a rational integer.

2 formatting; added 7 characters in body

Frictionless Jellyfish deleted his useful (but snide) answer, so here is some elaboration. Define

$r[q] = f2[q]^{24} + 2^{12}/f2[q]^{24} = q^{-1} - 24 + 4372q + O(q^{2})$,

where f2[.] is a Weber function. The Weber function satisfies functions satisfy a quadratic polynomial polynomials for Class 2 numbers such as 232:

$r[e^{-\pi \sqrt{58}}] = 64(((5 + \sqrt{29})/2)^{12} + 2^{12}/(5 + \sqrt{29})^{12}) = 24591257728 ~ e^{\pi sqrt{232}} \sqrt{232}} - 24$24 + ...$. Note that r[q] is specifically engineered to cancel square-roots and be an exact rational integer. Alternatively we could work directly with$2^{12}/f2[q]^{24} = q^{-1} - 24 + 276q + O(q^{2})$:$2^{12}/f2[e^{-\pi \sqrt{58}}]^{24} = 64(((5 + \sqrt{29})/2)^{12}$. The number (5 + sqrt{29})/2) \sqrt{29})/2) is a Pisot number because its conjugate is less than unity, so the twelfth power is approximately a rational integer. 1 Frictionless Jellyfish deleted his useful (but snide) answer, so here is some elaboration. Define$r[q] = f2[q]^{24} + 2^{12}/f2[q]^{24} = q^{-1} - 24 + 4372q + O(q^{2})$, where f2[.] is a Weber function. The Weber function satisfies a quadratic polynomial for Class 2 numbers such as 232:$r[e^{-\pi \sqrt{58}}] = 64(((5 + \sqrt{29})/2)^{12} + 2^{12}/(5 + \sqrt{29})^{12}) = 24591257728 ~ e^{\pi sqrt{232}} - 24$. Note that r[q] is specifically engineered to cancel square-roots and be an exact rational integer. Alternatively we could work directly with$2^{12}/f2[q]^{24} = q^{-1} - 24 + 276q + O(q^{2})$:$2^{12}/f2[e^{-\pi \sqrt{58}}]^{24} = 64(((5 + \sqrt{29})/2)^{12}\$.

The number (5 + sqrt{29})/2) is a Pisot number because its conjugate is less than unity, so the twelfth power is approximately a rational integer.