The discussion centers on proving that the expression \(\frac{a^2+b^2+6}{ab}\) is a perfect cube given that \(a\) and \(b\) are positive integers. Participants confirm the validity of the proof, emphasizing the necessity of integer properties in the analysis. The conclusion drawn is that under the specified conditions, the expression indeed results in a perfect cube.
PREREQUISITESMathematicians, students studying number theory, and anyone interested in algebraic proofs and properties of integers.
anemone said:The numbers $a,\,b$ and $\dfrac{a^2+b^2+6}{ab}$ are positive integers. Prove that $\dfrac{a^2+b^2+6}{ab}$ is a perfect cube.
Albert said:let k=$\dfrac{a^2+b^2+6}{ab}---(1)$
then: $2+\dfrac{6}{ab}\leq k\leq \dfrac {a^2+b^2+6ab}{ab}=\dfrac {b}{a}+\dfrac {a}{b}+6$
for each $a,b\in N , max (2+\dfrac {6}{ab})=8,(a=b=1)$
and $min(\dfrac {b}{a}+\dfrac{a}{b}+6)=8 (for \,\, each \,\, a=b>0)$
if $a,b,k \in N, \,\,then\,\, k=8$
$\therefore k=8 $ is a perfect cube