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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.

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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
    
Needing to know how something will help before you try it is a crippling disability in mathematics. –  Hurkyl 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. –  Hurkyl 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
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5 Answers

up vote 1 down vote accepted

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 $(x, y, z)$.

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

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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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I'm still surprised none has flagged it has duplicate of this Pythagorean Triplets. This can be done in many ways. Follow the link.

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An obvious explanation is that the question is not about Pythagorean triples, for which all of the exponents are "2". And I was surprised to find the question you linked isn't about Pythagorean triples either! –  Hurkyl Apr 18 '13 at 15:19
    
@Hurkyl: It was related to Pythagorean triplets, and the title of the question itself contained it.:) –  user63477 Apr 18 '13 at 15:21
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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).

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