1

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
  • 99,526
Math_Illiterate
  • 975
  • 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
  • 2
    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

4 Answers4

2

\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.

1

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

Easy
  • 4,485
1

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
  • 99,526
1

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
  • 68,864
  • 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. – Piquito Mar 30 '24 at 16:01