<s>[INST] Proof that every number ≥ $8$ can be represented by a sum of fives and threes. Can you check if my proof is right?
Theorem.  $\forall x\geq8, x$ can be represented by $5a + 3b$ where  $a,b \in \mathbb{N}$.
Base case(s): $x=8 = 3\cdot1 + 5\cdot1 \quad \checkmark\\
x=9 = 3\cdot3 + 5\cdot0 \quad \checkmark\\
x=10 = 3\cdot0 + 5\cdot2 \quad \checkmark$
Inductive step:
$n \in \mathbb{N}\\a_1 = 8, a_n = a_1 + (x-1)\cdot3\\
b_1 = 9, b_n = b_1 + (x-1)\cdot3 = a_1 +1 + (x-1) \cdot 3\\
c_1 = 10, c_n = c_1 + (x-1)\cdot3 = b_1 + 1 + (x-1) \cdot 3 = a_1 + 2 + (x-1) \cdot 3\\
\\
S = \{x\in\mathbb{N}: x \in a_{x} \lor x \in b_{x} \lor x \in c_{x}\}$
Basis stays true, because $8,9,10 \in S$
Lets assume that $x \in S$. That means $x \in a_{n} \lor x \in b_{n} \lor x \in c_{n}$.
If $x \in a_n$ then $x+1 \in b_x$,
If $x \in b_x$ then $x+1 \in c_x$,
If $x \in c_x$ then $x+1 \in a_x$.
I can't prove that but it's obvious. What do you think about this?
 [/INST] Write $x=8n+k$ for $0\leq k<8$ and $n\geq 1$. Because $8n=(3+5)n$, the problem is clear if $k=0,3,5$. Let's consider the remaining 5 cases:
\begin{align*}
k=1 & \implies 8n+1=5(n-1)+3(n+2),\\
k=2 & \implies 8n+2=5(n+1)+3(n-1),\\
k=4 & \implies 8n+4=5(n-1)+3(n+3),\\
k=6 & \implies 8n+6=5n+3(n+2),\\
k=7 & \implies 8n+7=5(n+2)+3(n-1).
\end{align*}
Note that $k=4$ is based on $k=1$ (just add one more $3$) and, similarly, $k=6$ and $k=7$ follow from $k=1$ and $k=2$, respectively. So in fact, we have really only considered 2 cases.
</s>