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  "body": "\n## Proof\n\nLet $a \u003e 1$ and $b \u003e 1$ be [[coprime|0GFKP8F8AG]] integers. Put $N = ab - a - b$. Here \"natural number\" includes $0$.\n\n**Part 1: $N$ is not of the form $ax + by$.**\n\n1. Suppose $N = ax + by$ with natural numbers $x$ and $y$. Then $ab = a(x + 1) + b(y + 1)$.\n2. So $b(y + 1) = a(b - x - 1)$, and $a$ divides $b(y + 1)$.\n3. Since $a$ and $b$ are coprime, [[the generalized Euclid's lemma|03BAREW1Z5]] gives $a \\mid y + 1$. As $y + 1 \\ge 1$, this forces $y + 1 \\ge a$.\n4. In the same way, $a(x + 1) = b(a - y - 1)$, so $b \\mid x + 1$ and $x + 1 \\ge b$.\n5. Then $ab = a(x + 1) + b(y + 1) \\ge ab + ab = 2ab$. This is false, since $ab \u003e 0$. So no such $x$ and $y$ exist.\n\n**Part 2: every integer $n \u003e N$ is of the form $ax + by$.**\n\n6. By [[Bézout's identity|0DQVHMYAGA]], there are integers $u$ and $v$ with $au + bv = 1$. So $a u \\equiv 1 \\pmod b$.\n7. Let $n \u003e N$ be an integer. By the [[division algorithm|0CHQKZFK5S]], there is an $x$ with $0 \\le x \\le b - 1$ and $x \\equiv nu \\pmod b$.\n8. Then $ax \\equiv anu \\equiv n \\pmod b$, so $y = (n - ax)/b$ is an integer, and $n = ax + by$.\n9. Since $x \\le b - 1$, we get $n - ax \\ge n - a(b - 1) \u003e N - ab + a = -b$.\n10. So $by = n - ax \u003e -b$, and $y \u003e -1$. As $y$ is an integer, $y \\ge 0$.\n11. So $x$ and $y$ are natural numbers with $n = ax + by$.\n\n## Idea\n\nFor the first part, a representation of $ab - a - b$ would force $x + 1 \\ge b$ and $y + 1 \\ge a$ by divisibility, which overshoots $ab$. For the second part, choose $x$ in $\\set{0, \\dots, b - 1}$ to make $n - ax$ divisible by $b$. The bound $n \u003e ab - a - b$ is exactly what makes the remaining $y$ nonnegative.\n",
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        "locator": "n = 2",
        "url": "https://en.wikipedia.org/wiki/Coin_problem"
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  "name": "Proof of the Frobenius coin problem for two coins",
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