Question

1.13. Let a1, a2, . . . , ak be integers with gcd(a1, a2, . ....

1.13. Let a1, a2, . . . , ak be integers with gcd(a1, a2, . . . , ak) = 1, i.e., the largest
positive integer dividing all of a1, . . . , ak is 1. Prove that the equation
a1u1 + a2u2 + · · · + akuk = 1
has a solution in integers u1, u2, . . . , uk. (Hint. Repeatedly apply the extended Euclidean
algorithm, Theorem 1.11. You may find it easier to prove a more general
statement in which gcd(a1, . . . , ak) is allowed to be larger than 1.)

Taken from "Introduction to Mathematical Cryptography" - Hoffstein (2nd ed)

Math   Advanced-Math   
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