x = a + f
(a^2 + 2af + f^2)*(2-a) = 1-f
2a^2 + 4af + 2f^2 - a^3 - 2a^2f - af^2 - 1 + f = 0
a^3 - 2(1-f)a^2 + (f^2 - 4f)a - (2f^2 + f - 1) = 0
Look for integer solutions.
Set f=0
a^3 - 2a^2 + 1 = 0
the only integer solution is
a = 1, f = 0 x=1 is one solution
Suppose f is not zero.
RHS is 1-{x} so 0 < RHS <= 1
x^2(2-[x])
0 < x^2(2-[x]) <= 1
There can be no solution with x >= 3 or x <= -1
but there could be a solution for -1 < x < 3
Try a = {-1, 0, 1, 2}
a= -1
-1 - 2(1-f) - (f^2 - 4f) - (2f^2 + f - 1) = 0
3f^2 - 5f + 2 = 0
(3f - 2)(f - 1) = 0 but f = {x} cannot be 1
this would be x = a + f = -1 + 1 = 0 which fails
a = -1; f = 2/3 or x = -1+2/3 = - 1/3 is a solution
a = 0
- (2f^2 + f - 1) = 0
(2f - 1)(f + 1) = 0
f cannot be -1; f = 1/2, a=0; x = 1/2 is a solution
a = 1
1 - 2(1-f) + (f^2 - 4f) - (2f^2 + f - 1) = 0
-f^2 - 3f = 0; f=0; x = 1+0= 1 is already found
a = 2
8 - 2(1-f)4 + (f^2 - 4f)2 - (2f^2 + f - 1) = 0
-f +7 = 0 no solution
Solution Set: -1/3, 1/2, 1
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Posted by Larry
on 2024-06-16 12:22:30 |
Solution
