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

Because (b−a)(b+a)=b2−a2(b-a)(b+a)=b^2-a^2 and a,ba,b are odd, b−ab-a and b+ab+a cannot both be divisible by 44. Thus 2m−12^{m-1} divides one of them.