How to get to $5^3 \geq n^3$ in the proof by contradiction?

Question

This is the same problem asked here. - Next step to take to reach the contradiction? Here is it again.

I understand the solution - how you want to get to the fact 100 divides n^2 and then go through all the possibilities to show that the integer k for which n^2 * k = 100 is not n + 1.

Here is my instructor's solution

The n^2 + n^3 > n^3 part of the proof makes sense to me. After all any positive integer squared will make an expression higher than itself. How did she get to 5^3 >= n^3 -> n < 5 though? I can't figure out the algebra step she took to get to that conclusion. I tried setting n^2 + n^3 equal to n^3 but that just got me zero. I am just to understand different ways of doing a proof

Answer 1

Let me write the proof she presents with the (so-called) intermediate steps:

Suppose the claim is true. Since $\forall n: n^2>0, 100=n^3+n^2>n^3$. So $5^3=125>100>n^3$, that is, $5^3>n^3$. Then $0<5^3-n^3=(5-n)(5^2+5n+n^2)$. The last factor is positive, so $(5-n)$ must be positive as well. Hence it suffices to check the first four positive integers.

Answer 2

The point of this proof is that it is impossible to have $n^2+n^3=100$ by splitting it up into two cases

CASE I: ($n \ge 5$) Since $5^2+5^3>5^3>100$ and our function increases as $n$ increases, no value of $n \ge 5$ can possibly work.

CASE II: ($n<5$) Since $4^2+4^3=80<100$ and our function decreases as $n$ decreases, then no value of $n<5$ can work either.

Thus no $n\in\mathbb{N}$ yields $n^2+n^3=100$.