Proving by induction that $n^2 - 7n - 2$ is divisible by $2$

Question

Now proving by induction is fairly simple. However, this is a multiple choice problem whose answers don't make any sense to me. The actual problem goes as follows:

To prove by induction that $n^2 - 7n - 2$ is divisible by $2$ is true for all positive integers $n$, we assume $k^2 - 7k - 2$ is divisible by $2$ is true for some positive integers $k$ and we show that $k^2 - 7k - 2 + A$ is divisible by $2$ where $A$ is:

Now I would've assumed that $A$ would be $(k+1)^2 - 7(k + 1) - 2$, but I checked the answer and it is actually $2(k-3)$ which makes no sense to me. I tried to factor, reduce, and anything I could think of but I can't get $(k+1)^2 - 7(k + 1) - 2$. Does anyone know where I am going wrong?

Answer 1

HINT: For the inductive step, we assume that $2$ divides $n^2-7n-2$, i.e., $n^2-7n-2 = 2M$. We now have \begin{align} (n+1)^2 - 7(n+1) - 2 & = n^2 + 2n + 1 - 7n - 7 +2 = n^2 - 7n -2 + 2(n-3)\\ & = 2(M+n-3) \end{align}

Answer 2

Use the fact that $$ n^2 - 7n - 2 - ( (n-1)^2 - 7(n-1) - 2)) = 2n - 1 - 7 = 2(n-4).$$

Answer 3

Hint $ $ If $\,f(0)\,$ is even and $\,f(n\!+\!1)-f(n)=2j\,$ is even then by induction $\,f(n)\,$ is even for all $\,n,\,$ the inductive step being: $\ \,f(n+1) = f(n) +2j = $ even + even = even.

Remark $\ $ mod $\,2\,$ it is simply: $ $ if $f\,$ is constant $\,f(n\!+\!1)\equiv f(n)\,$ then $\,f(n)\equiv f(0)\,$ for all $\,n$.

Answer 4

In the problem statement "we show that $k^2 - 7k - 2 + A$ is divisible by $2\dots$". I think that what is intended here is that $k^2 - 7k - 2 + A$ is the next case. In other words $$k^2 - 7k - 2 + A = (k+1)^2 -7(k+1)-2$$ Do the math, and I'm sure that you'll come up with $$A = 2(k-3)$$ To finish off the problem, show that $$k^2 - 7k - 2 + \color{blue}{2(k-3)} = (k+1)^2 -7(k+1)-2$$