If I have an equation say $$3(1+x+x^2)(1+y+y^2)(1+z+z^2)+1=4x^2y^2z^2 \quad (1)$$ and I know a non negative integer solution $x=4, y=64, z=262144$ then no odd positive integer solutions can possibly exist. I know that one way to show it for this example would be to compute but I am looking for a proof that does not rely on computation like using wolfram as I want to generalize this. I will happily award the bounty to anyone who can prove it. What I have tried is minimal. I suppose that $p,q,r \in \mathbb N$ and they satisfy equation $(1)$. I think if we let $p$ be the smallest integer of the solution $p<4$ is impossible so assume that $p\ge 5$ but this might lead to a contradiction since the coefficient of $4p^2q^2r^2 \quad $ is $4<5$ and that might be impossible?
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You have asked a few questions on similar-looking equations, and always about odd positive integer solutions. Is there a common motivation for these problems? – Aravind May 18 '20 at 17:32
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Aravind I noticed this interesting pattern and am just totally stuck why only even positive integer solutions exist and cannot for the life of me figure out how to show it. Mostly just driven to know why – argamon May 18 '20 at 17:41
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@argamon do these equations only have one solution? if so, what's the big deal if it's even and not odd... – mathworker21 May 18 '20 at 19:19
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there are other solutions but the only one which is all odd that I know of is $-1,-1,-1$ but those are negative – argamon May 18 '20 at 19:21
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@argamon u shud write "[at sign]mathworker21" so i get a notification that u responded. also, does "odd solution" mean all of $x,y,z$ are odd? – mathworker21 May 19 '20 at 14:18
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@Mathworker yes they are all odd – argamon May 19 '20 at 16:35
1 Answers
The given equation is $3(1+x+x^2)(1+y+y^2)(1+z+z^2)=4x^2y^2z^2-1$.
Let $x \leq y \leq z$. Then $x=1$ has no solution as in this case the LHS is at least $9y^2z^2>4y^2z^2$. Thus, let $x \geq 3$. Also note that $3$ divides the LHS so $3$ does not divide $xyz$; hence $x \geq 5$.
Divide the original equation by $x^2y^2z^2$ on both sides and upper-bound each sum by an infinite geometric series to get:
$3\dfrac{xyz}{(x-1)(y-1)(z-1)}>4-\dfrac{1}{x^2y^2z^2}>4-\dfrac{1}{xyz(x-1)(y-1)(z-1)}$.
Thus, $3xyz\geq 4(x-1)(y-1)(z-1)$, so that $$\left(1-\dfrac{1}{x}\right)\left(1-\dfrac{1}{y}\right)\left(1-\dfrac{1}{z}\right)\leq\dfrac{3}{4}.$$
From the above, the minimum of $x,y,z$ must be at most $9$, otherwise the LHS is at least $(10/11)^3>3/4$. This means that $x \in \{5,7\}$.
If $x=5$, then $y \leq 29$ and if $x=7$, then $y \leq 15$, both upper bounds obtained from the previous inequality.
When $x=5$, considering the cases modulo 5, we must have $y \equiv 0/2/4$ (mod 5). Thus, we are finally left with the following pairs of $(x,y)$ to check and eliminate:
$(5,5),(5,7),(5,17),(5,19),(5,25),(5,29),(7,7),(7,13),(7,15)$.
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