Integral roots for f ( x ) = 41 if f ( x ) = 37 has 5 distinct integral roots.Given a polynomial f(x) with integral coefficients and f ( x ) = 37 has 5 distinct integral roots, find the number of integral roots of f ( x ) = 41?My Approach: Say f ( x ) = ( x − r 1 ) ( x − r 2 ) ( x − r 3 ) ( x − r 4 ) ( x − r 5 ) g ( x ) + 37, where ri are the distinct integers.Now for f ( x ) = 41 we have ( x − r 1 ) ( x − r 2 ) ( x − r 3 ) ( x − r 4 ) ( x − r 5 ) g ( x ) = 4, so the factors can be ± 1 , ± 2 or ± 4. Given r i are distinct at most two of them will give ± 1, then there can be both of ± 2 or one of ± 4. This is where I get lost, since even if I use all of ± 1 , ± 2, I will be still be left with one x − r i factor. What about that? Does it matter that 37 and 41 are primes, or is it just a coincidence?

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Integral roots for   f  (  x  )  =  41 if   f  (  x  )  =  37 has 5 distinct integral roots.Given a polynomial f(x) with integral coefficients and   f  (  x  )  =  37 has 5 distinct integral roots, find the number of integral roots of   f  (  x  )  =  41?My Approach: Say   f  (  x  )  =  (  x  −      r    1    )  (  x  −      r    2    )  (  x  −      r    3    )  (  x  −      r    4    )  (  x  −      r    5    )  g  (  x  )  +  37, where ri are the distinct integers.Now for   f  (  x  )  =  41 we have   (  x  −      r    1    )  (  x  −      r    2    )  (  x  −      r    3    )  (  x  −      r    4    )  (  x  −      r    5    )  g  (  x  )  =  4, so the factors can be   ±  1  ,  ±  2 or   ±  4. Given       r    i   are distinct at most two of them will give   ±  1, then there can be both of   ±  2 or one of   ±  4. This is where I get lost, since even if I use all of   ±  1  ,  ±  2, I will be still be left with one   x  −      r    i   factor. What about that? Does it matter that 37 and 41 are primes, or is it just a coincidence?

Antwan Wiley · Accepted Answer

Using the fact about the Laplace transform L that  L  (  f  ∗  g  )  =  L  (  f  )  L  (  g  )  =  F  (  s  )  G  (  s  )    ⟹    (  f  ∗  g  )  (  t  )  =      L          −      1        (  F  (  s  )  G  (  s  )  )  .In our case, given   H  (  s  )  =      1          (      s      +      1      )            s          (              s        2            +      4      )        F  (  s  )  =      1          s      +      1          ⟹    f  (  t  )  =      e          −      t        ,    G  (  s  )  =      s                  s        2            +      4          ⟹    g  (  t  )  =  cos  ⁡  (  2  t  )  .Now, you use the convolution as  h  (  t  )  =      ∫          0              t            e          −      (      t      −      τ      )        cos  ⁡  (  2  τ  )  d  τ  .

odcizit49o · Accepted Answer

Step 1Finish by applying the followingKey Idea  The possible factorizations of a polynomial   ∈      Z    [  x  ] are constrained by the factorizations of the integer values that the polynomial takes. For a simple example, if some integer value has few factorizations (e.g. a unit     ±  1 or prime p) then the polynomial must also have few factors, asssuming that that the factors are distinct at the evaluation point. More preciselyStep 2If     f  (  x  )  =      f    1    (  x  )  ⋯      f    k    (  x  )   satisfy             f      i        (    n    )    ≠          f      j        (    n    )     for    i  ≠  j  , all     n  ∈      Z          f    (    n    )    =    ±    1      ⇒    k  ≤  2    else 1 would have 3 distinct factors         f    1    (  n  )  ,      f    2    (  n  )  ,      f    3    (  n  )    f  (  n  )  =  ±  p    ⇒    k  ≤      4      since a prime p has at most 4 distinct factors     ±  1  ,  ±  p

Given a polynomial f(x) with integral coefficients and f(x)=37 has 5 distinct integral roots, find the number of integral roots of f(x)=41?

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