diophantine question parity and sign

Question

If I have an equation say $$3(1+x+x^2)(1+y+y^2)=4x^2y^2-1 \quad (1)$$ and I rewrite it as $$4+3(x+y+xy+x^2+y^2+xy^2+x^2y)-x^2y^2=0$$ and I find a positive integer solution where both x and y are even, can I show that no odd positive integer solutions exist?

Answer 1

The equation is equivalent to $$3\left(\frac 1{x^2}+\frac 1x+1\right)\left(\frac 1{y^2}+\frac 1y+1\right)=4-\frac 1{x^2y^2},\quad (1)$$ since clearly $x\neq 0,y\neq 0$.

It appears that the only integers solutions are $$\{(-2,2),(-1,-1),(-1,4),(2,-2),(4,-1),(4,64),(64,4)\},$$ in particular your assertion is correct.

Let me just mention briefly how you can prove it. First of all, the equation is symmetric in $x$ and $y$, $x^2+x+1$ is positive definite, and so is $y^2+y+1$.

For the case in you question, you may reduce to a finite computation by observing the situation for $x=1,2,3,$ and $x\geq 4$. More precisely,

$$x=1\Rightarrow 9\left(\frac 1{y^2}+\frac 1 y+1\right)=4-\frac 1{y^2},$$ hence the equation cannot hold if $y$ is sufficiently large. This reduces to a finite computation. The other cases are similarly. And you can handle the negative solutions in a similar way.

Note. For the case $x\geq 4$ above (still assuming $y>0$), one has in (1), $${\rm LHS}\leq \frac {63}{16}\left(\frac 1{y^2}+\frac 1y+1\right)=:f(y)$$ and $${\rm RHS}\geq 4-\frac 1{16y^2}=:g(y).$$ Since $$\lim_{y\rightarrow \infty}f(y)=\frac{63}{16}<4=\lim_{y\rightarrow \infty}g(y),$$ one has ${\rm LHS}<{\rm RHS}$ if $y$ is sufficiently large.

(In fact, if $x\geq 4,y\geq 65$, then $${\rm LHS}\leq \frac {63}{16}\left(\frac 1{65^2}+\frac 1{65}+1\right)=\frac{270333}{67600}$$ and $${\rm RHS}\geq 4-\frac 1{16(65)^2}=\frac {270399}{67600},$$ so ${\rm LHS}<{\rm RHS}.$)

Answer 2

Since the equation is symmetric in $x$ and $y$, we only need to consider the case $x\le y$.

Dividing the both sides of $$3(1+x+x^2)(1+y+y^2)=4x^2y^2-1\tag1$$ by $x^2y^2$ gives $$3\bigg(1+\frac 1x+\frac{1}{x^2}\bigg)\bigg(1+\frac 1y+\frac{1}{y^2}\bigg)=4-\frac{1}{x^2y^2}\tag2$$

Suppose here that $x\ge 8$.

Then, we have $$ \text{(LHS of $(2)$)}\le 3\bigg(1+\frac 18+\frac{1}{8^2}\bigg)\bigg(1+\frac 18+\frac{1}{8^2}\bigg)=\frac{15987}{4096}$$ and $$\text{(RHS of $(2)$)}\ge 4-\frac{1}{8^4}=\frac{16383}{4096}$$ This is a contradiction. So, we have $x\le 7$.

Now, $(1)$ can be written as $$(3+3x-x^2)y^2+3(1+x+x^2)y+4+3x+3x^2=0$$ Seeing this as a quadratic equation on $y$ and considering the discriminant, we get $$D=9(1+x+x^2)^2-4(3+3x-x^2)(4+3x+3x^2)$$ It is necessary that $D$ is a square number.

For $x=1$, we get $D=-119$ which is not a square number.

For $x=3$, we get $D=1041$ which is not a square number.

For $x=5$, we get $D=11281$ which is not a square number.

For $x=7$, we get $D=46441$ which is not a square number.

Therefore, $(1)$ has no solution $(x,y)$ such that both $x$ and $y$ are positive odd integers.

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Added :

For $x=2$, $D$ is a square number, but $y=-2$ is not positive.

For $x=4$, $D$ is a square number, and $y=-1,64$.

For $x=6$, $D$ is not a square number.

Therefore, $$\color{red}{(x,y)=(4,64),(64,4)}$$ are the only positive integer solutions.