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It is well known that the pell-equation $$a^2-n\cdot b^2 =1$$ has a non-trivial solution for every positive integer $n$ not being a perfect square. And the fundamental solution can be very large.

What about the analogue equation $$a^3-n\cdot b^3=1$$ ?

I checked the first $30$ convergents of the non-cubes upto $10^5$ and only found the following solutions with $b\ge 10$ :

? for(n=1,10^5,s=sqrtnint(n,3);if(s^3<>n,for(j=1,30,w=contfracpnqn(contfrac(n^(1
/3)),j);[a,b]=[w[1,j],w[2,j]];if(a^3-b^3*n==1,if(b>=10,print(n,"  ",a," ",b)))))
)
635  361 42
5080  361 21
17145  361 14
18745  1036 39
32042  667 21
48949  4097 112
?

So, a "spectaculary" equation as $$4097^3-112^3\cdot 48949=1$$ seems to occur rather rarely. I would like to know something about how much such solutions exist, in particular can $b$ get arbitary large ?

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    For sure, if $a^3-n b^3=1$ then $\frac{a}{b}$ has to be a convergent of the continued fraction of $\sqrt[3]{n}$. In particular, we are looking for $n$s such that the continued fraction of $\sqrt[3]{n}$ has an exceptionally large element. – 2017-02-28
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    @JackD'Aurizio My best result yet is $$59320^3-273^3\cdot 10259247=1$$ – 2017-02-28
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    @JackD'Aurizio And that very large entry must come early enough but not too early, which makes it more difficult to find an example. – 2017-02-28
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    @JackD'Aurizio By the way, it is often claimed that convergents are best-approximations. This is not necessarily true for the first convergent, right ? Similar, for the semiconvergents, we need at least two entries , right ? – 2017-02-28
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    I wonder if it makes sense to go in the opposite direction, i.e. to construct few terms of a continued fraction in order to meet such constraints, then consider the natural number closer to the cube of such continued fraction. – 2017-02-28
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    @JackD'Aurizio This might well be a good approach! We could increase the first entry until we are close to a cube-root. I will try it! – 2017-02-28
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    However, that is not granted to be effective, since by Lagrange's theorem the continued fraction of $\sqrt{n}$ has a nice palyndromic structure, but I am not aware of similar results for the continued fraction of $\sqrt[3]{n}$. – 2017-02-28

1 Answers 1

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Well, you could just, for example, pick $b$ to be your favourite large prime with $b \equiv 1 \mod{3}$. Then there are three cube roots of unit modulo $b^3$. Choose a nontrivial one of these and you have an example. For instance, if $b=1000003$, we have $$ 333357166884000946^3 - 37044648719139238281882107242355205 \cdot 1000003^3=1. $$