2

For instance with two variables: $ax + by = c$, where x and y are variables.

I found these two threads [1, 2], where the solution is equal to $\binom{n+p-1}{p-1}$, where n is the desired sum and p is the number of variables, so for the case above it would be $\binom{c+2-1}{2-1}$. This is then divided by the product of the numbers multiplying the variables, so in this case by $a*b$. If the result is not an integer, it's rounded down. All in all: $\lfloor\frac{\binom{c+2-1}{2-1}}{ab}\rfloor$.

This works for many equations, but I have found one where it doesn't, and I have no idea why and how to solve it. The problematic equation is the following: $$54x+177y=81630.$$

Here the number of solutions should be 26, the solution above however gives 8. How do I get to 26?

Drejk
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  • Would it help to first divide by gcd of coefficients? – coffeemath Nov 16 '18 at 23:08
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    @coffeemath Yes it would, must have misunderstood that part in the linked threads. Thanks. – Drejk Nov 17 '18 at 00:06
  • I also noticed that if a and b are equal, and the number of solutions comes out greater than 1, it will be equal to 1. – Drejk Nov 17 '18 at 01:37
  • The equation $2x+3y=6$ has two nonnegative solutions, $(3,0),(0,2).$ But the formula gives floor of $7/6$ which is $1.$ Also in the link 1 only positive solutions were considered, and here there are none. – coffeemath Nov 17 '18 at 10:16
  • @coffeemath I think somebody here mentioned that it's for non-negative, but I can see it doesn't always work out. I think you could solve this by adding 1 to the result if a solution where x or y is equal to 0 exists (which is easy to check). – Drejk Nov 18 '18 at 04:11
  • Drejk-- I doubt the rule of when to add or subtract 1 (for either positive or non-negative solution count) is anything as simple as your comment suggests (I may be wrong but would like to see statement and proof if possible, say for the non-negative solution case.) – coffeemath Nov 18 '18 at 04:29
  • @coffeemath I don't really know how I would go about proving it, just trying many different equations. Here is a link to pastebin with a JavaScript function to print all possible results and result count. It does seem to make my results statistically more accurate, though it does make the algorithm work on all equations. It does seem to work on all which contain a solution where $x=0$ or $y=0$. – Drejk Nov 18 '18 at 04:51
  • @coffeemath Some examples:

    $5x+9y=600$ for instance has 14 different solutions, including x=120,y=0. Using the algorithm with dividing the equation by gcd(a,b), then using binomial coefficient and rounding down, you get 13. If you add 1 becouse a solution for x or y == 0 exists, you get 14, which is correct. It also works for $18x+59y=27210*18$.

    It does not help with $18x+59y=27210$, that does not contain such solution though.

     – Drejk Nov 18 '18 at 04:53
  • @coffeemath Actually, I've done some more testing and it seems this might be slightly more accurate: 1) Divide equation by gcd(a,b); 2) $\frac{\binom{c+1}{1}}{ab}$; 3) If equation had $(0,y)$ or $(x,0)$ solution, round up, else round down – Drejk Nov 18 '18 at 06:31
  • I think one has to round up, usually. E.g. $3x+5y=8$ has a solution and the formula gives $9/(3 \cdot 5)$ which if rounded down gives $0.$ The same goes for all right sides from $8$ to $15.$ This versio0n (round up) seems to work whenever $c$ mod $ab$ has a representation $ax+by=c.$ And when this 'reduced equation' haws no solution one must subtract one from the rounded up $(c+1)/(ab),$ i.e. that's when one rounds down. I didn't find this could be stated in terms of when one of $x,y$ is zero. – coffeemath Nov 19 '18 at 21:32
  • Drejk-- I looked at your example which you said didn't work, namely $18x+59y=27210.$ How many solutions did you get by actually checking with software? By my method, round up if reduced equation has solution, I got the number of solutions as $26.$ – coffeemath Nov 19 '18 at 22:41
  • @coffeemath 26 is correct. Always rounding up also won't work in every case though. Check out the updated answer below, it's pretty cool. – Drejk Nov 24 '18 at 19:46
  • Drejk-- Rounding up only works if the reduced equation (replace $c by its remainder mod $ab$) has a solution. Otherwise round down. [And this is for non negative values of $x,y.$] – coffeemath Nov 24 '18 at 20:10

1 Answers1

1

You're not supposed to divide by $ab$ but by the least common multiple $\operatorname{lcm}(a,b)$.

Alternatively you can divide the equation by the $\gcd(a,b)=3$ first, which yields $18x+59y=27210$. The formula works now.

Edit: The formula can be off by $1$. We can see this by considering that e.g. $3x+2y=5$ has $1$ solution, $3x+2y=6$ has $2$ solutions, but $3x+2y=7$ has $1$ solution again.

We can fix it for 2 variables (linear Diophantine equation) by first finding a solution $(x_0,y_0)$ that may contain a negative number. Let's assume that $\gcd(a,b)=1$, which we can always achieve by dividing the equation by the $\gcd$. Then consequently the set of solutions is: $$\{ (x_0+kb,y_0-ka) \mid x_0+kb \ge 0 \land y_0-ka\ge 0 \}$$ Solving this for $k$, we find: $$\left\lceil -\frac{x_0}{b} \right\rceil \le k \le \left\lfloor \frac{y_0}{a}\right\rfloor$$ Therefore the number of solutions is: $$\left\lfloor \frac{y_0}{a}\right\rfloor - \left\lceil -\frac{x_0}{b} \right\rceil + 1 \approx \frac{x_0}{b} + \frac{y_0}{a} = \frac{ax_0+by_0}{ab} = \frac c{ab} \approx \left\lfloor\frac{c+1}{ab}\right\rfloor$$ To find the initial solution $(x_0, y_0)$ we can use the Extended Euclidean algorithm.

Klaas van Aarsen
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  • Aha, I must have misunderstood this part in the linked threads. Thanks a ton. – Drejk Nov 17 '18 at 00:05
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    Actually, plugging the numbers I get only 25 (1 less). At first I thought I might have had the wrong reference result, but I checked the count by going through all the options (using loops, I can post the JavaScript function which prints all results if interested), and it should in deed be 26. Rounding up would give 26, but then it wouldn't work on other equations. – Drejk Nov 17 '18 at 01:15
  • @Drejk Which other equations would the formula not work on? – coffeemath Nov 17 '18 at 04:10
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    Yeah @Drejk, it's one of those $\pm 1$ problems, and it's not trivial. Note that $3x+2y=5$ has 1 solution, $3x+2y=6$ has 2 solutions, but $3x+2y=7$ has 1 solution again. To resolve it, I believe we need to first find an actual solution $(x_0,y_0)$, and then see how many cycles of $\operatorname{lcm}(a,b)$ fit within the domain. – Klaas van Aarsen Nov 17 '18 at 14:47
  • @IlikeSerena Yes I found the same thing... The number of solutions off $ax+by=c$ with $\gcd(a,b)=1$ is not typically, for fixed $a,b,$ weakly monotone increasing with $c$ so no formula based on stars and bars, by adjusting by division by $ab$ [or any constant] and then taking floor or ceiling, can be right, since a formula of the latter type is weakly increasing with $c.$ – coffeemath Nov 17 '18 at 17:48
  • @IlikeSerena I was thinking you could solve this by adding 1 to the result if a solution where x or y is equal to 0 exists (which is easy to check). This did made my results statistically more accurate, it does not however work with $18x+59y=27210$. – Drejk Nov 18 '18 at 04:27
  • @Drejk, I've edited my answer to find the exact number of solutions (for 2 variables). – Klaas van Aarsen Nov 19 '18 at 21:57
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    @IlikeSerena That's awesome, I really appreciate the effort. – Drejk Nov 24 '18 at 19:44