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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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2m−1∣(a+b)2^{m-1}\mid(a+b)
Detailed analysis

If it divided b−ab-a, then b−a≥2m−1b-a\ge2^{m-1}, so both bb and cc would exceed 2m−12^{m-1}, contradicting b+c=2mb+c=2^m. Thus it divides a+ba+b. Since a+b<b+c=2ma+b<b+c=2^m, we get a+b=2m−1a+b=2^{m-1}.