Value Sets miss an APFor any polynomial p ∈ Z [ X ], the set { p ( n ) : n ∈ Z } is called the value set of p.Problem:- Let p ∈ Z [ X ] be such that deg p &gt; 1. Then there exists an infinite arithmetic sequence none of who terms can be expressed as p(x) for some x ∈ Z .Progress:-For this problem, I first took example which was P ( x ) = x 2 − 1 then we get the value set as {0,3,4,15,24,…}. Since, x 2 − 1 ≡ − 1 , 0 mod 4.. So clearly the AP 2,6,10,… never come.I assumed the contrary. Hence for any n,d there will exist a x such that P ( x ) ≡ n mod d .. If there doesn't then we can select the AP n , n + d , n + 2 d , …Now we know that when p ∈ [ X ], and m ≡ n mod d ⟹ P ( m ) ≡ P ( n ) mod d .. (*)So, we can say that p ( n ) , p ( n + 1 ) , … , p ( n + d − 1 ) forms a complete residue set, because if it doesn't then by PHP, there will be an y ∈ 1 , 2 , … d not being a residue in p(x), when n ≤ x ≤ n + d − 1.Now, take any d consecutive numbers,say l , l + 1 , … , l + d − 1 , by (*), residues mod d of { P ( l ) , … , P ( l + d + 1 ) } = { p ( n ) , p ( n + 1 ) , … , p ( n + d − 1 ) } .This is what I have got till now.

Question

Value Sets miss an APFor any polynomial   p  ∈      Z    [  X  ], the set   {  p  (  n  )  :  n  ∈  Z  } is called the value set of p.Problem:- Let   p  ∈  Z  [  X  ] be such that deg   p  &amp;gt;  1. Then there exists an infinite arithmetic sequence none of who terms can be expressed as p(x) for some   x  ∈  Z  .Progress:-For this problem, I first took example which was   P  (  x  )  =      x    2    −  1 then we get the value set as {0,3,4,15,24,…}. Since,       x    2    −  1  ≡  −  1  ,  0    mod      4.. So clearly the AP   2,6,10,… never come.I assumed the contrary. Hence for any n,d there will exist a x such that   P  (  x  )  ≡  n    mod      d  .. If there doesn&#039;t then we can select the AP   n  ,  n  +  d  ,  n  +  2  d  ,  …Now we know that when   p  ∈      [    X  ], and   m  ≡  n    mod      d    ⟹    P  (  m  )  ≡  P  (  n  )    mod      d  .. (*)So, we can say that   p  (  n  )  ,  p  (  n  +  1  )  ,  …  ,  p  (  n  +  d  −  1  ) forms a complete residue set, because if it doesn&#039;t then by PHP, there will be an   y  ∈      1    ,    2    ,    …    d   not being a residue in p(x), when   n  ≤  x  ≤  n  +  d  −  1.Now, take any d consecutive numbers,say   l  ,  l  +  1  ,  …  ,  l  +  d  −  1  , by (*), residues mod d of   {  P  (  l  )  ,  …  ,  P  (  l  +  d  +  1  )  }  =  {  p  (  n  )  ,  p  (  n  +  1  )  ,  …  ,  p  (  n  +  d  −  1  )  }  .This is what I have got till now.

erlent00s · Accepted Answer

Step 1Short outline of a proof.Choose   x  ,  y  ∈      Z   and   n  ∈      N   such that   p  (  x  )  ≡  p  (  y  )    mod      n. (&amp;lt;- This requires a detailed argument.) Then the map      p    ¯    :      Z        /    n      Z    →      Z        /    n      Z    ,      x    ¯    ↦            p      (      x      )        ¯  is not injective.Step 2Because its a map on a finite set we have       p    ¯   injective     ⟺        p    ¯   surjective. From here we can construct an arithmetic sequence that doesn&#039;t met p(x).

For any polynomial p in Z[X],the set {p(n):n in Z} is called the value set of p.

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