2. First apply the Euclidean Algorithm to 105 and 56 to find gcd(105, 56).
105 |
= |
56 · 1 + 49 | (equation 1) |
| 56 | = | 49 · 1 + 7 | (equation 2) |
| 49 | = | 7 · 7 + 0 | (equation 3) |
Thus, gcd(105, 56) = 7. Since 7 | -14, we know that a solution exists to the linear Diophantine equation 105m + 56n = -14.
Now work backwards through the equations of the Euclidean Algorithm to find a solution.
From equation 2 we see that:
| 7 | = | 56 - 49 · 1 |
| = | 56 - [105 - 56 · 1] · 1 | |
| = | 56 · 2 - 105 · 1 |
Therefore, 105(-1) + 56(2) = 7. So, we multiply both sides of the equation by -2 to get: 105(2) + 56(-4) = -14.
Hence m = 2 and n = -4 is one solution to the linear Diophantine equation 105m + 56n = -14.