| Set X = x+1
Y = Y+1
A = a-4
B = b-5
And we get the equivalent problem, to find A & B such that
X^3 + XY^2 + Y^3 + AX + BY + (4-A-B) = 0.
Since the expression is a cubic in X, if it factors then it has a
linear factor, so write it as:
(X+k)(X^2+lX+m)
where k,l&m lie in C(Y).
Comparing the co-efficients, it's obvious that l = -k & m = k�+Y�+A.
We are left with the unit term (i.e. X is gone):
k(k^2+Y^2+A) = Y^3+BY+(4-A-B).
Regard k as a poly in Y, and compare the coefficients again. k = pY,
where p^3+p-1=0 (three complex solutions).
We are left with:
pA = B
0 = 4-A-B
whence the solution.
|
| Solution by the proposer.
If the cubic is to factor in C[x,y], then one of the factors must be
linear. Without loss of generality, we may assume this factor is of
the form x-py-q, where p and q are complex numbers.
Substituting x=py+q in the original cubic, we get a polynomial in y
that must be identically 0. Thus each of its coefficients must be 0.
This gives us the four equations:
1 + p + p^3 = 0
4 + 2p + 3p^2 + q + 3 p^2 q = 0
3 + aq + 3q^2 + q^3 = 0
b + ap + 2q + 6pq + 3 p q^2 = 0
Solving these equations simultaneously, yields a=4 and b=5.
As a check we note that the resulting polynomial can be written as
(x+y+2)(y+1)^2 + (x+1)^3. The reducibility of this polynomial will not
change if we let x=X-1 and y=Y-1. This produces the polynomial X^3 +
XY^2 + Y^3. Letting z=X/Y shows that this polynomial factors over
C[x,y] since z^3+z+1 factors of C[z].
Rabinowitz also remarks that,
"except for a few exceptional cases, if f(x,y) is a cubic polynomial in
C[x,y], there will be unique complex constants a and b such that
f(x,y)+ax+by factors over C[x,y]."
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