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For how many integer values of $a$ does the equation $x^2 + ax + 5a = 0$ have integer solutions for $x$?

Guest Jan 18, 2018
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x^2   +  ax  +  5a  = 0

 

Let m, n  be the integer roots...this implies that....

 

(x - m) (x - n)  =  0

 

x^2  - (m + n)x  + mn = 0

 

So....equating terms.....we have this

 

-m - n  = a   ⇒  -5m -5n   =  5a

mn  =  5a

 

So this implies that

 

-5m - 5n  =  mn

 

-5n - mn  =  5m

 

-n ( 5 + m)  =  5m

 

n  =   -5m / ( m + 5)

 

Note that, as m → ± inf,   n →  -5

Also.....m   must be  a multiple of 5  since  -5m  is a multiple of 5

 

So  we can  test some values

 

m       n         -m - n  =   a

0        0                          0

20     -4                       -16

All positive values of m > 20  means that -5 < n < -4   so no other integer values are possible for in this direction

-10   -10                       20

-30   -6                         36

All negative values of m < -30  means that  -6 < n < -5 so no other integer values are possible for n at all

 

So the possible values for "a" are :

 

  -16, 0, 20  and 36

 

 

cool cool cool

CPhill  Jan 19, 2018
edited by CPhill  Jan 19, 2018

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