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

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

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.

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