Polynomials and Arithmetic: If p ( x i ) = 7 for four integers, then p ( z ) ≠ 14Consider the polynomial p ( x ) = a 0 + a 1 x + a 2 x 2 + · · · + a n x n where a 0 , a 1 , . . . , a n ∈ Z . Show that if p ( x i ) = 7 for 4 distinct integers x 0 , x 1 , x 2 , x 3 , then p ( z ) ≠ 14 for any z ∈ Z .How do I start this question?

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Polynomials and Arithmetic: If   p  (      x    i    )  =  7 for four integers, then   p  (  z  )  ≠  14Consider the polynomial  p  (  x  )  =      a    0    +      a    1    x  +      a    2        x    2    +  ·  ·  ·  +      a    n        x    n  where       a    0    ,      a    1    ,  .  .  .  ,      a    n    ∈      Z  . Show that if   p  (      x    i    )  =  7 for 4 distinct integers       x    0    ,      x    1    ,      x    2    ,      x    3  , then   p  (  z  )  ≠  14 for any   z  ∈      Z  .How do I start this question?

HadoHaurrysap3w · Accepted Answer

Step 1Let   q  (  x  )  =  p  (  x  )  −  7. Now the problem becomes:Show that if   q  (      x    i    )  =  0 for 4 distinct integers       x    0    ,      x    1    ,      x    2    ,      x    3    , then   q  (  z  )  ≠  7 for any   z  ∈      Z  Step 2But if   q  (      x    i    )  =  0 for   i  =  0  ,  …  ,  3  , then  q  (  x  )  =  (  x  −      x    0    )  (  x  −      x    1    )  (  x  −      x    2    )  (  x  −      x    3    )  r  (  x  )for some integer polynomial r(x). Suppose   q  (  z  )  =  7 for some   z  ∈      Z  . Can you see the contradiction?

Polynomials and Arithmetic: If p(x_i)=7 for four integers, then p(z) ne 14

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