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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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a(d−c)<c(d−c)⟹a+d>b+c⟹k>ma(d-c)<c(d-c)\Longrightarrow a+d>b+c\Longrightarrow k>m
Detailed analysis

Since a<ca<c and ad=bcad=bc, we have a(d−c)<c(d−c)a(d-c)<c(d-c), hence b−a<d−cb-a<d-c. Therefore a+d>b+ca+d>b+c, so 2k>2m2^k>2^m and k>mk>m.