Proof that every number ≥ $8$ can be represented by a sum of fives and threes.

Proof by induction.
For the base case $n=8$ we have $8=5+3$.
Suppose that the statement holds for $k$ where $k\gt 8$. We show that it holds for $k+1$.

There are two cases.

1) $k$ has a $5$ as a summand in its representation.

2) $k$ has no $5$ as a summand in its representation.

For case 1, we delete "that $5$" in the sum representation of $k$ and replace it by two "$3$"s ! This proves the statement for $k+1$.

For case 2, since $k\gt 8$, then $k$ has at least three "$3$"s in its sum representation. We remove these three $3$'s and replace them by two fives! We obtain a sum representation for $k+1$. This completes the proof.

You are almost done, actually you can prove it even without induction:

If $\forall x \ge 8$ and you must prove that $x \in \mathbb{N} \land x = 5a + 3b$

Let $x_1$ be 8 and consider your base cases:

$x_1 = 8 = 3 ⋅ 1 + 5 ⋅ 1$

$x_2 = 9 = 3⋅3 + 5 ⋅ 0$

$x_3 = 10 = 3⋅0 +5⋅2$

This is true for the first 3 $n$ (1, 2 and 3) of $x_n$.

Now, consider the following:

$x_4 = 11 = x_1 + 3$

$x_5 = 12 = x_2 + 3$

$x_6 = 13 = x_3 + 3$

$x_7 = 14 = x_4 + 3$

$x_8 = 15 = x_5 + 3$

$...$

$x_n = x_{n-3} + 3 \,\,\,\,\,\,\forall n \gt 3$

As you can see from the pattern, it's obvious that you are summing only multiples of 3 basing on your previous results (which were also sums of multiples of threes and multiples of fives).

And you may also note that as soon as you sum 5 threes ($3+3+3+3+3 = 15 = 3⋅5$) you can safely replace them with 3 fives ($5 + 5 + 5 = 15 = 3⋅5$)

The previous equation leads to a recurrence relation and you can say with certainty that your statement is true $\forall x \ge 8$