MathLabs

Problem 1

Find all pairs (a,b)(a,b) of positive integers such that a3a^3 is a multiple of b2b^2, and b−1b-1 is a multiple of a−1a-1. (Here nn is called a multiple of mm if n=kmn=km for some integer kk.)
Step 5 of 5: Conclude
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(a,b)∈{(n,n):n≥1}∪{(n,1):n≥1}(a,b)\in\{(n,n):n\ge1\}\cup\{(n,1):n\ge1\}
Detailed analysis

Combining both cases, the only solutions are pairs with a=ba=b, or with b=1b=1 and aa arbitrary; both families were already verified to satisfy the two divisibility conditions.