$q^{3}\leq aq^{2}+bq+c$ 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 $\forall q \epsilon \mathbb{N}$
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 &amp;\ge  (a+b+c)^3 \\
&amp; \ge a(a+b+c)^2 + b(a + b + c)^2 + c(a+b+c)^2 \\
&amp; \ge a(a+b+c)^2 + b(a + b + c)^1 + c(a+b+c)^0 \text{, because $a+b+c &gt; 1$} \\
&amp; \ge aq^2 + bq + c \text{, by substitution, $q=a+b+c$} \\
\end{align}
At first,  we assume $a&gt;0$.

we have

$\displaystyle{\lim_{q\to +\infty} \frac{ q^3   }{  aq^2+bq+c }  =+\infty }$

so for enough large $q$ ($q\geq$ than a certain $q_0$),

$$\frac{ q^3}{ aq^2+bq+c  }&gt;1$$

in other terms

$\forall q\geq q_0   \frac{ q^3 }{ aq^2+bq+c }&gt;1$.

this answers your question.

if $a&lt;0$, the denominator goes to $-\infty$ and we use the same reasonning.

if $a=0$, you apply this to $b$ and to $c$ if $b=0$.