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
(1) Yes, e.g.
(2) No, see explanation.
Explanation:
(1) Let
Then:
a+b = 1 is rational
When
a^n+b^n = pi^n+(1-pi)^n
is a non-trivial polynomial in
(2) Suppose
If
If
(a^3+b^3)/(a^2+b^2) = a+b" " is rational
Otherwise suppose
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
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
(1) Let
Explanation:
Here's an alternative answer for (1) using simple irrational numbers:
Let
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 forn >= 0 -
The sequences begin:
1, 1, 3, 7, 25, 57, 139,...
0, -1, -2, -9, -16, -41, -98,...
-
p_n > 0 for alln >= 0 and is always odd. -
q_n < 0 for alln >= 1 and alternates between odd and even values.
We can also define sequences for the
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 forn >= 0 -
These sequences begin:
1, 0, 2, 0, 4, 0, 8,...
0, 1, 0, 2, 0, 4, 0,...
Note that when
When
When
Hence in all cases, when
sqrt(2)^n + (1-sqrt(2))^n = (p_n+r_n) + (q_n+s_n)sqrt(2)
with