Design a Mathematica program to find the smallest integer solution for the linear equation 1234567x – 7654321y = 1

To solve the linear Diophantine equation (1234567x – 7654321y = 1) using Mathematica, we can utilize the built-in function ExtendedGCD to find a particular solution and then determine the minimal integer solution. Here’s how:

Step-by-Step Explanation

Check Solvability:
The equation (ax + by = c) has integer solutions if and only if (\gcd(a, b)) divides (c). For our equation, (a = 1234567), (b = -7654321), and (c = 1).
Compute GCD:
Use GCD[1234567, 7654321] to check if the equation is solvable. If the result is 1, proceed; otherwise, no solution exists.
Find Particular Solution:
Utilize ExtendedGCD to find integers (x) and (y) such that (1234567x – 7654321y = 1).
Determine General Solution:
The general solution can be expressed as:
[
x = x_0 + \frac{b}{d}t \
y = y_0 + \frac{a}{d}t
]
where (d = 1), and (t) is an integer.
Find Minimal Positive Solution:
Adjust (t) to find the smallest positive integers (x) and (y).

Mathematica Code

(* Step 1: Check if GCD divides 1 *)
gcd = GCD[1234567, 7654321];
If[gcd != 1,
    Print[“No solution exists.”],
    (* Step 2 and 3: Find particular solution using ExtendedGCD *)
    {g, {x0, y0}} = ExtendedGCD[1234567, 7654321];
    (* The equation is 1234567x – 7654321y = 1, so adjust signs *)
    xParticular = x0;
    yParticular = -y0;

    (* Step 4: General solution form *)
    generalX[t_] := xParticular + (7654321)/gcd * t;
    generalY[t_] := yParticular + (1234567)/gcd * t;

    (* Find the smallest positive integers x and y *)
    (* We need to find t such that both generalX[t] and generalY[t] are positive and minimal *)
    (* Let’s set t such that generalY[t] > 0 and generalX[t] is minimized *)
    (* Since gcd=1, we solve for t in terms of positivity constraints *)

    (* Compute the smallest t where y is positive *)
    tMin = Ceiling[(-yParticular)/1234567];

    (* Calculate x and y for this t *)
    xMinimal = generalX[tMin];
    yMinimal = generalY[tMin];

    Print[“The minimal positive solution is:”];
    Print[“x =”, xMinimal, “, y =”, yMinimal]
}

(* Step 1: Check if GCD divides 1 *)

gcd = GCD[1234567, 7654321];

If[gcd != 1,

Print[“No solution exists.”],

(* Step 2 and 3: Find particular solution using ExtendedGCD *)

{g, {x0, y0}} = ExtendedGCD[1234567, 7654321];

(* The equation is 1234567x – 7654321y = 1, so adjust signs *)

xParticular = x0;

yParticular = –y0;

(* Step 4: General solution form *)

generalX[t_] := xParticular + (7654321)/gcd * t;

generalY[t_] := yParticular + (1234567)/gcd * t;

(* Find the smallest positive integers x and y *)

(* We need to find t such that both generalX[t] and generalY[t] are positive and minimal *)

(* Let‘s set t such that generalY[t] > 0 and generalX[t] is minimized *)

(* Since gcd=1, we solve for t in terms of positivity constraints *)

(* Compute the smallest t where y is positive *)

tMin = Ceiling[(–yParticular)/1234567];

(* Calculate x and y for this t *)

xMinimal = generalX[tMin];

yMinimal = generalY[tMin];

Print[“The minimal positive solution is:”];

Print[“x =”, xMinimal, “, y =”, yMinimal]

}

Explanation

Step 1: We compute the GCD of 1234567 and 7654321. If it’s not 1, the equation has no solution.
Step 2 & 3: Using ExtendedGCD, we find a particular solution ((x_0, y_0)). Adjust signs to match our original equation.
Step 4: Express the general solutions for (x) and (y).
Step 5: Determine the smallest positive integers by finding the appropriate (t), ensuring both (x) and (y) are positive.

This approach efficiently finds the minimal solution using built-in Mathematica functions.