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Suppose we are given an equation in $ax+(a + 1)y=c$

Now we have to find for how many values of $c$ where $c \in [1,\infty)$ will have no positive integral solution.

I'm new to diophantine equation, so I can't think of any approach. But can it be found mathematically?

Till now my approach is based on programming/brute force

I'm using a small function to check for all possible values.

void bruteforce(int a, int b, int n) 
{ 
    for (int i = 0; i * a <= n; i++) { 

        if ((n - (i * a)) % b == 0) { 
              if((i)>0 && ((n - (i * a)) / b)>0){
                 cout << "x = " << i << ", y = " << ; 
              }
            return; 
        } 
    } 

    cout << "Not Possible"; 
}

But how can i find it more mathematically?

Example -

$3x+4y$ This equation won't have any positive integer solution for $c∈\{1,2,5\}$

$4x+5y$ this equation won't have any positive integer solution for $c ∈ \{1,2,3,6,7,11\}$ so answer would be $6$

So answer comes as $^3C_2$ in first case and $^4C_2$ in second.

Chain Markov
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  • Show us your attempts – Michael Rozenberg Jun 09 '19 at 05:20
  • See https://math.stackexchange.com/questions/880744/diophantine-equation-ax-by-c-has-an-integer-solution-x-0-y-0-if-and-onl – nmasanta Jun 09 '19 at 05:31
  • I've added my attempt. Also the question is for how many values of $c$ there won't be any positive integer solution. –  Jun 09 '19 at 05:39
  • there are infinite numbers of such $c$ for example any prime numbers $p>\gcd(a,b).$ – Leox Jun 09 '19 at 05:54
  • Sorry i forgot to add one more point. It's given that $b= a+1$, i.e $a$ and $b$ are consecutive, from my brute force method i find the answer to be coming as aC2 –  Jun 09 '19 at 06:14
  • I'm giving an example.. . 3x+4y This equation won't have any positive integer solution for {1,2,5} so the answer would be 3 . 4x+5y this equation won't have any positive integer solution for {1,2,3,6,7,11} so answer would be 6 . So answer comes as 3C2 in first case and 4C2 in second –  Jun 09 '19 at 06:23
  • Can $a$ be non-positive ? –  Jun 09 '19 at 07:59
  • How do you achieve $3x+4y=4$ ??? –  Jun 09 '19 at 08:07
  • It is unclear what is being asked. The question as posed asks "For a chosen positive integer $c$, can we always find $a,x,y$ that satisfy the relationships?" The answer to that, provided by many respondents, is YES. But the examples seem to imply that the question OP is really interested in is "For chosen positive integers $c,a$, can we always find $x,y$ that satisfy the relationships?" The answer to that question, by OP examples, is NO. – Keith Backman Oct 28 '19 at 16:06

4 Answers4

1

The condition for the existence of integral solutions to $ax + by = c$ is $gcd(a, b) \; | \ c$. As the set $\mathbb{N}$ is infinite so we can always find infinite numbers which aren't multiples of $gcd(a, b)$.

Deepam Sarmah
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  • With the latest edit, $\gcd(a,b)=\gcd(a,a+1)=1$ – Shubham Johri Jun 09 '19 at 06:43
  • Yes the GCD of consecutive number is 1, please look at the example which i added. There are some values of c such that the equation won't have any positive integer solution like 3x+4y=5 –  Jun 09 '19 at 06:58
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My approach solves the equation over the integers, then tries to find a positive solution :

$x=-c$ and $y=c$ is a solution.

as $gcd(a,b)=1$, and $ab-ba=0$, all solutions to the inital equation are of the form $(-c-kb,c+ka)$ for an integer $k$.

assuming $a$ positive, both $x$ and $y$ are positive iff:

$$ \dfrac{-c}{a}< k <\dfrac{-c}{a+1} $$

So a positive solution exists iff there is an integer between $\dfrac{c}{a+1}$ and $\dfrac{c}{a}$

I think your examples allow non-negative solutions (eg $4x+5y=5$ works for $x=0$ and $y=1$), but you might live (like myself) in a country where positive means $\geq 0$ (then change the $<$'s in my solution by $\leq$'s).

imj
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The equation is

$$ax+(a+1)y=a(x+y)+y=az+y=c$$ where $z\ge2,y\ge1$.

Assuming $a\ge0$, There are solutions for all $c\ge2a+1$.

0

If $x \ge 1$ and $y \ge 1$, then $c = ax+(a + 1)y \ge a(1)+(a+1)(1) = 2a+1$ Why does your program include $x=0$ and $y=0$ since $0$ is not a positive integer?

Steven Alexis Gregory
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