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Show that if positive integers $a$ and $b$ are relatively prime, then every integer $c > ab$ has the form $ax + by = c$, where $x$ and $y$ are non-negative integers.

According to the common way of solving such a Diophantine equation $ax + by = c$, if $x_0$, $y_0$ is already a solution, then the pairs are also a solution: $$ x = x_0 + \frac{bn}{d}, y = y_0 - \frac{an}{d},\quad (n\in Z) $$ where $d$ is the greatest common divisor of $a$ and $b$.

However, although a pattern of periodicity is known, it is still not certain there are definite chances that $x$ and $y$ can be both positive.

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marked as duplicate by YuiTo Cheng, Community Jun 1 at 4:12

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Your problem is basically a special case of the Coin problem, with the n = 2 section including a proof that just requires the somewhat less restrictive condition of $c \ge (a-1)(b-1)$. Adjusted to use your variables, and paraphrased somewhat by myself, it says that for any such $c$, as $a$ and $b$ relatively prime, then all of the integers $c - jb$ for $0 \le j \le a - 1$ are mutually distinct modulo $a$. Thus, there's a unique value of $j$, say $j = k$, and a non-negative integer $n$, such that $c = na + kb$. Note that $n \ge 0$ because $na = c - kb \ge (a - 1)(b - 1) - (a - 1)b = -a + 1$.

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  • $\begingroup$ Thanks for your answer! It seems that the average of $x/b$ and $y/a$ can be used to prove that there is a positive combination, while the $c = ab$ is quite a limit. This result can't be further reduced to $(a-1)(b-1)$. $\endgroup$ – Wenkuei P'ei Apr 18 at 5:46
  • $\begingroup$ @WenkueiP'ei You are welcome. Thank you for updating your question to provide some extra context. If you haven't done so, I suggest you go through the linked Wikipedia page as it has other interesting, potential useful, information. $\endgroup$ – John Omielan Apr 18 at 5:49

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