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Answers to Exercises; Chapter (^1 251)
5.1. Homogeneous: (a), (b), (e), (f); Symmetric: (c).
5.2. The “only if’ part is straightforward. Suppose the result holds. Then
for each x, y, z, the equation can be written in the form
tdfd(? Y,z> + xtkfk(x, % z) = id+, Y, z>
kfd
where fk is homogeneous of degree k. By Exercise 1.12, fd = f, fk = 0,
whence the result follows.
5.3. CXk.
5.4. Each homogeneous polynomial can be obtained by collecting up like
terms in the given polynomial.
5.5. Every symmetric polynomial in two variables can be expressed as a
sum of polynomials of one or other of the types cxkyk = cst or c(xayb +
xbya) = csgpg-,, whereu<bandpi=x’+yifori=1,2,....Showby
induction, using the expansion of (x + y)‘, that pi is a polynomial in sr and
S2.
5.7. The sum of the zeros is equal to -p. The zeros are in arithmetic
progression if and only if their average is equal to one of them, i.e. if and
only if -p/3 is a zero. This yields the condition 2p3 + 27r = 9pq.
5.8. ST - 3SlS2 + 353; SlS; - S2S3 - 2s:~; (sl - z)(sl - y)(sl - x) =
sf - s& + SlS2 - s3 = SlS2 - s3.
5.12. Consider the polynomial 21(x, y) = g(x, y)-g(y, x) = uo(y)+ai(y)x+
U2(Y)X2 + f.. + u,(y)x". For each value of y, this polynomial vanishes
identically in x, so that the polynomial coefficients ai vanish identically
in y. But this means that each ai is the zero polynomial, so that u itself is
the zero polynomial.
6.1. gcd(20119, 34782) = 341; the quotients are 1, 1, 2, 1, 2, 5.
6.2. (b) (iv) 341 = (34782)(11) - (20119)(19). The pencil and paper table
is
-2 -1 -2 -1 -1
1 -2 3 -8 11 -19
6.3. (a) ,If a is a multiple of p, then gcd(a,p) = p # 1. On the other hand,
ifg=gcd(a,p)#1,then,sinceg~p,g=p,sop~u.
(b) Use the general result enunciated in Exercise 2.
(c) Suppose p does not divide a. For some integers x, y, we have that
1 = ux +py, whence b = ubx + bpy. Since p divides both terms of the right
side, p must divide b.
(d) The result holds for n = 2. Assume it holds for each number less
than n. Let pl be the smallest positive divisor of n exceeding 1; then pl
must be prime (why?). Then either n = ~1, in which case this is the only