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lfdahl
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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.
Prove, that $T$ is finite.
lfdahl said: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.
T in this statement refers to a set of numbers that satisfy the given conditions, namely a and b belong to this set and the expression a^2−ab+b^2 divides a^2b^2.
To prove that T is finite, we need to show that there are only a finite number of values for a and b that satisfy the given conditions. This can be done by using mathematical proofs such as contradiction or induction.
Yes, for example, let T be the set of all positive integers. If we take a=2 and b=3, then a^2−ab+b^2=4−6+9=-2 does not divide a^2b^2=4*9=36. Therefore, (2,3) does not belong to T. This shows that T is a finite set.
The fact that a^2−ab+b^2 divides a^2b^2 means that for every pair of numbers a and b in T, the result of this expression will always be a factor of a^2b^2. This limits the possible values of a and b that can belong to T, making it a finite set.
No, the given statement specifically refers to a set T that satisfies the given conditions. The proof may vary for different sets, depending on the conditions given. It is not applicable to all sets in general.