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Assume $a,b,c, N$ as positive integers, let primitive be $\gcd(a,b,c) = 1$ and,

$$a^2+b^2 = c^2\tag{1}$$

Supposing you want to know how many solutions there are with $c$ less than a bound $N$. Lehmer showed that, define the number of primitive Pythagorean triples as $\Delta N$, then,

$$\lim_{N\to\infty}\frac{\Delta N}{N}=\frac{1}{2\pi}$$

Thus, for example, for $N = 10^5$, then $\Delta N \approx \frac{1}{2\pi}10^5 \approx 15915.5$. (The exact number is $15919$.)

I wondered about $(1)$'s cubic analogue. The closest seems to be,

$$a^3+b^3 = c^3\pm1\tag{2}$$

and there are exactly $133$ solutions with $c<10^6$.

Question: How many solutions for $c<10^7$, or $c<10^8$? In general, just like Lehmer's result, is there a way to estimate how many solutions $(2)$ will have for $c<N$?

P.S. A question of mine on the asymptotics of the analogous $a^2+b^2 = c^2\pm1$ was ably answered by mercio in this post.

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Tito Piezas III
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  • The good thing about $a^2+b^2 = c^2$ was that it had genus $0$ so the curves $a^2+b^2 = c^2+k$ had lots of automorphisms. From one solution you could obtain a lot more (in a kind of a tree shape), and you could compute the density subtree by subtree. I once tried looking at $a^2 + b^2 = z(z+k)(z+l)$, but I couldn't predict anything, not even how the density distributed along the congruence classes of $z$. $a^3+b^3 = c^3$ is an elliptic curve, so it's totally different thing. Also, it seems harder computationally, I don't know any easy way to count the solutions of $x^3+y^3 = n$ – mercio Oct 01 '13 at 16:31
  • A key difference between (1) and (2) is that in (2) the number of terms on the left is one less than the power. For sufficiently large N I would expect $\Delta N$ for (2) to approximate to $kH_N$ where k is a constant and $H_N$ is the Nth harmonic number. If so the limit of $\Delta N/N$ will be zero. – Adam Bailey Oct 13 '13 at 22:45

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