How can one proves that the polynomial $$1-xy+xy^2+x^3y^3-x^3y^2+x^2y^3-x^2y^2+x^2y$$ that is equal to $$1+ xy(xy+y-1)(xy-x+1)$$ does not have rational zeros?
2 Answers
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The polynomial defines a genus 2 curve. Your curve is birational to the genus two curve $C$ given by $y^2=x^5+7x^4+18x^3+24x^2+16x+4$. This curves has two rational Weierstrass points. These points correspond with the two points at infinity on your curve.
There are standard techniques to bound the number of rational points on such curves. E.g., you can consider first the Jacobian of this curve and try to find a basis for the Mordell-Weil group. If this group is small enough you may be able to find all rational points on $C$. On the web page of Michael Stoll you can find many papers and slides on this subject.
answered Jul 3, 2013 at 12:35
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4$\begingroup$ it has two rational Weierstrass points. Also, the substitution x <- x -1 transforms it to the simpler looking curve $y^2 = x^5 + 2x^4 + 2x^2 - x = x(x^2 + 1)(x^2 + 2x - 1)$. $\endgroup$ Commented Jul 3, 2013 at 13:23
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1$\begingroup$ Also two pairs of rational points at $x = \pm 1$; and no other $x = m/n$ with $|m|,|n| \leq 10^6$ according to Stoll's ratpoints ("ratpoints '0 -1 0 2 2 1' 1000000" takes less than a second). The curve has good reduction away from $2$, and an automorphism $(x,y) \mapsto (-1/x, x/y^3)$ whose square is the hyperelliptic involution $(x,y) \leftrightarrow (x,-y)$. But neither of these observations seems to make the problem much easier. $\endgroup$ Commented Jul 3, 2013 at 18:11
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1$\begingroup$ Stoll has code on his webpage for doing this kind of calculation: faculty.jacobs-university.de/mstoll/magma $\endgroup$ Commented Jul 3, 2013 at 22:17
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This is just a simple reduction to integer:
So we seek to solve $1+xy(xy+y-1)(xy-x+1)=0.$ Assume there is such a solution. We can then put the numbers on least common denominator $d$, $x=a/d,y=b/d.$
Substituting and clearing denominator gives $d^4+ab(ab+bd-d^2)(ab-ad+d^2)=0$ which we wish to find integer solutions to.
Here, one can try to look for solutions modulo $d$ or $d^2$ and use that it is the least common denominator, but I have not put much thought into it.
For example, we get that $ab|d^4$ (considering modulo ab), and this feels sort of strange...
answered Jul 3, 2013 at 12:24