<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] I'm attempting to answer the question intuitively:
Every third number is a multiple of 3, and can definitely be written as a sum of 3s and 5s:
$3x$ + $5$ x $0$.
Now we need to deal with numbers $y$ for which $y mod  3 = 1$ and $y mod  3 = 2$
For $y mod  3 = 2$:
We know that $5 mod 3 = 2$, so 5 x 1 + 3$x$ should give us all numbers $y$ where $y mod 3 =2$
For $y mod  3 = 1$:
We know that $5$ x $2 mod 3 = 1$, so 5 x 2 + 3$x$ should give us all numbers $y$ where $y mod 3 =1$
The smallest case in which we can use this proof is 5 x 2 + 3 x 0 = 10
We can show it is possible for 8 and 9 as 5+3 =8 and 3 x 3 =9
Hence, we can show that is possible for all numbers greater than or equal to 8.
This is also proves that the maximum multiplier we would need for 5 is 2.
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