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

[anemone]

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 $p^{2014}+q^{2014}+r^{2014}$.

 

[Opalg]

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]

 

[anemone]

[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.

Spoiler

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 $\therefore p^{2014}+q^{2014}+r^{2014}=3$