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Consider the three equations of whom two only have trivial solutions in $\mathbb{Z}$. Determine a non-trivial solution of the third.

$3x^2+5y^2=7z^2$

$5x^2+7y^2=3z^2$

$3x^2+7y^2=5z^2$

My Approach:

I would use the Hilbert-Symbol, but the Hilbert-Symbol only says something about solutions in $\mathbb{Q}_p$.

This exercise is in the Chapter about Hilbert-Symbols, so I believe there should be some trick to it. Also although I know the Hilbert Symbol I do not know how to find solutions.

Bill Dubuque
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NTc5
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  • The solution is trivial. Just take $x=y=1$, so that the middle equation has a non-trivial integer solution with $z=2$. – Dietrich Burde Jun 27 '24 at 14:41
  • @DietrichBurde Thank you. But how did you find out the second equation has solutions? Calculating Brute Force would be a lot of work (and not the purpose of this exercise). – NTc5 Jun 27 '24 at 14:42
  • Brute force is not the right word for putting $x=y=1$. The main exercise is to show that there are no integer solutions to the other two equations - by using Lagrange's theorem with quadratic residues. Actually, Lagrange applies to $ax^2+by^2=cz^2$ and is an if and only if. – Dietrich Burde Jun 27 '24 at 14:46
  • @DietrichBurde But how does one know with which equation to start when trying solutions? In that case the solution may be easy to find, but in general it would be difficult. – NTc5 Jun 27 '24 at 14:49
  • As I said, compute the Legendre symbols $(bc/a)$, $(ac/b)$ and $(-ab/c)$. Then you are done by Lagrange. A solution in integers exists iff all of them are equal to $1$. See here. – Dietrich Burde Jun 27 '24 at 14:50
  • You could calculate the Hilbert Symbol of those $3$ equations and if you are lucky you could rule out the two equations that have no solutions in $\mathbb{Z}$. I.e. If only one of the equations has a Hilbert-Symbol equal to $1$, then you know where to search. – Peter Jun 27 '24 at 14:51
  • You could use the Hilbert-Symbol modulo $3,5,7$. – Peter Jun 27 '24 at 15:19
  • What book are you using? – Will Jagy Jun 27 '24 at 16:35

1 Answers1

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HINT.-Because of by q.r.l. $(\dfrac{-1}{p})=(-1)^{\frac{p-1}{2}}$ which is negative for $p=3,7$, we have that $-1$ is not a square modulo $p$ for $p=3,7$. It follows $3x^2+5y^2=7z^2$ and $3x^2+7y^2=5z^2$ imply respectively $$3x^2+5y^2\Rightarrow x^2\equiv-(5y)^2\equiv0\pmod7\\7y^2\equiv5z^2\Rightarrow y^2\equiv-z^2\pmod3$$ which is impossible excepting for values $0$. This leaves as only soluble equation $5x^2+7y^2=3z^2$ which since $20+7$ is a cube that $(2,1,3)$ is a non trivial solution.

Ataulfo
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