> Dear Sirs,
 >
 > The December 2003 issue of your journal, Southwest Journal of Pure
 > and Applied Mathematics, publishes the article "Advanced Polynomial
 > Factorization" by James Harris. This note is to inform you that the
 > article's main result, the claim that the factorization of the
 > polynomial
 >
 > 65 x^3 - 12 x + 1 = (a_1 x  + 1)(a_2 x  + 1)(a_3 x + 1)
 >
 > one of the a's is coprime to 5, is in error.
 >
 > I'll expand on the proof of this error in my postscript.
 >
 > My sole intent in this note is one of information. I have no
 > expectations one way or the other regarding how you treat this
 > information, but you are certainly welcome to determine its
 > correctness.
 >
 > If you check the Usenet record (via Google, for instance), you'll
 > quickly get the picture that this has been a topic of no small amount
 > of heated discussion on sci.math over the past year, at least (and if
 > extended to ancillary topics, over the past several years. I would
 > not, for instance, recommend becoming entangled in this by now
 > fruitless discussion on sci.math
 >
 > Kindest regards,
 >
 > W. Dale Hall (wdh...@alum.mit.edu)
 >
 > PS.
 >
 > Several proofs of this error have appeared online in the Usenet
 > newsgroup sci.math.
 >
 > One such article was written by me in the following article:
 >
 >
 > http://groups.google.com/groups?selm=3F1C3F01.7010501%40farir.com&oe=-
 > UTF-8&output=gplain
 >
 > I will apologize in advance for whatever intemperate language appears
 > in various articles written by me and others in this and related
 > threads.
 >
 > The gist of the demonstration is the explicit factorization of the
 > a's in the following fashion:
 >
 > Let
 >
 > r(a) = 8 a^2 - 4 a - 45
 >
 > Note first that r(a) is (1) an algebraic integer for any algebraic
 > integer "a", (2) a divisor of both "a" and 5 for a = -(any of the
 > ai's in the above factorization of 65x^3 - 12x + 1)
 >
 > The relevant factorizations claimed in (2) follow:
 >
 > q(a) = 8 a^2 - 76 a - 185
 > r(a) = 8 a^2 -  4 a -  45
 > s(a) = 4 a^2 - 37 a - 104
 >
 > Whenever a is a root of x^3 - 12 x^2 + 65 (that is, a is the
 > *negative* of any of the ai's of the above factorization of
 > 65x^3 - 12x + 1), the following factorizations hold:
 >
 > q(a) r(a) = 5                (*)
 > r(a) s(a) = a.
 >
 > These factorizations can be established by elementary methods: for
 > instance, multiplying the above polynomials in the variable x, and
 > dividing the result by the polynomial p(x) =  x3 - 12 x2 + 65 will
 > yield the remainders given on the right sides of the above equations
 > (*).
 >
 > First, here are the products that I'm making claims about:
 >
 > q(x)*r(x) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8325
 > r(x)*s(x) = 32 x^4 - 312 x^3 -  864 x^2 + 2081 x + 4680
 >
 > Next, a couple of products of p(x) = x^3 - 12 x^2 + 65 with
 > polynomials of degree 1:
 >
 > (64 x + 128)*(x^3 - 12 x^2 + 65) =
 >                      64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8320
 >
 > (32 x + 72)*(x^3 - 12 x^2 + 65) =
 >                      32 x^4 - 312 x^3 -  864 x^2 + 2080 x + 4680
 >
 > Finally, we compare the results and see this:
 >
 > q(x)*r(x) = (64 x + 128)*p(x) + 5
 > r(x)*s(x) = (32 x +  72)*p(x) + x,
 >
 > Note that, for any value xo that makes p(xo) = 0, that same value xo
 > will make q(xo)r(xo) = 5, so r(xo) is a factor of 5.
 >
 > That value of xo also makes r(xo)*s(xo) = xo, so r(xo) is a factor of
 > xo.
 >
 > In short, r(xo) becomes a factor of *both* xo and 5.
 >
 > Since r(x) is a polynomial with integral coefficients, r(xo) is an
 > algebraic integer whenever xo is.
 >
 > In fact, the minimal polynomial of this number (r(a) for -a = any of
 > the above ai's) is given as:
 >
 > MinPoly(r) = x^3 - 969 x^2 + 315 x + 5
 >
 > The above facts prove that this "a" has a non-unit algebraic integer
 > as a factor, and thus cannot be a unit in the ring of algebraic
 > integers.
 >
 >