proof verification - Are we properly using mathematical induction?

I recently had a lecture in a Computer Science course involving the use of induction. The main discrepancy lies in that a few students and I think our proof should suffice, but our instructor argues that there are gaps in our logic. We really just want to get a full understanding of why there may be gaps in our logic.

Problem: Prove that for $K\geq8$, the value $K$ can be made up of the sum of two different types of coins with value $3$ and $5$.

Our Proof: We show that this is true for $K=8,9,10$.

Let $m$ denote a $3$ coin and $n$ denote a $5$ coin.

$K=8=3+5=m+n$

$K=9=3+3+3=3m$

$K=10=5+5=2n$

So we have established that this holds for $K=8,9,10$. Let $N>10$. Assume that our hypothesis hold for $8\leq K\leq N$. We wish to show this implies it holds for $K=N+1$.

So, $K=N+1=(N-2)+3$, and by our inductive hypothesis, $N-2$ can be made up of 3-coins and 5-coins, (since $N>10$), and clearly $3$ is simply one 3-coin., thus $K=N+1$ can be made up of 3-coins and 5-coins.

We have established our base case ($K=8,9,10$) and shown that assuming $K=N$ holds for $N\geq10$, this implies that $K=N+1$ holds, and thus by the principle of mathematical induction, any value $K\geq8$ can be made up of 3-coins and 5-coins.

Is our argument unsound? Our professor's disagreement was that in order to prove it in this way, we actually must prove for 3 different sequences, that is, 3 separate sequences in which our $K$ value is $8,11,14,...$, and $9,12,15,...$, and $10,13,16,...$ respectively. He states this is because induction must rely solely on the immediately prior term, i.e. $A_N$ implies $A_{N+1}$ implies $A_{N+2}$ ... but that our proof only relates $A_{N+1}$ to $A_{N-2}$. Is this really how induction works?

We think this is maybe because of how induction and sequences are defined in our course, but based on our prior math courses, we think this should suffice.

Answer

First we have this principle: take any (mathematically well-formed) property $P(n)$, depending on $n$ if you can prove that

$$P(0)$$

$$\forall n \in \mathbb{N}\, P(n) \implies P(n+1)$$

from this you may conclude $$\forall n \in \mathbb{N}\, P(n)$$

From this induction principle you can derive many other induction principles: finite induction, backwards induction etc but we only need two of the following:

You may start from any other natural number other than $0$:

$$P(k)$$

$$\forall n \in \mathbb{N_{\ge{k}}}\, P(n) \implies P(n+1)$$

$$\forall n \in \mathbb{N_{\ge{k}}}\, P(n)$$

and a principle of strong induction:

Suppose you prove

$$\forall n \in \mathbb{N_{\ge{k}}}\,\Bigl(\forall k \in \mathbb{N_{\ge{k}}}\, [k < n \implies P(k)]\Bigr) \implies P(n)$$

then you may conclude

$$\forall n \in \mathbb{N_{\ge{k}}} P(n)$$

In words, the first principle of mathematical induction works like this:
You prove the base case (it is absolutely necessary to do this!) and after that you prove that the property holds for some number $n$ by knowing only that it holds for the previous number $n-1$!
And if you work with the principle of strong induction not only that you don't need to prove the base case - you can also consider that the property holds for all of the previous numbers! How cool is that?
And this is exactly what you did in your proof by strong induction:

Let's suppose that $P(n)$ means than $n$ can be rewritten as sum of $2$'s and $3$'s.

So you take any number $n \ge 8$. You assume that forall $8 \le k < n$ that property holds and then you prove that it also holds for $n$. You do this by cases:

if $n = 8$ then $P(n)$

if $n = 9$ then $P(n)$

if $n = 10$ then $P(n)$

if $n > 10$ then $8 \le n - 3 < n$ and thus by assumption $P(n-3)$ from this follows $P(n)$ since it only has one more $3$.

So you see that $P(n)$ holds in all cases thus by principle of strong induction you may conclude $\forall n \in \mathbb{N_{\ge{8}}}\, P(n)$

Now let's see how you could proceed by using first (ordinary) induction:

The idea is to use ordinary induction 3 times to prove that the property holds for 3 different sequences of natural numbers, that is for
$8 + 3n$, $9+3n$, $10+3n$ $n \ge 8$ from this you will be able to conclude that it in facts hold for any natural number greater than $8$, since it'll definitely be in one of those sequences.

I'll show how to do that for the first one, the other two are proved verbatim.

We prove $$\forall n \in \mathbb{N} \mbox{ we have } P(8+3n)$$

$$n = 0, \mbox{ we certainly have } P(8)$$

Now assume we have $P(8+3k)$ for some $k \in \mathbb{N}$ then we need to prove that $P[8 + 3(k+1)]$.

But $8+3(k+1) = (8 + 3k) + 3$ and since $8 + 3k$ could be rewritten as a sum of $3$'s and $2$'s then $(8+3k)+3$ can be also rewritten this way since it is the same but has one more 3.
Thus, we have $P[8 + 3(k+1)]$ and thus by the ordinary induction you may conclude $$\forall n \in \mathbb{N}\, P(8+3n)$$