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?

Any number is of the form $5k+1$, $5k+2$, $5k+3$, $5k+4$ and $5k+5$.
If $n=5k+3$ then nothing to prove. 
$n=5k+1 \implies n=5(k-1)+2\cdot3$
$n=5k+2 \implies n=5(k-2)+4\cdot3$
$n=5k+4\implies n=5(k-4)+8\cdot3$
$n=5k+5\implies n=5(k-2)+5\cdot 3$
Hence for all $k\ge5$ the result holds.
Hence for all $n\ge 25$ the result hold. 
What is left is to check the remaining cases and that is also true.
Hence the proposition is true.
