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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b2−a2=2m(b−2k−ma)b^2-a^2=2^m(b-2^{k-m}a)
Detailed analysis

Substituting c=2m−bc=2^m-b and d=2k−ad=2^k-a into bc=adbc=ad gives b2−a2=2m(b−2k−ma)b^2-a^2=2^m(b-2^{k-m}a).