Suppose that a, b \in \mathbb{Z}a, b∈Z, not both 00, and let d=\gcd(a, b)d=gcd(a, b). Bezout's theorem states that dd can be written as a linear combination of aa and bb, that is, there exist integers m, n \in \mathbb{Z}m, n∈Z such that d = am + bnd=am+bn. Prove that, on the other hand, any linear combination of aa and bb is divisible by dd. That is, suppose that t = ax + byt=ax+by for some integers x, y \in \mathbb{Z}x, y∈Z. Prove that d \, | \, td∣t.