<s>[INST] Can this ratio be an integer? Is it true that the number
$$\frac{b^3+1}{ab-1}$$
is never an integer for positive integers $a$ and $b$ such that $a\geq 6$ and $a\geq b$?
 [/INST] Here is my approach:
By definition you can say that $b^2+{2 \over b} \ge a \ge b$ because $(b^2+{2 \over b})b-1 = b^3+1$ . But because  ${2 \over b}$ is only an integer for $b=1$ or $b=2$ your interval for $a$ is effectively $[b, b^2]$ . This means you can write $a=b\cdot b^x$ with $0 \le x \le 1$ or $a=b \sqrt[y]{b}$ with $y = {1 \over x}$ . If you combine this with  the original term you get ${b^3+1 \over b^2 \sqrt[y]{b}-1}=n, \quad n \in N \quad \Rightarrow b^3+(1+n) = n b^2 \sqrt[y]{b}$ . This concludes that if you can show that for some $b $ if  $n\sqrt[y]{b} \in N$ , $b^2\sqrt[y]{b} \in N$ and $nb^2\sqrt[y]{b} \in N$ lead to a contradiction  then for this $b$ ${b^3+1 \over ab-1}$  is not an integer for every $a$.
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