MHB Evaluate p^{2014}+q^{2014}+r^{2014}

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The discussion revolves around evaluating the expression p^{2014}+q^{2014}+r^{2014} given the polynomials P(x) and Q(x) with specific properties. It is established that the roots of Q(x) are the squares of the roots of P(x), leading to the conclusion that Q(x^2) equals the product of P(x) and P(-x). By comparing coefficients, the relationships between p, q, and r are derived, resulting in p = -1, q = 1, and r = -1. Ultimately, the value of p^{2014}+q^{2014}+r^{2014} is calculated to be 3, confirming that the condition P(1) = 0 is not necessary for solving the problem.
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Let $P(x)=x^3+px^2+qx+r$ and $Q(x)=x^3+qx^2+rx+p$, where $p,\,q,\,r$ are integers with $r\ne 0$. Suppose $P(1)=0$ and the roots of $Q(x)$ are squares of the roots of $P(x)$. Find the value of $p^{2014}+q^{2014}+r^{2014}$.
 
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anemone said:
Let $P(x)=x^3+px^2+qx+r$ and $Q(x)=x^3+qx^2+rx+p$, where $p,\,q,\,r$ are integers with $r\ne 0$. Suppose $P(1)=0$ and the roots of $Q(x)$ are squares of the roots of $P(x)$. Find the value of $p^{2014}+q^{2014}+r^{2014}$.
[sp]If the roots of $Q(x)$ are the squares of the roots of $P(x)$, then the roots of $Q(x^2)$ are the roots of $P(x)$ together with their negatives. The (monic) polynomial whose roots are the negatives of those of $P(x)$ is $P(-x)$. Therefore $Q(x^2) = P(x)P(-x)$. Thus $$x^6 + qx^4 + rx^2 + p = (x^3+px^2+qx+r) (x^3-px^2+qx-r) = x^6 + (2q-p^2)x^4 + (q^2-2pr)x^2 - r^2.$$ Compare coefficients of powers of $x$ to see that $$q=p^2, \qquad r = q^2-2pr, \qquad p=-r^2.$$ Hence $p=-r^2$, $q = r^4$ and $r^8+2r^3 - r = 0.$ But $r\ne0$, so $r^7 + 2r^2 - 1 = 0.$ The only integer solution of that is $r=-1$, so that $p=-r^2 = -1$ and $q = p^2 = 1.$ Finally, $p^{2014}+q^{2014}+r^{2014} = 1+1+1 = 3.$

[Note: It follows that $P(1) = 1+p+q+r = 1 - 1 + 1 - 1 = 0$. So the condition $P(1) = 0$ is automatically satisfied, and it seems that this condition was not needed in the statement of the problem.][/sp]
 
Opalg said:
[sp]If the roots of $Q(x)$ are the squares of the roots of $P(x)$, then the roots of $Q(x^2)$ are the roots of $P(x)$ together with their negatives. The (monic) polynomial whose roots are the negatives of those of $P(x)$ is $P(-x)$. Therefore $Q(x^2) = P(x)P(-x)$. Thus $$x^6 + qx^4 + rx^2 + p = (x^3+px^2+qx+r) (x^3-px^2+qx-r) = x^6 + (2q-p^2)x^4 + (q^2-2pr)x^2 - r^2.$$ Compare coefficients of powers of $x$ to see that $$q=p^2, \qquad r = q^2-2pr, \qquad p=-r^2.$$ Hence $p=-r^2$, $q = r^4$ and $r^8+2r^3 - r = 0.$ But $r\ne0$, so $r^7 + 2r^2 - 1 = 0.$ The only integer solution of that is $r=-1$, so that $p=-r^2 = -1$ and $q = p^2 = 1.$ Finally, $p^{2014}+q^{2014}+r^{2014} = 1+1+1 = 3.$

[Note: It follows that $P(1) = 1+p+q+r = 1 - 1 + 1 - 1 = 0$. So the condition $P(1) = 0$ is automatically satisfied, and it seems that this condition was not needed in the statement of the problem.][/sp]

Well done Opalg! Your approach made it very obvious that we don't need the condition where $P(1)=0$ to solve for the problem, bravo! But, I think if we solved it using another route, then that condition is kind of needed.:o

Note that $P(1)=Q(1)=0$, so 1 is a root of both $P(x)$ and $Q(x)$. Let $m$ and $n$ be the other two roots of $P(x)$, so $m^2$ and $n^2$ are the other two roots of $Q(x)$.

We then get $mn=-r$ and $m^2n^2=-p$ so $p=-r^2$. Also, $(-p)^2=(m+n+1)^2=m^2+n^2+1+2(mn+m+n)=-q+2q=q$.

Therefore $q=r^4$. Since $P(1)=0$, we therefore get $1+r-r^2+r^4=0$. Factorizing, we get $(r+1)(r^3-r^2+1)=0$. Note that $r^3-r^2+1=0$ has no integer root and hence $r=-1$, $q=1$, $p=1$ and $\therefore p^{2014}+q^{2014}+r^{2014}=3$
 
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