Homework 4 Solutions
Math 318, Spring 2016
Problem 1.
Part (a)
√
√
Proposition. The number 3 2 + 3 4 is irrational.
√
√
Proof. Let r = 3 2 + 3 4. Then it is easy to check that
√
√
3
3
r3 = 6 + 6 2 + 6 4 = 6 + 6r.
Thus r is a root of the polynomial x3 − 6x − 6. Since r is not an integer, it follows from the
rational root theorem that r is irrational.
Part (b)
Proposition. The number cos(π/9) is irrational.
Proof. Recall the cosine of triple angle formula
cos 3θ = 4 cos3 θ − 3 cos θ.
Substituting θ = π/9 gives
4 cos3 (π/9) − 3 cos(π/9) = cos(π/3) =
1
.
2
It follows that r = 2 cos(π/9) is a root of the polynomial x3 − 3x − 1. Since r is not an
integer, the rational root theorem implies that r is irrational, and hence cos(π/9) is irrational
as well.
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Problem 2.
Part (a)
Proposition. If a, b, and c are relatively prime positive integers, then there do not exist
positive integers x, y, and z such that
abx + acy + bcz = 2abc.
Proof. Suppose to the contrary that such an x, y, and z exist. Then
abx = 2abc − acy − bcz = c(2ab − ay − bz)
so c | abx. Since c is relatively prime to a and b, it follows that c | x, so x ≥ c. Similar
arguments show that y ≥ b and z ≥ a, so
abx + acy + bcz ≥ abc + abc + abc > 2abc,
a contradiction.
Part (b)
Proposition. If a, b, and c are relatively prime positive integers and d > 2abc, then there
exist positive integers x, y, and z such that
abx + acy + bcz = d.
Proof. Consider the equation
abx + cw = d.
Since ab and c are relatively prime, there exist integers x and w that satisfy this equation,
and indeed there exists a solution for which 1 ≤ x ≤ c. Then
cw = d − abx > 2abc − abc = abc
so w > ab. Since a and b are relatively prime and w > ab, there exist positive integers y
and z so that
ay + bz = w.
Then
abx + acy + bcz = abx + c(ay + bz) = abx + cw = d.
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