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I'm looking for a way to show that an equation like $$15(1+x+x^{2})(1+y+y^{2})=16(xy)^{2}-1$$ has no solutions with x and y odd positive integers. The 16 and 15 can be changed but I'm trying to generalize it. Any help is appreciated.

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  • $\begingroup$ I would like to understand how to address the odd and positive requirement $\endgroup$
    – argamon
    May 10, 2020 at 18:48

2 Answers 2

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Only comment.

$15(1+x+x^{2})(1+y+y^{2})=16(xy)^{2}-1\implies\\ \Bigl(2 y (15 + 15 x - x^2) + 15 (1 + x + x^2)\Bigr)^2 = -735 - 1410 x - 1061 x^2 - 390 x^3 + 285 x^4$

Equation $z^2=-735 - 1410 x - 1061 x^2 - 390 x^3 + 285 x^4$ in Magma Calculator with code IntegralQuarticPoints([285, -390, -1061, -1410, -735],[-1,17]); have only solutions

[x,z]=[[-1,+-17],[-4,+-293],[4,+-157],[16,+-4097],[4096,+-283184657]]

In source equation $x$ and $y$ is symmetric, and then they is not possible odd positive.


$a (1 + x + x^2) (1 + y + y^2) = b (x y)^2 - 1 \implies\\ \Bigl(2 y (a (1 + x + x^2) - b x^2) + a (1 + x + x^2)\Bigr)^2 =\\ -a (4 + 3 a) - 2 a (2 + 3 a) x + (4 b (1 + a) - a (4 + 9 a)) x^2 + 2 a (2 b - 3 a) x^3 + a (4 b - 3 a) x^4$


Examples odd positive x,y:

(a,b,x,y)=

(1,10,1,1)
(2,3,3,27)
(2,3,27,3)
(2,7,1,7)
(2,7,7,1)
(4,5,5,125)
(4,5,125,5)
(4,13,1,13)
(4,13,13,1)
(6,7,7,343)
(6,7,343,7)
(8,9,9,729)
(8,9,729,9)
(10,11,11,1331)
(10,11,1331,11)
(12,13,13,2197)
(12,13,2197,13)
(14,15,15,3375)
(14,15,3375,15)
(23,48,3,3)
(53,82,3,15)
(53,82,15,3)
(79,100,5,49)
(79,100,49,5)
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  • $\begingroup$ Could we prove x and y not odd and positive? $\endgroup$
    – argamon
    May 12, 2020 at 16:19
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Factorization should be useful, but another idea is to note that if the equation is $a(x^2+x+1)(y^2+y+1)=b(xy)^2-1$ and $a<b$, then $min(x,y)$ is upper-bounded by a function of $a,b$ and it is sufficient to examine those values (and solve for the other).

Specifically, here is one (not necessarily sharp) bound: $min(x,y) \leq \dfrac{\sqrt{b}}{\sqrt{b}-\sqrt{a}}$.

To prove the bound, we write: $a(1+\dfrac{1}{x}+\dfrac{1}{x^2})(1+\dfrac{1}{y}+\dfrac{1}{y^2})=b-\dfrac{1}{x^2y^2}$; thus $a\dfrac{xy}{(x-1)(y-1)}>b-\dfrac{1}{xy(x-1)(y-1)}$, and hence $axy>b(x-1)(y-1)-1$, so that $\left(1-\dfrac{1}{x}\right)\left(1-\dfrac{1}{y}\right)\leq\dfrac{a}{b}$.

So if $x,y \geq n$, then $1-1/n \leq \dfrac{\sqrt{a}}{\sqrt{b}}$ and thus $n \leq \dfrac{\sqrt{b}}{\sqrt{b}-\sqrt{a}}$.

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  • $\begingroup$ could we generalize the bound to more variable say our equation became $a(1+x+x^2)(1+y+y^2)(1+z+z^2)=b(x^2y^2z^2)-1$ $\endgroup$
    – thestar
    Dec 7, 2021 at 19:10

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