In $ n < (n^2(s-2)-n(s-4))/2$ can I make it so that one side becomes completely know when I give a value to $s$

Question

If I have $$ n < \cfrac{n^2(s-2)-n(s-4)}{2}$$ can I make it so that one side becomes completely known when I give a value to $s$, for example(obv. not true) $$123n^2+456n+789<987s+654$$ So that I can get a maximum value for $n$

Also $n,s$ are positive integers

Answer 1

$$n < \frac{n^2(s-2)-n(s-4)}{2}\Rightarrow (s-2)n^2-(s-2)n>0\Rightarrow n(s-2)(n-1)>0$$ Because $n>0$, $$(s-2)(n-1)>0\Rightarrow s>2,n>1$$

So by fixing $s$ there is no maximum value for $n$.

Answer 2

Sure, if I am not mistaking, you can simplify to: $$n < \frac{1}{2}(n^2(s-2)-n(s-4)) \\ \Leftrightarrow 2n < n^2(s-2)-n(s-4) \\ \Leftrightarrow 2 < n(s-2)-(s-4) \\ \Leftrightarrow \frac{2}{(s-2)} < n-\frac{(s-4)}{(s-2)} \\ \Leftrightarrow \frac{2}{(s-2)} +\frac{(s-4)}{(s-2)} < n \\ \Leftrightarrow \frac{s-2}{s-2}<n$$

Answer 3

You can do it as follows: $$2n < n^2(s-2)-n(s-4)$$ or, $$ 0 < n^2(s-2)-n(s-2)$$ or, $$ 0 < n(n-1)(s-2)$$