SOLUTIONS
The gcd of X2+1 and X3+1.
Euclid's algorithm for polynomials gives
X3 + 1 = (X2 + 1)X +
(-X + 1)
(X2 + 1) = (-X + 1)(-X-1) + 2
Since (-X + 1) is divisible by 2, the gcd is 2. Note that the gcd of polynomials is determined up to a scalar multiple. So we might equally well say that the gcd is 1.
If we want to express the gcd as a combination of the starting polynomials
we have to keep track at every step how the remainder can be expressed in
such a way. So in this example we get:
(-X+1) = (X3+1) + (-X)(X2+1)
2 = (X2+1) + (-X+1)(X+1) =
(X2+1) + ((X3+1) +
(-X)(X2+1))(X + 1)
= (X+1)(X3+1) + (-X2-X+1)
(X2+1)
Since we claimed above that we could equally well say that 1 is the
gcd we should be able to express it as a combination of
the polynomials. Well here it is:
1 = (1/2)(X+1)(X3+1) +
(1/2)(-X2-X+1)(X2+1)
We find the gcd of
X2+1 and
X3+1 in Z/2Z[X] in the same
manner. Euclid's algorithm gives
X3+1 = (X2+1)X+X+1
X2+1 = (X+1)2
So the gcd is (X+1). We can express it as
X3+1+(X2+1)X = X+1
Finally, applying Euclid's algorithm to find the gcd of
X2-X+1
and X3+X+2
in Z/3Z[X] gives:
X3+X+2 = (X2-X+1)
+ X+1
X2-X+1 = (X+1)2
Hence the gcd equals X+1
These polynomials satisfy, a = qb + r. It is given that a and b have integer coefficients, moreover b is monic (e.g. the leading coefficient is 1). We will prove that both q and r have integer coefficients.
First we make the observation that the leading coefficient of q
is an integer.
Since the degree of r is strictly lower than that of b
we have:
(leading coefficient of a) =
(leading coefficient of q) · (leading coefficient of b).
However,
the leading coefficient of b is 1 and the leading coefficient
of a is an integer, hence
the leading coefficient of q must be an integer.
We will now prove that all coefficients of q are integers
by using induction on the degree of q.
If q is of degree 0, then it only has one coefficient,
which by the above remark must be an integer.
Now suppose that q is of degree n and that we have
proven the claim for the cases where the degree of q is less
than n. We know that the leading coefficient of q
is an integer so we can express q
as (cXn+q'), for some
integer c.
We can now
write:
(a-cXn· b) =
q'b + r
Since the degree of q' is less than n we already know
that it has integer coefficients thus the same holds for q.
Finally we have to prove that r has integer coefficients. The expression a-qb has only integer coefficients, and it is equal to r. This finishes the proof.
-
Let d1 be the gcd on the left-hand side
and d2 the gcd on the right-hand side.
Then d1 is a common divisor of gcd(a,b)
and c. But then d1 is a common divisor
of a, b, c and hence a divisor of
d2. So d1 divides d2.
Since d2 is a common divisor of a, b, c, it is certainly a common divisor of a and b and therefore a divisor of gcd(a,b). But then d2 is a common divisor of gcd(a,b) and c and we conclude that it is a divisor of d1. Hence d2 divides d1.
As d1 and d2 have leading coefficient 1, they must be equal.
-
Suppose a, b, c are relatively prime.
The previous item implies that gcd(a,b) and c
are relatively prime. By the extended Euclidean algorithm there exist polynomials
p1 and r such that
p1gcd(a,b) + rc = 1. Again by the extended Euclidean algorithm there exist polynomials p2 and p3 such that
gcd(a,b) = p2a + p3b. By substitution we then find
pa + qb + rc = 1, with p = p1p2 and q = p1p3.Conversely, suppose there exist polynomials p, q, r such that pa + qb + rc = 1. If d is a common divisor of a, b, c, then d is also a divisor of the combination pa + qb + rc. Since this combination equals 1, we conclude that gcd(a,b,c) = 1.
Consider the division of X2+1 by 2X+1 in Z/4Z[X].
Suppose that f(n) = p is a prime. We claim that p is a divisor of f(n+kp), for any integer k. Let f equal a0 + a1X + ··· + amXm. We look at the difference f(n)-f(n+kp); it is the sum of factors of the form aini - ai(n+kp)i. Applying Newton's binomium, we see that the factor aini drops out and that the rest is divisible by p.
The only way the numbers f(n+kp) can all be prime is when they are all equal to p. But in that case the polynomial would be constant, contradicting the assumption that the degree of f is at least 1.
A polynomial is irreducible if it has no non-trivial divisors. An important observation is that a polynomial that has a zero cannot be irreducible unless it has degree 1 (since it will have a linear factor). Note that the reverse is also true for polynomials of degree 2 and 3 but not for polynomials of higher degree.
- X3+8 in Q[X],
Reducible since X = -2 is a zero (and hence X+2 is a factor) - X3+X+1
in Z/2Z[X],
Irreducible. Neither X = 0 nor X = 1 is a zero. Since the degree is 3 it must have a linear factor if reducible. - X4+X+1
in Z/2Z[X],
Irreducible. It has no zero's, so it can not have linear factors. Hence were it reducible it must be the product of two polynomials of degree 2. Checking all 4 possibilities, it turns out that the only irreducible polynomial of degree 2 is X2+X+1 (mod 2). But (X2+X+1)(X2+X+1)= X4+X2+1. - X4+ 2X2+1 in C[X],
Reducible, X4+2X2+1 = (X2+1)2. - (X2+X+1)3 +
(X7-1)3
in Q[X],
Reducible, X = 0 is a zero. - (X2-1)
97+(X3-1)
78 in Z/5Z[X].
Reducible, both X = 0 and X = 1 are zero's.
Let p be a prime, we will prove the claim using induction on the degree of the polynomial. If the degree is 0, it is obvious. In that case both sides of the equations are exactly equal. Suppose the claim is valid for all f of degree less than n. We can write f = anXn + f', where f' is of degree less than n.
We have to prove that anXn + f'p = anpXnp + f'p. Then the rest follows by the induction hypotheses. We can expand the power on the left using Newton's binomium. The coefficients that we get equal p!/(i!(p-i)!) for i = 0, ..., p. Note that all these terms are divisible by p and hence they are zero mod p. Except when i is 0 or p. These correspond to the terms anpXnp and f'p. This completes the proof.
Write the polynomial p as
p(x) =
ai x2i +
bi x
2i+1.
Suppose we are working in the ring Q[X] the equation
p(x) = p(-x) gives:
2 bi
x2i+1 = 0,
for all x. From this it follows that all the bi
are zero. For suppose some are nonzero. Then a polynomial over Q
would have an infinite number of zeroes, a contradiction.
Now suppose we are working over Z/6Z, we then get the
same expression
2 bi x2i+1 = 0, which we rewrite as:
2x( bi
x2i) = 0.
This expression is always even hence we can suffice
to consider this expression modulo 3. If x is 0 modulo 3 then
we are done. So suppose it is not, in that case
x2i
equals 1. From this we get a condition on the coefficients of p, viz.
bi =
0 mod 3.
Vice versa it is straightforward to check that if this condition is satisfied
then the polynomial has the property that p(x) =
p(-x)
for any x.
If we are working over Z/2Z then always x = -x hence also p(x) = p(-x).
1. Let r be root of
a = a0+ ··· +an-1Xn-1+anXn.
Substitute r for x and rewrite
to get:
a0 =
-(a1r+ ··· +an-1rn-1+anrn). Since r divides the right hand side, it
must also divide the left hand side.
2. Now suppose that r/s is a root of a, with
r and s relatively prime. Of course s cannot be
zero. Again we substitute this in a, now multiply with
sn. We get:
0 =
a0sn+a1sn-1r+
··· +an-1s
rn-1+anrn.
Now all terms are divisible by s except for the term
anrn. Since r and
s are relatively prime, this proves that s must divide
an. On the other hand, all terms are
divisible by r except for the term
a0sn; this means that
s must divide a0.
3. If r/s is a rational root of 15-32X+3X2+2X3, then r must divide 15 and s must divide 2. Therefore there are only 32 cases to consider. Trying them all, we find that there are three rational roots: 3, 1/2, -5.
1. A polynomial in Z/3Z[X] of degree n has the form a0+ ··· +an-1Xn-1+anXn. The coefficient an can be 1 or 2, all the other coefficients can be 0, 1 or 2. In total there are 2 · 3n choices.
2. If a polynomial of degree 2 or 3 is reducible, it must have a factor of degree 1; hence it must have a root. So to check whether a given polynomial (of degree 2 or 3) is irreducible it suffices to check if 0, 1 or 2 is a zero.
We will only do the degree 2 case in detail. First, note
that we may presume that a2 = 1 (if a is an
irreducible with a2 = 2 then -a is also
irreducible), Next we may also presume that a0
is not 0, (since in that case we will have 0 as a root). The list
of remaining polynomials is:
| X2+0·X+1, irreducible | X2+0·X+2, x=1 is a zero |
| X2+1·X+1, x=1 is a zero | X2+1·X+2, irreducible |
| X2+2·X+1, x=2 is a zero | X2+2·X+2, irreducible |
The above three polynomials and minus these three polynomials together are all 6 irreducible polynomials of degree 2.
The same strategy (only with some more work) yields all irreducible polynomials of degree 3. The results of that analysis is:
| X3+2·X+1 | X3+2·X+2 |
| X3+X2+2 | X3+X2+X+2 |
| X3+X2+2·X+1 | X3+2·X2+1 |
| X3+2·X2+X+1 | X3+2·X2+2·X+2 |
Expanding the left-hand side of the identity yields
X4 - 2X2 + 1 + 4X2 = X4 + 2X2 + 1,
Substituting p/q in the identity and multiplying through by q4 produces the identity of integers
(p2 - q2)2 + (2pq)2 = (p2 - q2)2.
Suppose f = a0 + a1X + ... + anXn, where all the ai are rational numbers. Expanding the product yields that -3a0, -3a1 + 4a0, -3a2 + 4a1, ..., 4an should all be integers. From these conditions but the last we conclude that a0 = b0/3 for some integer b0, a1 = b1/9, for some integer b1, ..., an = bn/3n + 1 for some integer bn. The last condition (4an is an integer) implies that an = c/4 for some integer c. We find that
bn/3n + 1 = c/4 or 4bn = 3n + 1c.
Proceeding in the same way, you can now prove that an - 1 
Z, etc. A formal way of doing
this is to use induction with respect to the degree of f.
The starting point of the induction is then to show that the statement in the exercise
is true for polynomials which are constants. This is easy and in fact
the same argument as just given for the top coefficient works here. The inductive hypothesis is that the statement in the exercise is true
for polynomials of degree at most n - 1. If f has degree
n, then the preceding argument shows that the top coefficient
an is an integer. Now apply the inductive
hypothesis to the polynomial f - anXn.
- Of course, since 5 is such a small number, one can easily check case by case. One finds the zeros 1 and 2. More elegantly, split off a square (X + 1)2 + 1 and note that -1 = 22 mod 5, so that (X + 1)2 - 22 factors as (X + 3)(X - 1). Now, a product of numbers is 0 modulo a prime if and only if at least one of its factors is 0 modulo the prime. So we find -3 mod 5 and 1 mod 5 as the zeros.
- Rewrite as X(X + 1) + 1. If you substitute an integer in X(X + 1), the result is always even. Calculating modulo 24, the result is always even modulo 24 and can never equal -1 modulo 24. Therefore there are no solutions to this equation.
- Whenever you substitute an integer you get the product of three consecutive integers. Being 0 mod 12 means that the product is divisible by 12 = 4 · 3. Among three consecutive integers there is always one that is divisible by 3. If the first and third integer are even, then the product is divisible by 4 · 3. If only the second integer is even, then the product is divisible by 12 if this even number is divisible by 4. Otherwise the product is not divisible by 12. In the range 0,1,..., 11 this happens precisely for the products 1 · 2 · 3, 5 · 6 · 7 and 9 · 10 · 11. So the `nonzeros' are 1, 5, 9 mod 12. The zeros are 0, 2, 3, 4, 6, 7, 8, 10, 11 mod 12.
- Since 2 and 33 are relatively prime, 2 is invertible in Z/33Z. Using 13 = -20 mod 33 and 9 = 42 mod 33, we can rewrite the equation 2X2 + 13X + 9 = 0 as X2 - 10X + 21 = 0. Next we split off a square: (X - 5)2 - 4 = 0. So we must solve (X - 5)2 - 22 = 0 or, equivalently, (X - 5 - 2)(X - 5 + 2) = 0, i.e., (X - 7)(X - 3) = 0. Two zeros are immediate: 3 and 7. But more zeros could arise since the product on the left-hand side is also zero modulo 33 if one of the factors is zero modulo 3 and the other zero modulo 11.
By the Euclidean algorithm, there exist polynomials r(X) and s(X) such that
r(X)f(X) + s(X)g(X) = d(X).
r(X + a)f(X + a) + s(X + a)g(X + a) = d(X + a).
The extended Euclidean algorithm yields the relation
X2 + X + 1 + (-X - 2)(X - 1) = 3.
Suppose p1, p2, ..., pn are distinct irreducible polynomials in R[X] and consider the polynomial
Note that this argument also shows that there are infinitely many irreducible polynomials with leading coefficient 1.
Rewrite (X2 + 1)p + (X + 2)q = pq as (X2 + 1)p = q(p - X - 2). Since p divides the left-hand side it divides the right-hand side. As p is irreducible, it divides q or p - X - 2; in the latter case it divides X + 2.
-
Case 1: p divides q. Then p = aq for
some rational number a. Substituting in the equation we find
q = X2 + 1 + (1/a)(X + 2). It follows that a = +1 or -1 because q has integer coefficients. This leads to two candidate solutions:- For a = 1 we find p = q = X2 + X + 3.
- For a = -1 we find p = -X2 + X + 1 and q = X2 - X - 1.
- Case 2: p divides X + 2. Then p = b(X + 2) for some integer b. Substituting in the equation we find q = b/(b - 1) · (X2 + 1). But this polynomial has integer coefficients only if b = 2. In conclusion, we find p = 2X + 4 and q = 2X2 + 2. Since both are irreducible and since they satisfy the equation we have found another solution.