MathLabs

Problem 6

Let a,b,c,d be odd integers with 0<a<b<c<d0<a<b<c<d and ad=bcad=bc. Prove that if a+d=2ka+d=2^k and b+c=2mb+c=2^m, then a=1a=1.
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b=2m−1−a,c=2m−1+a,d=2k−ab=2^{m-1}-a,\quad c=2^{m-1}+a,\quad d=2^k-a
Detailed analysis

The two sum equations now give b=2m−1−ab=2^{m-1}-a, c=2m−1+ac=2^{m-1}+a, and d=2k−ad=2^k-a.