Prove that m − n divides pLet a 1 , a 2 , a 3 be a non constant arithmetic Progression of integers with common difference p and b 1 , b 2 , b 3 be a geometric Progression with common ratio r. Consider 3 polynomials P 1 ( x ) = x 2 + a 1 x + b 1 , P 2 ( x ) = x 2 + a 2 x + b 2 , P 3 ( x ) = x 2 + a 3 x + b 3 . Suppose there exist integers m and n such that g c d ( m , n ) = 1 and P 1 ( m ) = P 2 ( n ) , P 2 ( m ) = P 3 ( n ) and P 3 ( m ) = P 1 ( n ). Prove that m − n divides p.I made the 3 equations but am unable to get the result from them. How should I manipulate them.

Question

Prove that   m  −  n divides pLet       a    1    ,      a    2    ,      a    3   be a non constant arithmetic Progression of integers with common difference p and       b    1    ,      b    2    ,      b    3   be a geometric Progression with common ratio r. Consider 3 polynomials       P    1    (  x  )  =      x    2    +      a    1    x  +      b    1    ,      P    2    (  x  )  =      x    2    +      a    2    x  +      b    2    ,      P    3    (  x  )  =      x    2    +      a    3    x  +      b    3  . Suppose there exist integers m and n such that   g  c  d  (  m  ,  n  )  =  1 and       P    1    (  m  )  =      P    2    (  n  )  ,      P    2    (  m  )  =      P    3    (  n  ) and       P    3    (  m  )  =      P    1    (  n  ). Prove that   m  −  n divides p.I made the 3 equations but am unable to get the result from them. How should I manipulate them.

Jovanni Salinas · Accepted Answer

Step 1The claim is not true. A counterexample is  m  =  5  ,    n  =  3  ,    p  =  3  ,    r  =      3    5    ,        a    1    =  −  11  ,        b    1    =      75    2    .In the following, I&#039;m going to write about how I figured out these values and about that the claim is true if       b    1   is an integer.We have      P    1    (  m  )  =      P    2    (  n  )    ⟺        m    2    +      a    1    m  +      b    1    =      n    2    +      a    2    n  +      b    2                      (1)                            ⟺                          m          2                +                  a          1                m        +                  b          1                =                  n          2                +        (                  a          1                +        p        )        n        +                  b          1                r                  P    2    (  m  )  =      P    3    (  n  )    ⟺        m    2    +      a    2    m  +      b    2    =      n    2    +      a    3    n  +      b    3                      (2)                            ⟺                          m          2                +        (                  a          1                +        p        )        m        +                  b          1                r        =                  n          2                +        (                  a          1                +        2        p        )        n        +                  b          1                          r          2                          P    3    (  m  )  =      P    1    (  n  )    ⟺        m    2    +      a    3    m  +      b    3    =      n    2    +      a    1    n  +      b    1                      (3)                            ⟺                          m          2                +        (                  a          1                +        2        p        )        m        +                  b          1                          r          2                =                  n          2                +                  a          1                n        +                  b          1                    From   (  2  )  −  (  1  ), we have                    (4)                    p        m        +                  b          1                (        r        −        1        )        =        p        n        +                  b          1                r        (        r        −        1        )            From   (  3  )  −  (  2  ), we have                    (5)                    p        m        +                  b          1                r        (        r        −        1        )        =        −        2        p        n        +                  b          1                (        1        −                  r          2                )            Now   (  4  )  −  (  5  ) gives                    (6)                              b          1                (        r        −        1        −                  r          2                +        r        )        =        3        p        n        +                  b          1                (                  r          2                −        r        −        1        +                  r          2                )                ⟺                p        n        =                  b          1                (        −                  r          2                +        r        )            Also   −  2  ×  (  4  )  −  (  5  ) gives                    (7)                    p        m        =                  b          1                (        −        r        +        1        )            Step 2Hence, from (6)(7), we have  p  m  r  =  p  n  ,i.e.  r  =      n    m    .Now from (7), we have                    (8)                    p        m        =                  b          1                          (          −                      n            m                    +          1          )                ⇒        p                  m          2                =                  b          1                (        m        −        n        )            Note here that   gcd  (      m    2    ,  m  −  n  )  =  1Hence, if       b    1   is an integer, then we have that   m  −  n divides p.By the way, from (3), we can have       a    1    =  −  m  −  n  −  p

Prove that m-n divides p. Let a_1,a_2,a_3 be a non constant arithmetic Progression of integers with common difference p and b_1,b_2,b_3 be a geometric Progression with common ratio r.

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