Given a,b∈T, a^2−ab+b^2 divides a^2b^2, Prove that T is finite

[lfdahl]

Let $T$ be a set of natural numbers such that for any $a, b \in T$, $a^2 − ab + b^2$ divides $a^2b^2$.

Prove, that $T$ is finite.

 

[kaliprasad]

Let $T$ be a set of natural numbers such that for any $a, b \in T$, $a^2 − ab + b^2$ divides $a^2b^2$.

Prove, that $T$ is finite.

For the above to be valid a and b should be co-prime or one of them 1 else if (a,b) is a solution then (na,nb) is also a sloution for integer N

 

[lfdahl]

Spoiler

Hi, kaliprasad!
In order to answer your comment, I´ll suppose, that $a,b \in T$ and $a$ and $b$ are not coprimes.

Let $d = gcd(a,b)$. Then we have:

$a = da_1$ and $b = db_1$, where $a_1$ and $b_1$ are coprimes.

Then: $a_1^2-a_1b_1+b_1^2$ divides $d^2a_1^2b_1^2$, but $gcd(a_1^2-a_1b_1+b_1^2,a_1b_1) = 1$. Hence, $a_1^2-a_1b_1+b_1^2$ divides $d^2$, i.e. $a^2-ab+b^2$ divides $d^4$.

Since $d \leq a$, we have $a^2-ab+b^2 \leq a^4$. If you fix any $a \in T$, $b$ can only take on a finite number of distinct values.