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 4 of 5: Rule in c≥a
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c≥a ⟹ b2c≥b2a≥a3=b2c ⟹ a=b=cc\ge a\ \Longrightarrow\ b^2c\ge b^2a\ge a^3=b^2c\ \Longrightarrow\ a=b=c
Detailed analysis

If instead c≥ac\ge a, then since b≥ab\ge a we get b2c≥b2a≥a2⋅a=a3=b2cb^2c\ge b^2a\ge a^2\cdot a=a^3=b^2c, so equality holds throughout: b2c=b2ab^2c=b^2a gives c=ac=a, and b2a=a3b^2a=a^3 gives b2=a2b^2=a^2, i.e. b=ab=a.