MHB Find all combinations of a,b,c

[Albert1]

let $f(x)=ax^2+bx+c,with \,\, a\neq 0$ here $a,b,c\in Z$
if the solutions of $ax^2+bx+c=0,\,\, also \,\, \in Z$
given :$f(8)=1$
please find all combinations of $a,b,c$

 

[kaliprasad]

let $f(x)=ax^2+bx+c,with \,\, a\neq 0$ here $a,b,c\in Z$
if the solutions of $ax^2+bx+c=0,\,\, also \,\, \in Z$
given :$f(8)=1$
please find all combinations of $a,b,c$

Spoiler

let $f(x) = a (x- p)(x-q)$
$f(8) = a(8-p)(8-q) = 1$
gives 4 choices
choice 1

$ a = 1, 8-p = 1, 8-q = 1 => p=q = 7$
or $f(x) = (x-7)^2 = x^2 - 14x + 49=> a= 1, b= - 14, c= 49$

choice 2
$ a = 1, 8-p = -1, 8-q = -1 => p=q = 9$
or $f(x) = (x-9)^2 = x^2 - 18x + 81=> a= 1, b= - 18, c= 81$
choice 3

$ a = -1 , 8-p = 1, 8-q = -1 => p=7, q = 9$
or $f(x) = -(x-7)(x-9) = -x^2 + 16x -63 => a= -1, b= 16, c= -63$

choice 4 is same as choice 3 except that p and q are interchanged
hence 3 sets $(1,- 14,49),(1,- 18,81),(-1,16,-63)$