Can the powers of 2 in the denominators of a fraction sum be used to find a contradiction?Let a 1 , a 2 , b 1 , b 2 , c 1 , c 2 , d 1 , d 2 , e 1 , e 2 be odd integers, with gcd ( a 1 , a 2 ) = gcd ( b 1 , b 2 ) = gcd ( c 1 , c 2 ) = gcd ( d 1 , d 2 ) = gcd ( e 1 , e 2 ) = 1 ,and assume n ≥ 3 and m ≥ 1 are integers such that ( ⋆ ) a 1 2 3 ( n − 1 ) m a 2 + b 1 2 3 ( ( n − 1 ) m + 1 ) b 2 + c 1 2 3 ( ( n − 1 ) m + 1 ) c 2 + d 1 2 3 n m d 2 + e 1 2 3 n m e 2 = 1. Is there a way to prove, using only the powers of 2 in the set of denominators, that (⋆) is impossible?

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Can the powers of 2 in the denominators of a fraction sum be used to find a contradiction?Let       a    1    ,      a    2    ,      b    1    ,      b    2    ,      c    1    ,      c    2    ,      d    1    ,      d    2    ,      e    1    ,      e    2   be odd integers, with  gcd  (      a    1    ,      a    2    )  =  gcd  (      b    1    ,      b    2    )  =  gcd  (      c    1    ,      c    2    )  =  gcd  (      d    1    ,      d    2    )  =  gcd  (      e    1    ,      e    2    )  =  1  ,and assume   n  ≥  3 and   m  ≥  1 are integers such that                              (                      ⋆                    )                                                  a            1                                              2                              3                (                n                −                1                )                m                                                    a              2                                      +                              b            1                                              2                              3                                  (                  (                  n                  −                  1                  )                  m                  +                  1                  )                                                                    b              2                                      +                              c            1                                              2                              3                                  (                  (                  n                  −                  1                  )                  m                  +                  1                  )                                                                    c              2                                      +                              d            1                                              2                              3                n                m                                                    d              2                                      +                              e            1                                              2                              3                n                m                                                    e              2                                      =        1.            Is there a way to prove, using only the powers of 2 in the set of denominators, that (⋆) is impossible?

Zichetti4b · Accepted Answer

The way it stands now,   (  ⋆  ) is obviously possible. Let   n  =  3  ,  m  =  2. Then the denominators of five fractions contain (correspondingly, in the same order)       2          12        ,      2          15        ,      2          15        ,      2          18        ,   and       2          18      . Now,            184313              45        ⋅                  2                      12                                +            1              9        ⋅                  2                      15                                +            1              1        ⋅                  2                      15                                +            1              15        ⋅                  2                      18                                +            1              1        ⋅                  2                      18                                =  …...guess what?

Can the powers of 2 in the denominators of a fraction sum be used to find a contradiction? Let

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