# Integer solutions to multivariate polynomial

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.

• I would like to understand how to address the odd and positive requirement May 10, 2020 at 18:48

## 2 Answers

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)

• Could we prove x and y not odd and positive? May 12, 2020 at 16:19

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 $$a, then $$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 $$n \leq \dfrac{\sqrt{b}}{\sqrt{b}-\sqrt{a}}$$.

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