why this Diophantine equation 2 k s = ( 5 t + 3 ) ( 16 t + 9 ) has always a solution for every k?I would like to solve the following Diophantine equation and show that it has always a solution; i.e. for every positive integer k, there exists an integer t such that the fraction is an integer: ( 5 t + 3 ) ( 16 t + 9 ) 2 k Any hint will be grateful.P.S.: Without considering cases for k ≡ 0 , 1 , 2 , 3 , 4 ( mod 5 )

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why this Diophantine equation   2  k  s  =  (  5  t  +  3  )  (  16  t  +  9  ) has always a solution for every   k?I would like to solve the following Diophantine equation and show that it has always a solution; i.e. for every positive integer   k, there exists an integer   t such that the fraction is an integer:            (      5      t      +      3      )      (      16      t      +      9      )              2      k      Any hint will be grateful.P.S.: Without considering cases for   k  ≡  0  ,  1  ,  2  ,  3  ,  4  (    mod      5  )

Judovh0 · Accepted Answer

Note that   t will have to be odd, say   2  w  +  1, so we want   (  5  w  +  4  )  (  32  w  +  25  ) to be divisible by   k. Let   k  =      2    a        5    b    l where   l is divisible neither by   2 nor by   5. We will succeed if we can find a value of   w such that   5  w  +  4 is divisible by       2    a   and   32  w  +  25 is divisible by       5    b    l. So we want to solve the system of congruences  5  w  ≡  −  4    (  mod        2    a    )  ,    32  w  ≡  −  25    (  mod        5    b    l  )  .By multiplying the first congruence through by the inverse of   5 modulo       2    a  , and the second congruence by the inverse of   32 modulo       5    b    l, we obtain a system of congruences of the shape   w  ≡  c    (  mod        2    a    ),   w  ≡  d    (  mod        5    b    l  ). By the Chinese Remainder Theorem, this system has a solution.

why this Diophantine equation 2 k s = ( 5 t + 3 ) ( 16 t + 9 ) has

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