MathLabs

Problem 4

Find all pairs (k,n)(k,n) of positive integers such that k!=(2n−1)(2n−2)(2n−4)⋯(2n−2n−1).k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}).
Step 6 of 6: Conclude the complete solution set
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(k,n)=(1,1)or(k,n)=(3,2)(k,n)=(1,1)\quad\text{or}\quad(k,n)=(3,2)
Detailed analysis

Step 4 rules out every n≥6n\ge6, and Step 5 rules out n=3,4,5n=3,4,5, while Step 1 verifies the two surviving cases. Therefore the complete set of pairs is (k,n)=(1,1)(k,n)=(1,1) and (k,n)=(3,2)(k,n)=(3,2).