Modular arithmetic polynomial questionIs it possible to find all polynomials of the form a n 2 + b n + c where a,b, and c are integers and such that a + b + c ≡ 31 ( mod 54 ) 4 a + 2 b + c ≡ 3 ( mod 54 ) 9 a + 3 b + c ≡ 11 ( mod 54 )

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Modular arithmetic polynomial questionIs it possible to find all polynomials of the form   a      n    2    +  b  n  +  c where a,b, and c are integers and such that  a  +  b  +  c  ≡  31    (  mod    54  )  4  a  +  2  b  +  c  ≡  3    (  mod    54  )  9  a  +  3  b  +  c  ≡  11    (  mod    54  )

Samsonitew7b · Accepted Answer

Step 1That you&#039;re making them coefficients of a polynomial is irrelevant: the problem is &quot;solve this linear system of three equations in three variables&quot;.That you&#039;re solving systems of equations modulo 54 rather than systems of equations of real numbers doesn&#039;t change things much: the only real difference is how to compute the inverse of a number, and that some non-zero numbers can fail to be invertible.Solving the first equation gives  c  ≡  31  −  a  −  bplugging into the other equations gives  3  a  +  b  ≡  −  28  ≡  26  8  a  +  2  b  ≡  −  20  ≡  34Solving the first of these gives  b  ≡  26  −  3  aplugging into the other equation gives  2  a  ≡  −  18  ≡  36Solving this is a little trickier: the solution space to  g  u  x  ≡  g  v    (  mod    g  w  )is the same as the solution space to  u  x  ≡  v    (  mod    w  )(Note if the equation was   g  u  x  ≡  v    (  mod    g  w  ) where v is not divisible by g, then there would be no solutions for x)So to solve  2  a  ≡  36    (  mod    54  )you factor out the 2  a  ≡  18    (  mod    27  )  a  ≡  18  ,  45    (  mod    54  )Step 2Now backsolving gives  a  ≡  18    (  mod    54  )    b  ≡  26    (  mod    54  )    c  ≡  41    (  mod    54  )  a  ≡  45    (  mod    54  )    b  ≡  53    (  mod    54  )    c  ≡  41    (  mod    54  )Of course, I could have used Gaussian elimination and its variants instead. Or I could have solved for a first, or any other variation.The big thing I didn&#039;t demonstrate is the Chinese remainder theorem approach.   54  =  2  ⋅      3    3  ; Often, it&#039;s easier to solve a problem modulo 2 then solve the problem modulo       3    3  , then use the Chinese remainder theorem to combine the solutions to modulo 54 than it is to solve it directly

bruinhemd3ji · Accepted Answer

Step 1So you want to solve the system of linear equations      (                            1                          1                          1                                      4                          2                          1                                      9                          3                          1                      )        (                            a                                      b                                      c                      )    =      (                            31                                      3                                      11                      )  over the ring Z/54Z. By the chinese remainder theorem, it is enough to solve this over Z/2Z and Z/27Z separately. Using the Gaussian algorithm over Z/2Z, we get      (                            1                          1                          1                          1                                      0                          0                          1                          1                                      1                          1                          1                          1                      )    ⇝      (                            1                          1                          0                          0                                      0                          0                          1                          1                      )    ,whose solutions are (0,0,1) and (1,1,1).Step 2Over Z/27Z the matrix is invertible (its determinant is −2 which is a unit modulo 27), so we get a unique solution, namely (18,26,14) by the Gaussian algorithm or by computing the inverse of the matrix.Putting the solutions together with the Chinese Remainder Theorem gives  (  a  ,  b  ,  c  )  ≡  (  18  ,  26  ,  41  )    mod      54(corresponding to (0,0,1)mod2) and  (  a  ,  b  ,  c  )  ≡  (  45  ,  53  ,  41  )    mod      54(corresponding to (1,1,1)mod2).

Modular arithmetic polynomial question. Is it possible to find all polynomials of the form an^2+bn+c where a,b, and c are integers and such that a+b+c equiv 31 (mod 54)

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