I am interested in the diophantine equation in $\mathbb Z[t]$:
$$6Z^2 + 5((t + 1)X + tY − 1)Z +((t + 1)X + tY − 1)^2+ XY = 0$$ (the unknown variables are $X,Y,Z$)
Can one determine ALL the solution in $\mathbb Z[t]$?
Thanks in advance
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Why? And why do you want all solutions? – Will Jagy Sep 03 '15 at 18:08
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This equation comes from the comptutation of the characteristic polynomial of a matrix with coefficients in $\mathbb Z[t]$. I am looking for all its eigenvalues in $\mathbb Z[t]$. I obtained this diophantine equation. – joaopa Sep 03 '15 at 18:29
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do you know any solution whatsoever? – Will Jagy Sep 03 '15 at 18:53
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The reason for asking about one solution is an approach specific to this dimension. It is pages 507-508 in Fricke and Klein book (1897) and written in an offhand way there. It is stated as a theorem in William Plesken, Automorphs of Ternary Quadratic Forms, pages 5-30 in Ternary Quadratic Forms and Norms, edited by Olga Taussky, 1982. It is theorem I.9 on page 15. It is at least possible that the same result holds over $\mathbb Q(t),$ which can then be adjusted to $\mathbb Q[t]$ and $\mathbb Z[t]$ – Will Jagy Sep 03 '15 at 19:36
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If there was not $1$ it would be possible to use the formula. http://math.stackexchange.com/questions/794510/curves-triangular-numbers Solutions of this equation are determined by the solutions of the Pell equations. This formula exactly right? Because such a cumbersome formula nobody likes. – individ Sep 04 '15 at 04:32
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Yes, the equation is absolutely correct. – joaopa Sep 04 '15 at 04:43
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The solution to the equation.
$$6Z^2+5((t+1)X+tY-1)Z+((t+1)X+tY-1)^2+XY=0$$
Can be written as an ordinary quadratic equation.
$$Z=\frac{-5((t+1)X+tY-1)\pm{D}}{12}$$
Where.
$$D^2=25((t+1)X+tY-1)^2-24((t+1)X+tY-1)^2-24XY$$
$$((t+1)X+tY-1+D)((t+1)X+tY-1-D)=24XY$$
The solution of this equation there. https://mathoverflow.net/questions/215197/quadratic-diophantine-equation-in-mathbb-zt
