Let $a,b,c,d$ be integers such that $GCD(a,b,c,d)=1$. Assume that the diophantine equation $ax^2+bxy+cxz+dyz-x=0$ has a non-zero solution.Can we assert that it admits infinitely many solutions?
Thanks in advance
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$\begingroup$ It's all in the coefficient of $x$. When he is alone, then infinitely many solutions. The solutions can be expressed using equations Pell. The formula is rather long. Here not welcome a long formula and don't know if there is a point lead. $\endgroup$– individCommented Oct 25, 2015 at 4:41
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$\begingroup$ Individ. I am interested in a formula. Can you give one? $\endgroup$– joaopaCommented Oct 25, 2015 at 6:45
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$\begingroup$ Below is a formula which proves that infinitely many solutions. Formally, the task is already solved. $\endgroup$– individCommented Oct 25, 2015 at 6:49
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1 Answer
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Let $k$ be integer and $f(x,y,z)=ax^2+bxy+cxz+dyz-x$.
Unless $b=1,d=0$ then $f(x,y,z)=0$ has infinitely many solutions via the parametrization
$$X= -dk,Y=adk-ckb+ck+1,Z=k(b-1)$$ and $f(X,Y,Z)=0$.
If $b=1,d=0$ parametrization is $y=-ax-cz+1$ for integer $x,z$.
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$\begingroup$ Funny! This answer is a joke probably? $\endgroup$– individCommented Oct 25, 2015 at 5:38
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$\begingroup$ @individ Can't you reproduce these are parametrizations? $\endgroup$– joroCommented Oct 25, 2015 at 5:43
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$\begingroup$ This answer is useful. Is there infinite solution such that $x$ divdes $yz$? $\endgroup$– joaopaCommented Oct 25, 2015 at 6:46
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$\begingroup$ @joaopa I think there is at for the first case. Substituting the parametrization, we must have $d$ divides expression depending on $k$. Solve it modulo $d$ and if it has solution, there will be infinitely many. $\endgroup$– joroCommented Oct 25, 2015 at 6:51
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1$\begingroup$ For the equation. $$ax^2+bxy+cxz+dyz-x=0$$ We will set the number $p$ . And then select $s$ so that the fraction was an integer. And then Your formula can be written in General form. $$x=-dp$$ $$y=cp+s$$ $$z=p(b+\frac{(bc-ad)p-1}{s})$$ $\endgroup$– individCommented Oct 25, 2015 at 8:14