If $\alpha$ and $\beta$ are the roots of the equation $x^2 px + q =0$ and $x^{2n} + p^n x^n + q^n =0$ and if $\left(\frac{\alpha}{\beta},\frac{\beta}{\alpha}\right)$ are the roots of $x^n + 1 + (x + 1)^n = 0$, then $n$ must be

a. Odd Integer c. Rational but not Integer.

b. Even Integer d. None of these.

Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2 px + q =0$ and $x^{2n} + p^n x^n + q^n =0$ and if $\left(\frac{\alpha}{\beta},\frac{\beta}{\alpha}\right)$ are the roots of $x^n + 1 + (x + 1)^n = 0$, then $n$ must be

a. Odd Integer               c. Rational but not Integer.

b. Even Integer             d. None of these.

anonymous · Answer

$x^2 + px +q = 0$ not $x^2 px + q = 0$

jamesj · Answer

That last equation is right?

anonymous · Answer

yeah

jamesj · Answer

Well, n can't be even then.  Now ...

anonymous · Answer

Wow, but How?

jamesj · Answer

From that last equation.  If n is even, every quantity (..stuff...)^n is non-negative and we have +1 as well, all of it = 0.  That's a contradiction.

jamesj · Answer

[  Formally.  If n = 2k, for some integer k, then
$$ 0 = x^{2k} + 1 + (x+1)^{2k} = (x^2)^k + 1 + ((x+1)^2)^k \geq 0 + 1 + 0 = 1 $$
contradiction.]

Now, from the first equation
$$ \alpha + \beta = -p  --- (1)$$
$$ \alpha . \beta = q ---(2) $$

From the second ... this is bit more subtle, think of it as a quadratic in x^n...
$$ \alpha^n + \beta^n = -p^n  --- (3) $$
$$ \alpha^n . \beta^n = q^n  --- (4) $$

(1) => (4), so that's nothing new.  But clearly (1) and (3) are different.

Looking now at (1) and taking the nth power of both sides and using n as an odd integer,
$$ (\alpha + \beta)^n = (-p)^n = -p^n  --- (5) $$
From the 3rd equation of the problem

$$ -[ (\alpha/\beta)^n + 1] = (\alpha/\beta+1)^n --- (6) $$
In equation (5), divide both sides by beta^n,
$$ (\alpha/\beta+1)^n = -p^n/\beta^n $$
By equation (6),
$$ -p^n/\beta^n =  -[ (\alpha/\beta)^n + 1] \implies p^n = (\alpha + \beta)^n$$
consistent with equation (1).

Therefore, in short, the system of equations are consistent for n being an odd integer.

jamesj · Answer

**correction, in that last deduction it should be 
$$ ... \implies p^n = \alpha^n + \beta^n $$
consistent with equation (3).

jamesj · Answer

Where's this question from?   It looks like it could be from the Math GRE.

anonymous · Answer

Thanks JamesJ


I don't know, my teacher gave it to me.

jamesj · Answer

I'd like to point out to the purists that while I have shown option A isn't an option, I haven't shown C and D cannot be options.  But I have shown B is a possibility and I'm now leaning on the fact that only one option is correct. 

I suspect it wouldn't be too hard to show that C and D lead to a contradiction; I can see how it would pivot around the one small result I used in equation (5): 
the treatment of (-p)^n.

**One other correction, just after equation 4:
"(2)  [*not (1)] => (4), so that's nothing new."