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I’m not sure how to approach this number theory problem I’ve been working on for a while. So basically I need to show that the Diophantine equation

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

has an infinite number of integral solutions. The hint in my textbook is to consider $x = n^3 – 3n$ and $y = 3n^2 – 1$ where $n \in \mathbb{Z}$, but I’m not sure how that helps me show an infinite number of solutions exist.

Michael Albanese
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Math_Illiterate
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  • I think there should be some requirement for $x,y,z$, otherwise $(x,y,z)=(0,n^3,n^2)$ always works. – Easy Apr 18 '13 at 14:43
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    Needing to know how something will help before you try it is a crippling disability in mathematics. –  Apr 18 '13 at 15:21
  • I am surprised this got flagged as duplicate, since an aspect of the question is to inquire about a specific technique that does not appear in the other question. –  Apr 18 '13 at 15:28
  • The technique is different from the other question and is well demonstrated in the answers. – Ross Millikan Apr 18 '13 at 16:13
  • I'm still surprised none has flagged it has duplicate of this Pythagorean Triplets. This can be done in many ways. Follow the link. – Inceptio Apr 18 '13 at 15:12

5 Answers5

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\begin{align}x^2+y^2&=(n^3-3n)^2+(3n^2-1)^2=n^6+9n^2-6n^4+9n^4+1-6n^2\\&=n^6+3n^4+3n^2+1=(n^2+1)^3=z^3\end{align}

so $\forall n\in\mathbb{Z},\quad (x,y,z)=(n^3-3n,3n^2-1,n^2+1)$ is a solution to the equation.

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$(n^3-3n)^2+(3n^2-1)^2=(n^2+1)^3$

Easy
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Using the hint we have

$$x^2 = (n^3 - 3n)^2 = n^6 - 6n^4 + 9n^2,$$

and

$$y^2 = (3n^2 - 1)^2 = 9n^4 - 6n^2 + 1,$$

so

$$x^2 + y^2 = n^6 + 3n^4 + 3n^2 + 1 = (n^2 + 1)^3.$$

So for $n \in \mathbb{Z}$, $x = n^3 - 3n$, $y = 3n^2 - 1$, and $z = n^2 + 1$ is a solution to the Diophantine equation $x^2 + y^2 = z^3$.

You should then check that this gives you infinitely many solutions - it may be the case that even though there are infinitely many possibilities for $n$, there might only be finitely many resulting triples $(x, y, z)$.

Michael Albanese
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Given any solution $(x, y, z)$, observe that $( A^3 x, A^3 y, A^2 z)$ (where $A$ is any integer) will also be another solution.

Thus, to generate infinitely many solutions, we need to start with a non-zero solution.

We know that $2^2 + 11^2 = 5^3$ (which is $n=2$ in your textbook).

Calvin Lin
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  • An additional problem that should not be omitted is whether the identity used provides all the solutions, as does the identity giving Pythagorean triples. Can you add this proof to your answer? Regards. – Ataulfo Mar 30 '24 at 16:01
  • @Piquito The problem doesn't ask about complete classification, and just about an infinite family of solutions. I think it's obvious that this approach will most likely not generate all solutions, esp with the given hint. I don't see why I should add that clarification to my solution, esp since none of the rest bring it up. – Calvin Lin Apr 02 '24 at 00:02
  • .- I know what you say. I asked you simply because it had appeared interesting my question to me. Thank you very much for your answer. – Ataulfo Apr 03 '24 at 02:50
  • @Piquito Using Gaussian integers, I believe that (but haven't checked the details) the solution to $x^2 + y^2 = z^3$ is given by $ x = m^3 - 3mn^2, y = 3m^2n - n^3, z = m^2+n^2$. $\quad$ However, that might not be the complete classification, because it relies on $ \gcd(x+iy, x-iy) = 1$, which need not be the case like if $x$ is a multiple of $y$. So maybe it's a classification of the "primitive" solutions. If you need more clarification, please post it as a new question. – Calvin Lin Apr 03 '24 at 16:02
  • I proposed the problem to you because I think it is mathematically relevant and because I appreciate your intellectual solvency for these things. Greetings. – Ataulfo Apr 04 '24 at 18:50
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We can generalize the given problem as fallows :

Let $ax^2 + by^2 = z^3$ such that $a,b,x,y,z$ are all Integers.

Two variables parameterization solution for $x,y,z$ in terms of $p,q$

$x = ap^3 - 3bpq^2$

$y = 3ap^2q - bq^3$

$z = ap^2 + bq^2$

Take $a =1$ & $b = 1$ in above equation then we get

Let $x^2+y^2 = z^3$ such that $x,y,z$ are all Integers.

Where

$x = p^3 - 3pq^2$

$y = 3p^2q - q^3$

$z = p^2 + q^2$

Guruprasad
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