Prove by induction $n^2 + n \geq 42$ when $n \geq 6$ and $n \leq -7$ (induction with two different intervals)

Question

I am asked the following:

Prove by induction $n^2 + n \geq 42$ when $n \geq 6$ and $n \leq -7$

The first part of the exercise seems ok:

Base case P(6): $\quad LHS = 36+6=42\geq 42$

Inductive hypothesis $\quad P(k) : k^2 + k \geq 42$

Inductive step $$ (k+1)^2 + (k+1) = k^2 + 2k + 1 + k + 1 = k^2 + k + (2k+2) > k^2 + k \geq 42$$

Now, for the second part:

Base case P(7): $\quad LHS = 49-7=42\geq 42$

Inductive hypothesis $\quad P(k) : k^2 + k \geq 42$

Inductive step $$\quad (k+1)^2 + (k+1) = \cdots$$ (not sure how to go from here)

My intuition says I need to use the fact that I have a negative $n$ and use it to my favor, however, I am not sure how.

Any help is highly appreciated.

Answer 1

First part is fine, for the second part to avoid negative number we can use that by $m=-n>0$ such that

$$n^2 + n \geq 42 \;\text{for}\; n\le -7 \iff m^2-m\ge 42 \;\text{for}\; m\ge 7$$

and proceed in a similar way as for the first part.

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As suggested by Robert Israel in the comments, also note that for $m\ge 7$ by $p=m-1\ge 6$ we have

$$ m^2-m\ge 42 \;\text{for}\; m\ge 7 \iff p^2+p\ge 42 \;\text{for}\; p\ge 6$$

which was already proved by the first part.

Answer 2

proving $n^2+n\geq42$ for $n\leq-7$

Since $f(n)=n^2+n-42$ is a quadratic, by taking reflection of $f(n)=n^2+n-42$ about y axis that is $f(-n)=n^2-n-42$, prove $n\geq 7$ for $n^2-n\geq 42$

step $1$: for $n=7$

$7^2-7=42$ is true.

step $2$: assume that it is true for $n=k$,

$k^2-k\geq 42$

then for $n=k+1$

$(k+1)^2-(k+1)=k^2+k > k^2-k\geq 42$, for $n=7$.