Solution for Section 3.10 Question 2

2. We know from the previous section that gcd(105, 56) = 7, and that m = 2 and n = -4 is one solution to the linear Diophantine equation
  105m + 56n = -14.

Let m₀ = 2, n₀ = -4, a = 105, b = 56 and d = 7, and apply Theorem 3.10.1.

Thus the general solution is given by	m = 2 + (56 / 7) · t	and	n = -4 - (105 / 7) · t	where t Z.
Hence	m = 2 + 8 · t	and	n = -4 - 15 · t	where t Z.

There will be no solutions in which both m and n are positive since 105 times a positive number plus 56 times a positive number will never give you a negative answer. Alternately you could solve the inequalities:

and see that there is no value of t which satisfies both these inequalities.

Back to Section 3.10