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The set of solutions of quadratic equation $a^2+b^2=c^2$ on $\mathbb{Z}$ can be described by Pythagorean triples up to multiplication. Can I use similar results on the ring of integer coefficient polynomials $\mathbb{Z}[x]$? More concretely,

(1) Is there a complete description of the set of solutions of $a^2+b^2=c^2$, $a,b,c\in\mathbb{Z}[x]$?

(2) In general, is there a theory on the class of equations $a^2+f(x)b^2=c^2$, where $f(x)\in\mathbb{Z}[x]$ is a given polynomial?

D. Lee
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  • It's a triviality. Pythagorean triples. In General, the formula looks like this. http://math.stackexchange.com/questions/738446/solutions-to-ax2-by2-cz2/738527#738527 For the case of. $a^2+tb^2=c^2$ The solution is simple $a=p^2-ts^2$ ; $b=2ps$ ; $c=p^2+ts^2$ – individ May 29 '15 at 07:11
  • Hmm, what if we consider Pell's equations? For example, if a=8, b=3, c=1, t=-7, then b cannot be even. – D. Lee May 29 '15 at 08:45
  • Everything can be. $p=3$ ; $s=1$ And to reduce common divisor 2. – individ May 29 '15 at 08:53
  • Ah, thanks! It's interesting. – D. Lee May 29 '15 at 09:19

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Pythagorean triples in a polynomial ring $R[x]$ can be classified as follows (see Theorem $5.1$ in K. Conrad's notes here):

Theorem $5.1$: All primitve Pythagorean triples $(f(t),g(t),h(t))$ with $f^2+g^2=h^2$ are given by the formulas \begin{align*} f(t) & = c(k(t)^2-\ell(t)^2),\\ g(t) & = \pm 2c\cdot k(t)\ell(t),\\ h(t) & = \pm c(k(t)^2+\ell(t)^2), \end{align*} where $c\in R^{\times}$ and $k(t),\ell(t)$ are two relative prime polynomials in $R[t]$.

Dietrich Burde
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