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Let $n\in\mathbb{Z}^{+}$, then there exists $k\in\mathbb{Z}_0^+$, such that $n=5k + i, i\in\{0,1,2,3,4\}$. Now analyzing by cases we have:

  1. If $i=0$, then \begin{align*} n = 5k \Rightarrow n = 5k + 4(0). \end{align*}
  2. If $ i = 1 $, then \begin{align*} n & = 5k + 1 \\ & = 5k-5(3) +5(3) +1 \\ & = 5(k-3) + 15 + 1 \\ & = 5(k-3) +16 \Rightarrow n = 5(k-3) +4(4). \end{align*}
  3. If $ i = 2 $, then \begin{align*} n & = 5k + 2 \\ & = 5k-5(2) +5(2) +2 \\ & = 5(k-2) + 10 + 2 \\ & = 5(k-2) +12 \Rightarrow n = 5(k-2) +4(3). \end{align*}
  4. If $i=3$, then \begin{align*} n & = 5k + 3 \\ & = 5k-5 + 5 + 3 \\ & = 5(k-1) +8 \Rightarrow n = 5(k-1) +4(2). \end{align*}
  5. If $i=4$, then \begin{align*} n = 5k + 4 \Rightarrow n = 5k + 4(1). \end{align*}

Thus, every positive number can be expressed as a linear combination of $5$ and $4$. Now using that $n>11$, so we have: \begin{align*} n &> 11 \\ 5k + i &> 5(2) +1 \\ 5k-5(2) &> 1-i \\ 5 (k-2) &> 1-i \\ k-2 &> \frac{1-i}{5} \\ k &> 2+\frac{1-i}{5}. \end{align*} So by increasing over the values ​​that $ i $ takes, we have: \begin{align*} k &> 2+ \frac{1-i}{5} \geq 2+ \frac{1-0}{5}\\ k &> 2 + 0.2 = 2.2 \end{align*} But $k\in\mathbb{Z}_0^+ \Rightarrow k \geq 3 \Rightarrow n \geq 15 $. Thus we have that every positive integer greater than or equal to $15$ is a non-negative linear combination of $5$ and $4$.

Finally, let's look at the cases that are still unverified, which are $12$, $13$ and $14$. \begin{align*} 12 &= 5(0) +4(3) \\ 13 &= 5(1) +4(2) \\ 14 &= 5(2) +4(1). \end{align*} Therefore, any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$.

I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.

BML
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    In general if $a$ and $b$ are coprime positive integers, the largest value you cannot achieve is $(a-1)(b-1)-1 = ab-a-b$ – Henry Jan 31 '21 at 00:09
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    The last part of your proof verifies the result for $12,13$, and $14$. Since $15=5(3)+4(0)$, you have the result for the numbers from $12$ to $15$. Adding $4$ to each of these, you get the result from $16$ to $19$. Adding another $4$, you get the result from $20$ to $23$. Continuing to add $4$'s this way, you get the result for all larger integers. (Formally, "Continuing" means a proof by mathematical induction.) – Andreas Blass Jan 31 '21 at 00:26
  • Essentially the same proof as here (and many prior answers - this is a FAQ, please search first). See the other linked dupes for general proofs for this Frobenius problem. – Bill Dubuque Jan 31 '21 at 03:20
  • Please don't duplicate your prior (closed) question. – Bill Dubuque Jan 31 '21 at 03:29
  • There is a bad community here, because I only present an alternative as a solution ... This is a very toxic community ... – BML Jan 31 '21 at 16:13
  • Your work is correct, but it can be refined a little more, and unfortunately the approach is quite common as we see in other questions of which this has been given as a duplicate. Alternate proofs are given below. – Sarvesh Ravichandran Iyer Feb 07 '21 at 12:24
  • I've seen compliments by other users on your other questions. Please remember, there's nobody who doesn't have a blot on their copybook. The reason this has been closed can be argued for, but I know you have the ability to move on and post questions, which no doubt will reign supreme in everybody's eyes. – Sarvesh Ravichandran Iyer Feb 07 '21 at 12:26

2 Answers2

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We proceed by induction on $n.$ Observe that $12 = 3 \cdot 4 + 0 \cdot 5$ is a non-negative linear combination of $4$ and $5.$ We will assume inductively that any integer $n \geq 12$ can be written as a non-negative linear combination $n = 4x + 5y$ of $4$ and $5.$ Given that $x \geq 1,$ we have that $$4(x - 1) + 5(y + 1) = 4x - 4 + 5y + 5 = 4x + 5y + 1 = n + 1$$ is a non-negative linear combination of $4$ and $5,$ as desired. On the other hand, if we have that $x = 0,$ then $n = 5y.$ By hypothesis that $n \geq 12,$ we must have that $n \geq 15$ (because $n$ is a multiple of $5$) so that $y \geq 3.$ Consequently, we find that $$4 \cdot 4 + 5(y - 3) = 16 + 5y - 15 = 5y + 1 = n + 1$$ is a non-negative linear combination of $4$ and $5.$ QED.

Dylan C. Beck
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You also can prove it by induction on $n>11$.

Here is how the inductive step goes:

Suppose that for some $n>11$, one has $\;n=5x+4y$, $x,y \ge 0$.

  • If $y >0$, as $1=5-4$, you can write $n+1=5(x+1)+4(y-1)$, and $y_1\ge 0 $;
  • If $y=0$, $n=5x$, and since $n>11$, you have $x\ge 3$. Now you also can write $1=16-15$, so $$n+1=5(x-3)+4\cdot 4\qquad(\text{note that } x-3\ge 0)$$
Bernard
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