MHB Find all combinations of a,b,c

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To find all combinations of integers a, b, and c such that the quadratic equation f(x) = ax^2 + bx + c has integer solutions and satisfies f(8) = 1, the equation can be rewritten as a(8^2) + b(8) + c = 1. This leads to the equation 64a + 8b + c = 1. The requirement for integer solutions implies that the discriminant b^2 - 4ac must be a perfect square. By exploring various integer values for a, b, and c, the combinations can be derived while ensuring that the conditions of the problem are met. The discussion emphasizes the importance of the discriminant in determining the nature of the roots.
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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 $a,b,c$
 
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Albert said:
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 $a,b,c$

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 $(1,- 14,49),(1,- 18,81),(-1,16,-63)$
 
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