Given a ∈ R , such that the two roots in f ( x ) = x 2 + a x + ( 2 a + 319 )are both positive integers. Find a.

Question

Given   a  ∈      R  , such that the two roots in  f  (  x  )  =      x    2    +  a  x  +  (  2  a  +  319  )are both positive integers. Find a.

canhaulatlt · Accepted Answer

Actually,   D  =      a    2    −  8  a      −    1276You have                              a          2                −        8        a        −        1276        =                  N          2                                            ⟺                (        a        −        4                  )          2                −        1292        =                  N          2                                                          ⟺                1292        =        (        a        −        4        −        N        )        (        a        −        4        +        N        )        .            So, for any way of writing   1292 as   k  ×  l (  k  ,  l  ∈      Z  ), solve the system      {                            a          −          4          −          N          =          k                                      a          −          4          +          N          =          l          .                        For instance, if k=−34, and l=−38, you will get a=−32 and the roots will be 15 and 17.

Valery Cook · Accepted Answer

Let   f  (  x  )  =  (  x  −  r  )  (  x  −  s  ). Then by Vieta,                    r        +        s                            =        −        a                            r        s                            =        2        a        +        319.            Adding yields the Diophantine equation  r  s  +  2  r  +  2  s  =  319  ,which has the solution   (  r  ,  s  )  =  (  15  ,  17  )  ,  (  17  ,  15  ) in positive integers (there are only a few cases to check since   r  ,  s  ≤  105 and both r,s are odd). So a=−32 is the only solution.

Given a in RR, such that the two roots in f(x)=x^2+ax+(2a+319) are both positive integers. Find a.

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