Prove by induction $n^2>7n+1$ for $n\geq8$

Question

I am trying to prove the following by induction.

$n^2>7n+1$ for all $n\geq8$

I have proved that the function works in the first possible case $$8^2>7(8)+1$$ $$64>57$$

Then subbed $k+1$ and set the inequality to $0$ giving:

$$(k+1)^2-7k-7>0$$ $$k^2+1-7k-7>0$$

Where do I need to go from here to finish proving this inequality?

Answer 1

As written, this is rather troubling: You're assuming the conclusion when you write any variant on $(k + 1)^2 - 7k - 7$. You should start with the assumption that $k^2 - 7k - 1 > 0$ and try to get something about $k + 1$. In particular, you can write something along the lines of \begin{align*} (k + 1)^2 - 7(k + 1) - 1 &= (k^2 - 7k - 1) + 2k + 1 - 7 \\ &> 2k - 6 \\ &> 0 \end{align*} because $2k - 6 \ge 16 - 6 =10$ for all $k \ge 8$. Notice that the inequality $... > 0$ is the conclusion, not the assumption.

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As a second issue, you have $7(k + 1)^2 - 7k - 7$ when it should be $-1$ for the last term. But this is a secondary thing.

Answer 2

If $k^2>7k+1$, then $(k+1)^2=k^2+2k+1>7k+2k+1>7k+16+1$, since $k\geqslant8$. But $7k+16+1>7k+8=7(k+1)+1$.

Answer 3

As an alternative

$$n^2>7n+1 \iff n^2-7n-1>0$$

and for $x^2-7x-1=0$ we obtain the roots

$$x_{1,2}=\frac{7\pm \sqrt{49+4}}{2} \implies x_2=\frac{7+ \sqrt{53}}{2}\approx 7.14$$