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 1 of 6: Name the product and record the first solutions
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Ln=∏i=0n−1(2n−2i),L1=1=1!,L2=6=3!L_n=\prod_{i=0}^{n-1}(2^n-2^i),\qquad L_1=1=1!,\qquad L_2=6=3!
Detailed analysis

Let Ln=∏i=0n−1(2n−2i)L_n=\prod_{i=0}^{n-1}(2^n-2^i). Directly, L1=1=1!L_1=1=1! and L2=6=3!L_2=6=3!, so (k,n)=(1,1)(k,n)=(1,1) and (k,n)=(3,2)(k,n)=(3,2) are solutions. We prove that no other positive-integer pair works.