B Integer solutions for equations

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The discussion focuses on finding integer solutions for equations involving positive integers, specifically in the context of a project related to prime numbers. The user is exploring how to derive a formula for C5 in the equation (C1 * X) + C2 = (C3 * Y) + C4 = (C1 * C3 * n) + C5, where C1, C2, C3, and C4 are constants. They reference the Chinese remainder theorem as a basis for existence and mention that the coefficients C1 and C3 are co-prime. While a direct closed formula for C5 may not exist, an algorithm can be employed to find solutions, particularly when limited to two equations. The discussion emphasizes the importance of not getting bogged down in complex calculations while working on the project.
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I am trying to understand how to find solutions for a problem when parameters are limited to positive integers.
Example:
30x+19= 7y+1 =a ; where x,y,a are positive integers

Wolframalpha outputs:
a = 210 n + 169, x = 7 n + 5, y = 30 n + 24, n element Z(integers)

30*7= 210 (obviously)
How do I calculate 169? I can do it by hand, but am likely missing something obvious generating a universal equation.

I am trying to find a formula to solve for C5 below:
(C1 * X) +C2 = (C3*Y) + C4 = (C1*C3*n) +C5 ; where C1,C2,C3,C4 are constants and X,Y,n are the set of positive integers. Looking for a formula for smallest integer solution C5

Any suggestions?
 
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The existence is a result of the Chinese remainder theorem, and with the formulas there you can also construct the smallest solution.

##30x+19=a## can be interpreted as ##a=19 \pmod{30}##.
 
MFB: Thanks for the link.
I forgot to specify, coefficients (C1&C3) of x and y are also defined as co-prime.

This is a small part of a (non-school) prime number related summer project I am working on with my 13&16 year old and we trying no to get bogged down on this equation. If it does not a have a formula solution for C5 (above), we may need to move onto a different path.
 
There is an algorithm to find it. It is not directly a nice closed formula.
 
MFB: It seems not too complex of an algorithm if limited to 2 equations. We may be able to use it and define the bounds needed.
Thanks
 
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