You'll find something interesting if you take the golden ratio, \( \phi = \frac{1 + \sqrt{5}}{2} \), to high powers:

\[ \phi^6 = 17.944 \] \[ \phi^{10} = 122.992 \] \[ \phi^{15} = 1364.0007 \]

It seems as if large exponents of \(\phi\) come arbitrarily close to integer values. In fact, \(\phi\) isn't the only number which does this. For example, \( (2+ \sqrt{3})^{5} = 723.999 \). This is no coincidence.

These numbers - known as Pisot-Vijayaraghavan numbers - produce "almost-integers" when raised to high powers. I'll give a quick sketch of why this is, as well as an easy way to find other numbers with this property:

We can take a number \(q = a + \sqrt{b}\), where \(a\) and \(b\) are integers, and consider how it behaves under an operation called Galois conjugation. Essentially, this is just a more general version of what you may already know as complex conjugation. To find the Galois conjugate of \(q\), simply change the sign of the square root term: \(q* = a - \sqrt{b}\). This is equivalent to setting \(q\) to be the root of the polynomial \(x^2 - 2ax + a^2 - b = 0\) and finding the second root, or \(q*\).

A few properties of Galois conjugation can be easily deduced, and are already implicit in the quadratic equation above:

\[ q + q* = 2a \] \[ qq* = a^2 - b \] \[ q^2 + q*^2 = 2a^2 + 2b \] \[ q^n + q*^n = \sum \binom{n}{k} a^{n-2k}b^{2k} \]

The last equality is important here. Notice especially that these four equations all produce integers results on the righthand side. Now, whenever \( |q*| < 1 \), the second term of \( q^n + q*^n \) becomes arbitrarily small for increasing \(n\). This immediately implies that

\[ q^n + q*^n \approx q^n \approx \sum \binom{n}{k} a^{n-2k}b^{2k} \qquad n \gg 1 \]

Thus, any number \(q\) whose Galois conjugate satisfies \(|q*| < 1\) will produce near-integer values when raised to high powers. Here are some examples of such numbers:

\[ (2 + \sqrt{6})^7 = 34528.0037 \] \[ (4 + \sqrt{21})^9 = 252672968.008 \] \[ (1 + \sqrt{3})^{15} = 3526400.0093 \]

*It's worth noting that in the case that \(q\) has a denominator of \(t\), as is the case with \( \phi\) where \(t=2\), then we also need the further restriction that \(t^2 | qq*\), or that \(a^2 -b\) is divisible by \(t^2\) so that \(qq*\) remains an integer.

Note by Levi Adam Walker
3 weeks, 2 days ago

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@Levi Adam Walker Yes, this is amazing right???!! I got to know about these numbers quite recently I must confess........But, check out "Heegner Number "...........Wikipedia has a page on it too...........when it comes to "Close" approximations, I prefer to quote them........:)

Aaghaz Mahajan - 3 weeks, 1 day ago

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Yeah, they're much cooler, but also much harder to explain :p

Levi Adam Walker - 3 weeks, 1 day ago

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Does this also work with cuberoots? I suppose \( q* \) has to be defined differently, right?

Henry U - 2 weeks, 4 days ago

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It won't generally work with this construction, since \(q^n + q*^n\) must give an integer and cube roots tend to act erratically when raised to arbitrary powers.

Levi Adam Walker - 2 weeks, 4 days ago

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Ah, right. Thanks!

Henry U - 2 weeks, 4 days ago

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