| (Simon Clinch) My answer follows:
Eliminating c from the two equations and factorizing we get:
(a-2)(a+b)(b-2) = 0
By symmetry an elimination of the other two variables will produce
the two equations:
(a-2)(a+c)(c-2) = 0
(b-2)(b+c)(c-2) = 0
Suppose 2 is not in {a,b,c}, then we have a+b=0, a+c=0 and b+c=0.
This implies that a=b=c=0 which contradicts the second of the two
initial equations, because none of a,b,c can be zero if
1/a+1/b+1/c=1/2, because 1/0 is undefined. So 2 must be in {a,b,c}.
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| I found "eliminating ...and factorizing" in 520.1 to be somewhat cryptic.
While trying to figure it out, I came across this other solution. (I
suspect that it's essentially the same as 520.1.)
John
** ** ** ** ** ** ** ** ** ** ** ** ** ** ** **
Lemma 1 (Means): if a + b = x + y
and ab = xy
then (a = x and b = y) or (a = y and b = x)
Proof: (a - b) ^ 2 = (a + b) ^ 2 - 4ab
= (x + y) ^ 2 - 4xy
= (x - y) ^ 2
So either (i) a - b = x - y
(a = x and b = y)
or (ii) a - b = y - x
(a = y and b = x)
Lemma 2: if a + b = x + y
and 1/a + 1/b = 1/x + 1/y
then (a = x and b = y) or (a = y and b = x)
Proof: 1/a + 1/b = 1/x + 1/y given
(a + b) / ab = (x + y) / xy regrouping
1 / ab = 1 / xy a + b = x + y
ab = xy reciprocals
(a = x and b = y) or (a = y and b = x) Lemma 1
End: the problem is Lemma 2, with: x = 2
y = -c
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