Prove by induction that $n^2>7n+1$ for all $n \ge 8$

Question

So in my math book I came across this proof by induction.

It states:

Example Prove that $n^2>7n+1$ $\forall n\ge8$

Solution The result is true when n=8 because $8^2=64$ and $7*8+1=57$.

Suppose it is true when $n$ is any number $k\ge8$ that is $k^2>7k+1$ Then.

$(k+1)^2 = k^2 +2k +1 > (7k+1)+2k+1=7(k+1)+1+(2k-6)$

Since $k\ge8$, $2k-6$ is a natural number and the last expression above is greater than $7(k+1)+1$. The induction step is verified, and so the result is true $\forall n\ge8$.

My question Can you just add natural numbers to the equation? Like in this case $(2k-6)$ just to make the statement true? Since

$7(k+1)+1+(2k-6)>7(k+1)+1$

I understand that we want use our induction hypothesis $k^2>7k+1$ in our induction step like this $k^2+2k+1>7k+1+2k+1$ and then prove that $7k+1+2k+1>7(k+1)+1$

Which is what the $(2k-6)$ gives us.

Answer 1

we did not add natural numbers to the equation. We can of course drop a positive quantity in $a=b+y > b$ if $y > 0$.

\begin{align} (7k+1)+2k+1 &=(7k+7-6)+2k+1 \\ &=(7k+7)-6+2k+1 \\ &=7(k+1) + (2k-6)+1 \\ &>7(k+1)+1, \text{since 2k-6 > 0} \end{align}

Answer 2

without induction you can write $$n^2-2\cdot \frac{7}{2}+\frac{49}{4}-\frac{53}{4}>0$$ and this is $$\left(n-\frac{7}{2}\right)^2>\frac{53}{4}$$ so we have $$n>\frac{7}{2}+\frac{\sqrt{53}}{2}$$

Answer 3

I think it is cleaner to argue without induction. If $n$ is a positive integer and $n\ge 8$, then $7n$ is both less than $n^2$ and a multiple of $n$, so at most $n^2-n$ and therefore $7n+1$ is at most $n^2-n+1<n^2$.

Answer 4

$n\geq 8\Rightarrow n^2\geq 8n = 7n +n \geq7n +8 >7n +1$

(why induction?)

Do I miss anything?