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Fearless Symmetry by Robert Gross, Avner Ash

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62 CHAPTER 6
In other words, p(x) = x
3
+ x 2 = (x 1)(x
2
+ x + 2). Now for
any integer a, p(a) = 0 if and only if (a 1)(a
2
+ a + 2) = 0, because
p(a) = (a 1)(a
2
+ a + 2). A product of two integers is 0 only if one
or both of them is 0. (This is true in any number system that we
will use in this book.) So if we try to solve (a 1)(a
2
+ a + 2) = 0,
we see that either a = 1 (which we already knew was a possibility)
or else a
2
+ a + 2 = 0.
The quadratic formula
6
will tell us what all the solutions of
the quadratic equation are: a =
1±
18
2
. But the negative number
7 has no real square root, let alone an integer square root.
So this option leads us nowhere in finding more integer solutions
to p(x) = 0.
But we have been exhaustive: If there were any other solutions,
we would have found them because p(x) = 0 if and only if x 1 = 0
or x
2
+ x + 2 = 0. In conclusion, we can confidently say that S(Z) =
{1}. In fact, for any number system A,
7
this same method will find
S(A) for us, and you can see that S(A) will have at least one element
(since every number system has to have the number 1 in it) and may
possibly have two or three elements, but no more.
EXERCISE: Find a number system A for which S(A)has
exactly two elements.
SOLUTION: Use trial-and-error on various F
p
’s. One answer
is F
2
: S(F
2
) ={0, 1}.AnotherisF
7
: S(F
7
) ={1, 3}.
Are There General Methods for Finding Solutions to
Systems of Polynomial Equations?
We were very lucky in our attempt to solve (6.2) because
1. we could guess one of the solutions;
6
The quadratic formula states that the solutions of the equation ax
2
+ bx + c = 0are
x =
b±
b
2
4ac
2a
.
7
As long as we are allowed to divide by 2 in A; otherwise, the quadratic formula does not
work.
EQUATIONS AND VARIETIES 63
2. we knew the quadratic formula;
3. the original equation was only degree 3 to start with
(i.e., the highest power of x appearing was x
3
).
Are there general methods to solve systems of polynomial
equations?
Answer: No, no, no! That difficulty is what makes this whole
subject so interesting. With some exaggeration, you can say that
this difficulty is the reason that whole chunks of mathematics such
as algebraic number theory and algebraic geometry have come into
being.
How do we know there is no general method of solving an
arbitrary system of polynomial equations? It all depends on what
is meant by “general. ”
First, we have to distinguish between exact solutions and approx-
imate solutions. Suppose that p(x) is a polynomial. Especially with
the advent of computers, there are ways of finding numbers a so
that p(a) is approximately 0. These wa ys may be general, meaning:
You give the computer the polynomial p(x) and you give it a positive
number, a “tolerance, as small as you like, say 0.001. Then the
computer will grind away and perhaps find numbers a so that p(a)
is between 0.001 and +0.001. There are two problems with this
from our more “exacting” point of view.
1. The computer may never report back, even if there is a
solution. Suppose we give it a gigantic p(x), of degree
100,000,000, with all kinds of large integer coefficients,
and suppose we make our tolerance very small, say
0.00000000000000001. Then perhaps the computer will
not find a solution within our lifetimes .
2. Even if the computer reports back, it is only giving
us an approximate solution.
8
That does not have to
bother us if we are building a bridge or doing our
taxes—approximations are fine if we ha ve set the
tolerances properly. But the game of number theory is
a game where we want to know about exact solutions.
8
We are assuming that the computer is using “floating point arithmetic.”

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