Use induction to prove that $n^ 2 > 5n − 7$ for all $n \in \mathbb{Z}, n ≥ 2$

Question

Use induction to prove that $n^ 2 > 5n − 7$ for all $n \in \mathbb{Z}, n ≥ 2$.

I can't seem to figure out how to prove this. This is what I did so far:

Base case:

Let $n=2$. Then $n^2=4$ and $5(2)-7=3$.

$n^2 - 5n + 7 = 4 - 10 + 7 = 1\in\Bbb{N}$

Hence $n^2 > 5n - 7$.

Inducton step:

Let $n\ge2$ and assume $n^2 > 5n - 7$.

We want to show $(n+1)^2 > 5(n+1) - 7$.

$(n+1)^2 - 5(n+1) + 7 = n^2 + 2n + 1 - 5n - 5 +7$

$(n^2 - 5n + 7) + 2n - 4$

I don't know how to finish the proof.

Answer 1

Hint: We have to prove that

$$(n+1)^2>5(n+1)-7$$ adding $$2n+1$$ to $$n^2>5n-7$$ we get

$$(n+1)^2>7n-6$$ but $$7n-6>5(n+1)-7=5n-2$$ and this is true if $$2n\geq 4$$ this means $$n\geq 2$$

Answer 2

Since Dr. Sonnhard Graubner already gave a hint on how to do the job inductively, I am giving a combinatorial proof. Consider an $n$-by-$n$ checkerboard, where $n\geq 2$. Since $n\geq 2$, the checkerboard has $4$ corners, $4$ sides, and $2$ diagonals. We are counting the number $N$ of squares on the boundary of the checkerboard and on one of the diagonals.

We have $4$ corner squares. Each side contains $n-2$ non-corner squares. A diagonal also has $n-2$ non-corner squares. Thus, $$N=4+4(n-2)+(n-2)=5n-6>5n-7\,.$$ Clearly, $N\leq n^2$, so $$n^2\geq N>5n-7\,.$$

Answer 3

Hint $\,f(n) = n^2\!-\!5n\!+\!7 > 0$ for $n=2.\,$ Note $\,f(n+1)-f(n) = 2(n-2)\,$ so for $\color{#c00}{n\ge 3}\,$

it yields induction $\ \color{#0a0}{f(n)> 0} \,\Rightarrow\, f(n+1) = \color{#0a0}{f(n)} + 2(\color{#c00}{n-2}) > 0\, $ by both summands are $\ge 0$

Remark $\ $ This is a special case of telescopic induction. This view yields more: it immediately shows we can write $f(n)$ as the sum of its first differences $\,f(n)-f(n-1)\,$ yielding the representation

$$ n^2 -5n + 7 \ =\ 1 +\sum_{\large k\,=\,3}^{\large n-1} 2(n-2)\qquad$$

which - being a sum of positive terms - is clearly positive. That's the key idea behind the induction.