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This is a look-a-like to Fermat's last theorem for $n=3$, but it has solutions! I believe that its solution requires knowledge of the techniques of algebraic or analytic number theory which I don't have.

Thomas Andrews
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mike brady
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  • It's not surprising that such things have solutions. And it certainly doesn't require any trick to show it has solutions - just find solutions. Or do you want to classify all solutions? – Thomas Andrews Mar 15 '15 at 03:52
  • Would like an initial solution and formula to generate infinitely many more. – mike brady Mar 15 '15 at 20:59
  • @mikebrady: Formula $(1)$ below gives an infinite more solutions. – Tito Piezas III Mar 16 '15 at 01:27
  • What about an initial solution? – mike brady Mar 16 '15 at 01:57
  • @mikebrady: Let me make it easier. Given a "seed" solution, then formula $(1)$ will yield an infinite more. For example, using $x,y,z = 17299, 25469, 9954$ as the "seed" yields, $$x,y,z = -661146496267328783, 684469533791312783, 112919729369578740$$ Iterating the process two more times finally yields positive $x,y,z$. Then repeat an infinite number of times to get an infinite supply of positive (as well as non-positive) solutions. – Tito Piezas III Mar 16 '15 at 03:57
  • It's obvious $a^3+b^3$ is divisible by 11. Less obvious, this implies the same thing for $a+b$. Not sure what that gets you, but narrows down trials a bit. – TurlocTheRed Aug 17 '22 at 04:50

3 Answers3

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By a theorem of Satgé, there are rational solutions to,

$$x^3+y^3 = 2p$$

if $p \equiv 2\; \text{(mod 9)}$, of which your $p=11$ qualifies. If there is an initial non-zero integer solution to,

$$x^3+y^3 = cz^3$$

then infinitely many subsequent ones can be found as,

$$(-x^4 - 2x y^3)^3 + (2x^3 y + y^4)^3 = c (-x^3z + y^3z)^3\tag1$$

Positive $x,y,z$ will appear every few iterations. However, this identity will not give all solutions. See this related post.

P.S. By the way, what's the initial solution to $x^3+y^3=22z^3$?

Community
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Tito Piezas III
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  • Yes, I would like to find an initial solution. – mike brady Mar 15 '15 at 18:11
  • Here is a solution but probably too large to be an initial one. x=17,299; y=25,469; z=9,954. – mike brady Mar 15 '15 at 21:05
  • @mikebrady: How did you find this? – Tito Piezas III Mar 16 '15 at 01:22
  • Good question but how is not science. Google Solutions to x^3+y^3=22z^3 and scroll to Fermat Last Theorem Look-alike Physics Forum and scroll to Marty Ellis. I was browsing book Riemann Hypothesis by Karl Sabbagh and on page 8 he references said formula which comes from paper by Louis de Branges titled Apology for proof of the Riemann Hypothesis which can be found on line. Branges says he solved it while a student. No idea where Ellis got the specific solution. – mike brady Mar 16 '15 at 02:24
  • Ah, you mean this. – Tito Piezas III Mar 16 '15 at 04:22
  • Reading Ellis blogs again he says how is exhaustive search on HP calculator. I will contact some professor friends to ask for initial solution and let you know. Thanks for help so far. – mike brady Mar 16 '15 at 13:52
  • The more I think about it, the solution I gave may be initial. x/z and y/z are certainly in reduced form since x and y are both prime. I'm still working on it. – mike brady Mar 17 '15 at 19:04
  • @mikebrady: It seems it is indeed the smallest. See Hisanori Mishima's list for $x^3+y^3 = Nz^3$ with $N<1000$. – Tito Piezas III Mar 18 '15 at 07:43
  • Great list Tito, with de Branges recollection this doe it for me. MIke – mike brady Mar 18 '15 at 16:03
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I believe that the Marty Ellis solution is in fact an initial solution. Not being an expert in Diophantine equations it just appeared large to me, but Louis de Branges, who found solutions years ago, confirmed to me that his original generation solution was indeed of this size. This along with Tito Piezas' formula for generation of infinitely more solutions satisfies me that my original questions are answered. Mike.

mike brady
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Might have found something to generate more from an initial solution.

Suppose $a^3+b^3=22c^3$. If want the smallest such $a$ and $b$, they have to be relatively prime to each other and to $c$. $a^3+b^3=(a+b)(a^2+b^2-ab)$. Via trial and error, $a^2+b^2-ab$ cannot be divisible by 11 unless both $a$ and $b$ are, which is not allowed. Combined, this tells us $a+b$ is divisible by $22$.

After substitution: $a^3+(22k-a)^3=22c^3$. After expanding, cancelling the term cubic in $a$ and finding the discriminant of the resulting quadratic, one finds $c^3=11^2k^3+3n^2k=k(11^2k^2+3n^2)$ where $a=11k+n$ and $b=11k-n$. So $k$ and $n$ must be relatively prime in order for $a$ and $b$ to be relatively prime. This leads us to consider two scenarios.

If $k$ is divisible by 3, it can be shown that it must also be divisible by 9. Then $c$ must also be divisible by 3. If $c=3p$ and $k=9q$, then $p^3=11^2\cdot27\cdot q^3+qn^2=q[3\cdot(33)^2q^2+n^2]$. Since $q$ and $n$ are relatively prime then q and the term in square brackets must also be relatively prime. If the product of two relatively prime numbers is a perfect cube, then the factors themselves must be perfect cubes.

This suggests there is a $x$ so that $x^3=3\cdot(33)^2q^2+n^2$.

The form of the equation remains largely unchanged upon multiplication by 27.

$27x^3=81\cdot(33)^2q^2+27n^2\implies (3x)^3=(33)^2(9q)^2+3(3\cdot n)^2$ Again we have an expression of the form a perfect cube is the sum of three times a perfect square plus some other perfect square. A similar situation occurs if $k$ is not taken to be divisible by 3.

By arguments above we also have that a perfect cube has that form if we don't take that 3 divides $k$.

So possible solutions have the form $x^3 = 3\cdot(33)^2r^6+n^2$ where $q=r^3$ or $x^3=11^2\cdot r^6+3n^2$ where $k=r^3$.

We have an example of the first where $r=6$ and $n=4085$. This leads to $a=17299, b=25579, c=9954$, a solution mentioned elsewhere.

TurlocTheRed
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