Do there exist real numbers a,b such that (1) a+b is rational and a^n+b^n is irrational for each natural n ge 2 (2) a+b is irrational and a^n+b^n is rational for each natural n ge 2?

2 Answers
Sep 11, 2017

(1) Yes, e.g. a=pi, b=1-pi

(2) No, see explanation.

Explanation:

(1) Let a=pi, b=1-pi

Then:

a+b = 1 is rational

When n >= 2

a^n+b^n = pi^n+(1-pi)^n

is a non-trivial polynomial in pi, hence irrational, since pi is transcendental.

color(white)()
(2) Suppose a^n+b^n is rational for all n >= 2

If a=b=0 then a+b is rational.

If a=0 and b!=0 or a!=0 and b=0 then:

(a^3+b^3)/(a^2+b^2) = a+b" " is rational

Otherwise suppose a, b != 0

Then:

(a^2+b^2)^2-(a^4+b^4) = 2a^2b^2

The left hand side is rational, so the right hand side must be.

Hence a^2b^2 is rational.

Then:

(a^2+b^2)(a^3+b^3)-(a^5+b^5) = a^2b^2(a+b)

The left hand side of this equation is rational and the multiplier a^2b^2 is also rational (and non-zero). So a+b is rational.

Sep 12, 2017

(1) Let a=sqrt(2) and b=1-sqrt(2)

Explanation:

Here's an alternative answer for (1) using simple irrational numbers:

Let a = sqrt(2), b=1-sqrt(2)

Then:

a+b = 1" " is rational

Note that:

(p+qsqrt(2))(1-sqrt(2)) = (p-2q)+(q-p)sqrt(2)

So consider the recursively defined sequences:

p_0 = 1

q_0 = 0

p_(n+1) = p_n-2q_n

q_(n+1) = q_n-p_n

Then:

  • p_n + q_nsqrt(2) = (1-sqrt(2))^n for n >= 0

  • The sequences begin:

1, 1, 3, 7, 25, 57, 139,...

0, -1, -2, -9, -16, -41, -98,...

  • p_n > 0 for all n >= 0 and is always odd.

  • q_n < 0 for all n >= 1 and alternates between odd and even values.

We can also define sequences for the a^n term, by:

r_0 = 1

s_0 = 0

r_(n+1) = 2s_n

s_(n+1) = r_n

Then:

  • r_n + s_nsqrt(2) = sqrt(2)^n for n >= 0

  • These sequences begin:

1, 0, 2, 0, 4, 0, 8,...

0, 1, 0, 2, 0, 4, 0,...

Note that when n >= 2, both r_n and s_n are even.

When n is even then s_n = 0, so q_n+s_n = q_n < 0 for n >= 2.

When n is odd then q_n is odd so q_n+s_n is odd and non-zero.

Hence in all cases, when n >= 2 we find:

sqrt(2)^n + (1-sqrt(2))^n = (p_n+r_n) + (q_n+s_n)sqrt(2)

with q_n+s_n != 0 and hence irrational.