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Finding multiple solutions to Diophantine equations

Suppose I need to find integer solutions to Ax+By=CAx+By=C. I know that there will also be solutions for every A(x+c)+B(x+d)=14A(x+c)+B(x+d)=14 such that Ac+Bd=0Ac+Bd = 0.

Suppose I already know the values of xx and yy. Is there a quick way to find values for cc and dd without having to solve another equation?
(edited 6 years ago)
Original post by MR1999
Suppose I need to find integer solutions to Ax+By=CAx+By=C. I know that there will also be solutions for every A(x+c)+B(x+d)=14A(x+c)+B(x+d)=14 such that Ac+Bd=0Ac+Bd = 0.

Suppose I already know the values of xx and yy. Is there a quick way to find values for cc and dd without having to solve another equation?


I'm puzzled, don't you just read solutions off from Ac+Bd=0Ac+Bd = 0? I'm assuming that as you've called this a Diophantine problem, A and B are integers.
Original post by MR1999
Suppose I need to find integer solutions to Ax+By=CAx+By=C. I know that there will also be solutions for every A(x+c)+B(x+d)=14A(x+c)+B(x+d)=14 such that Ac+Bd=0Ac+Bd = 0.

Suppose I already know the values of xx and yy. Is there a quick way to find values for cc and dd without having to solve another equation?


Original post by Gregorius
I'm puzzled, don't you just read solutions off from Ac+Bd=0Ac+Bd = 0? I'm assuming that as you've called this a Diophantine problem, A and B are integers.
I think the question has been badly paraphrased (in particular, what has 14 got to do with the problem as stated...?)

But I think what the OP actually wants to know is (with me giving concrete examples):

Suppose we have a simple Diophantine equation such as 18x+15y = 15.

It's obvious that x = 0, y = 1 is a solution, but the general form x = 5n, y=1-6n is less obvious.

But (to the OP), it's still pretty straightforward: it's clear that 18(15) + 15(-18) = 0, (which would give x = 15n, y = 1-18n). For full generality we divide by the highest common factor of (18, 15), which is 3.
Original post by DFranklin
I think the question has been badly paraphrased (in particular, what has 14 got to do with the problem as stated...?)

But I think what the OP actually wants to know is (with me giving concrete examples):

Suppose we have a simple Diophantine equation such as 18x+15y = 15.

It's obvious that x = 0, y = 1 is a solution, but the general form x = 5n, y=1-6n is less obvious.

But (to the OP), it's still pretty straightforward: it's clear that 18(15) + 15(-18) = 0, (which would give x = 15n, y = 1-18n). For full generality we divide by the highest common factor of (18, 15), which is 3.


Ah, thanks!
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Original post by DFranklin
I think the question has been badly paraphrased (in particular, what has 14 got to do with the problem as stated...?)

But I think what the OP actually wants to know is (with me giving concrete examples):

Suppose we have a simple Diophantine equation such as 18x+15y = 15.

It's obvious that x = 0, y = 1 is a solution, but the general form x = 5n, y=1-6n is less obvious.

But (to the OP), it's still pretty straightforward: it's clear that 18(15) + 15(-18) = 0, (which would give x = 15n, y = 1-18n). For full generality we divide by the highest common factor of (18, 15), which is 3.


Yeah I guess the question was very badly worded. The given equation was 826x+350y=14 but I wanted to make it more general.

So given the solution pair (-11,26), we can say that 826(-11+25n) +350(26-59n) = 14, for n an integer?
Original post by MR1999
Yeah I guess the question was very badly worded. The given equation was 826x+350y=14 but I wanted to make it more general.

So given the solution pair (-11,26), we can say that 826(-11+25n) +350(26-59n) = 14, for n an integer?
Looks right.
I'm not sure but this may be of help for you:

https://www.youtube.com/watch?v=TIk3ujphMfk

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