 Solution

ID: past-exam-of-the-short-options-course-of-the-university-of-oxford/2014/1/b/i/solution

a = 21 r

, the second condition becomes

r \equiv 4 (mod 5)

b = 15 s

, its second condition becomes

s \equiv 6 (mod 7)

. Consequently

a = 84 + 105 m, b = 90 + 105 n, m, n \in Z_{\geq 0}

gives an infinite set of positive integer solutions. Conversely these are the positive solutions in the indicated congruence classes, by the Chinese remainder theorem.

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