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.