Prove that $n^2>n+1$ for all $n\geq 2\in\mathbb{Z}$

Question

I'm a bit unsure of my reasoning here:

Clearly the proposition $P_2$ which states that $2^2>2+1$ is true.

Suppose that it holds true for $k^2>k+1$ where $k>2$.

We define $d_k=k^2-(k+1)=k^2-k-1$

Suppose we also have $(k+1)^2>(k+1)+1$.

As before we define $d_{k+1}=(k+1)^2-((k+1)+1)$

Now we have

$\begin{align} d_{k+1}=k^2+k-1&>k^2-k-1=d_k\\ 2k&>0 \end{align}$

which must be true since $k>2$.

My intuition was that by showing that the difference between the terms is increasing with each iteration, it would prove that the inequality must always hold. However I feel a bit uneasy that I assumed that both $P_k$ and $P_{k+1}$ were true. Is this line of reasoning valid?

Any advice helps, thanks!

Answer 1

$P_k$ is the statement that $d_k >0$. You showed that $d_{k+1} >d_k$ so if you assume $P_k$ you get $d_{k+1} >d_k>0$ which shows that $P_{k+1}$ is true.