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# 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
#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
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