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±

√

1−8

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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