If #alpha, beta# are the roots of #x^2+px+q=0# and also of #x^(2n) + p^nx^n + q^n# = 0 and if #alpha/beta, beta/alpha# are the roots of #x^n + 1 + (x+1)^n = 0 ,# then prove that n must be an even integer?

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Avery Alexander · Accepted Answer

See below

From #x^n + 1 + (x+1)^n = 0# we have #2x^n+(n-1)x^{n-1}+cdots+2=0# and also #x^n+(n-1)/2x^{n-1}+cdots+1=0#. This polynomial obeys #(alpha/beta)^{n_1}(beta/alpha)^{n_2} = 1# with #n_1# and #n_2# integers such that #n_1+n_2=n# Calling now #y = (alpha/beta)# we have #y^{n_1-n_2} = y^{2n_1-n} = 1->2n_1-n=0->n=2r_1# so #n# is an even integer. Note. #alpha ne beta# because #x=pm1# is not root for #x^{2n} + 1 + (x+1)^{2n} = 0#

Kennedy Adams · Answer

undefined To prove that $ n $ must be an even integer, we'll use the given information and apply Vieta's formulas.

Given:
1. $ \alpha, \beta $ are the roots of $ x^2 + px + q = 0 $, thus their product is $ q $ and their sum is $ -p $.
2. $ \alpha, \beta $ are also the roots of $ x^{2n} + p^n x^n + q^n = 0 $, implying $ \alpha^n $ and $ \beta^n $ are also roots.
3. $ \frac{\alpha}{\beta} $ and $ \frac{\beta}{\alpha} $ are the roots of $ x^n + 1 + (x+1)^n = 0 $.

Firstly, we find the sum and product of the roots of the equation $ x^n + 1 + (x+1)^n = 0 $.
Let $ \frac{\alpha}{\beta} = A $ and $ \frac{\beta}{\alpha} = B $.

Using Vieta's formulas:
1. Sum of roots, $ A + B = -1 $.
2. Product of roots, $ AB = \left(\frac{\alpha}{\beta}\right) \left(\frac{\beta}{\alpha}\right) = 1 $.

For $ x^2 + px + q = 0 $, the sum of roots is $ \alpha + \beta = -p $ and the product of roots is $ \alpha \beta = q $.

Now, $ \alpha^n $ and $ \beta^n $ are also roots of $ x^{2n} + p^n x^n + q^n = 0 $.
Using Vieta's formulas again:
1. Sum of roots, $ \alpha^n + \beta^n = -(p^n) $.
2. Product of roots, $ (\alpha^n)(\beta^n) = (q^n) $.

From the given conditions, we have:
$ A + B = -1 $ and $ AB = 1 $.

From Vieta's formulas for $ \alpha^n $ and $ \beta^n $:
$ \alpha^n + \beta^n = -(p^n) $ and $ (\alpha^n)(\beta^n) = (q^n) $.

Substituting $ A + B = -1 $ and $ AB = 1 $ into $ \alpha^n + \beta^n = -(p^n) $ and $ (\alpha^n)(\beta^n) = (q^n) $, we get:

$ \alpha^n + \beta^n = -(p^n) = -(A + B)^n $
$ (\alpha^n)(\beta^n) = (q^n) = (AB)^n $

Using binomial expansion, we have:
$ -(p^n) = -(A^n + nA^{n-1}B + ... + nB^{n-1}A + B^n) $
$ (q^n) = (A^nB^n) $

From the equation $ (q^n) = (A^nB^n) $, we can deduce that $ n $ must be even because an odd power cannot result in a positive number when squared. Therefore, $ n $ must be an even integer.