<s>[INST] $q^{3}\leq aq^{2}+bq+c$ finite amount of q problem Im a bit confused why this is true it's simple but i seem to have some problems with it $q^{3}\leq aq^{2}+bq+c$ is only true for a finite amount of $\forall q \epsilon \mathbb{N}$
 [/INST] First, you can assume that $a, b, c \ge 0$, for if they're negative, then $q^3 \le aq^2 + bq + c \le |a|q^2 + |b|q | c$, and if there are at most finitely many solutions the the latter, then there are to the former as well. For the same reason, we can assume $c \ge 1$, so that $a+b+c \ge 1$.
Claim: For $q \ge a+b+c$, the left hand side is larger than the right hand side. That shows that there are at most $a+b+c$ solutions. 
Why is the claim true? Because
\begin{align}
q^3 &\ge  (a+b+c)^3 \\
& \ge a(a+b+c)^2 + b(a + b + c)^2 + c(a+b+c)^2 \\
& \ge a(a+b+c)^2 + b(a + b + c)^1 + c(a+b+c)^0 \text{, because $a+b+c > 1$} \\
& \ge aq^2 + bq + c \text{, by substitution, $q=a+b+c$} \\
\end{align}
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