Let $ \varphi(x)$ be a cubic polynomial with integer coefficients. Given that $ \varphi(x)$ has have 3 distinct real roots $u,v,w $ and $u,v,w $ are not rational number. there are integers $ a, b,c$ such that $u=av^2+bv+c$. Prove that $b^2 -2b -4ac - 7$ is a square number .
Problem
Source: China Shanghai ,Mar 12, 2017
Tags: number theory, algebra, polynomial, China TST
rafayaashary1
12.03.2017 20:27
By the condition, $\varphi(x)$ is the minimal polynomial of its roots over $\mathbb Q$ (down to monicity of course). Let $\rho(x)=ax^2+bx+c$, and note that we have that $\text{deg}\left(\gcd(\varphi\circ \rho,\varphi)\right)>0$, since $v$ is a root of both. Specifically, we must have $\varphi\mid \varphi\circ\rho$; otherwise by Euclid's Algorithm there is a contradiction to the assumption that $\varphi$ is the minimal polynomial of its roots. So we know that $\rho$ is closed over the roots of $\varphi$. Actually, we can know something more precise. Consider the roots of $\rho\circ\rho-x$ and $\rho-x$. The minimality of $\varphi$ means that its roots are either all present or all absent from another polynomial in $\mathbb Q[x]$. Specifically, no digraph of $\rho$ over $\{u,v,w\}$ can satisfy either of the above polynomials. Hence $\rho(v)=u$, $\rho(u)=w$, and $\rho(w)=v$. So it suffices to show that if for integer-coefficient quadratic $\rho(x)=ax^2+bx+c$ the polynomial $\rho\circ\rho\circ \rho-x$ has an irreducible cubic factor, then $b^2-2b-4ac-7$ is a perfect square. And here I'm stuck :/
sqing
13.03.2017 10:20
rafayaashary1 wrote:
By the condition, $\varphi(x)$ is the minimal polynomial of its roots over $\mathbb Q$ (down to monicity of course). Let $\rho(x)=ax^2+bx+c$, and note that we have that $\text{deg}\left(\gcd(\varphi\circ \rho,\varphi)\right)>0$, since $v$ is a root of both. Specifically, we must have $\varphi\mid \varphi\circ\rho$; otherwise by Euclid's Algorithm there is a contradiction to the assumption that $\varphi$ is the minimal polynomial of its roots. So we know that $\rho$ is closed over the roots of $\varphi$. Actually, we can know something more precise. Consider the roots of $\rho\circ\rho-x$ and $\rho-x$. The minimality of $\varphi$ means that its roots are either all present or all absent from another polynomial in $\mathbb Q[x]$. Specifically, no digraph of $\rho$ over $\{u,v,w\}$ can satisfy either of the above polynomials. Hence $\rho(v)=u$, $\rho(u)=w$, and $\rho(w)=v$. So it suffices to show that if for integer-coefficient quadratic $\rho(x)=ax^2+bx+c$ the polynomial $\rho\circ\rho\circ \rho-x$ has an irreducible cubic factor, then $b^2-2b-4ac-7$ is a perfect square. And here I'm stuck :/
Thanks.
angiland
15.03.2017 23:58
As pointed out above, the polynomial $\varphi$ is irreducible and is the minimal polynomial for $u, v, w$. Since $\varphi(ax^2+bx+c)$ has $v$ as its root, $u, w$ must be the roots as well. Hence, $aw^2 + bw + c \in \{u, v, w\}$. We can rule out $aw^2 + bw + c = w$ because $w$ has degree $3$. Moreover if $aw^2 + bw + c = u = av^2 + bv + c$, then $(v-w) (a(v+w) + b) = 0$, which implies $v + w$ is rational. But then $u$ would be rational, contradicting the assumption. Thus we are left with $v = aw^2 + bw + c$ and similarly $w = au^2 + bu + c$. Next, to simply the problem, we may add a common rational number to $u, v, w$ and assume wlog that $u + v + w = 0$. We may also consider $au, av, aw$ in place to $u, v, w$ to assume that $a = 1$. Finally, it will turn out to be helpful to work with $2c$ instead of $c$. With these transformations we have three equations: $$ u = v^2 + bv + 2c $$$$ v = w^2 + bw + 2c$$$$ w = u^2 + bu + 2c$$Summing them and using $u + v+ w = 0$ gives $u^2 + v^2 + w^2 = -6c$, and so $$uv + vw + wu = 3c.$$But using the RHS of these equations, we can also obtain $$uv + vw + wu = \sum u^2v^2 ~+~ b \sum u^2v ~+~ 4c \sum u^2 ~+~ b^2 \sum uv ~+~ 4bc \sum u ~+~ 12 c^2.$$We have $\sum u^2v^2 = (\sum uv)^2 - 2uvw(u+v+w) = 9c^2$ and $\sum u^2v = (u+v+w)(uv+vw+wu) - 3uvw = -3uvw$. Simplifying, the above equation gives us the key expressions $$uvw = \frac{b^2c - c^2 - c}{b}.$$ Next, consider the difference between the first two equations: $u - v = (v-w)(v+w+b) = (v-w)(b-u)$. Multiplying such relations yields $$(b-u)(b-v)(b-w) = 1.$$Plugging in the expressions derived earlier, we conclude that $$c^2 + (1+2b^2)c + (b^4-b) = 0.$$Simple factorization gives $c = - (b^2 + b + 1)$ or $c = - (b^2 - b)$. In the former case, $b^2 - 2b - 8ac - 7 = 9b^2 + 6b + 1 = (3b+1)^2$ (note that we have $8ac$ here because we are working with $2c$). The latter case gives $uv + vw + wu = -3(b^2-b)$ and $uvw = -2b^3 + 3b^2 - 1$. Thus $\varphi(x) = x^3 - 3(b^2-b)x + (2b^3-3b^2+1)$, which admits a rational root $x = b - 1$ and leads to a contradiction. The proof is completed.
somepersonoverhere
16.03.2017 02:48
rafayaashary1 wrote:
By the condition, $\varphi(x)$ is the minimal polynomial of its roots over $\mathbb Q$ (down to monicity of course). Let $\rho(x)=ax^2+bx+c$, and note that we have that $\text{deg}\left(\gcd(\varphi\circ \rho,\varphi)\right)>0$, since $v$ is a root of both. Specifically, we must have $\varphi\mid \varphi\circ\rho$; otherwise by Euclid's Algorithm there is a contradiction to the assumption that $\varphi$ is the minimal polynomial of its roots. So we know that $\rho$ is closed over the roots of $\varphi$. Actually, we can know something more precise. Consider the roots of $\rho\circ\rho-x$ and $\rho-x$. The minimality of $\varphi$ means that its roots are either all present or all absent from another polynomial in $\mathbb Q[x]$. Specifically, no digraph of $\rho$ over $\{u,v,w\}$ can satisfy either of the above polynomials. Hence $\rho(v)=u$, $\rho(u)=w$, and $\rho(w)=v$. So it suffices to show that if for integer-coefficient quadratic $\rho(x)=ax^2+bx+c$ the polynomial $\rho\circ\rho\circ \rho-x$ has an irreducible cubic factor, then $b^2-2b-4ac-7$ is a perfect square. And here I'm stuck :/
How is that a bad habit? I would think it's a good habit to write our your progress.
MathPanda1
16.03.2017 05:14
Dear angiland, I was just wondering, what is the motivation for your extraordinary proof? Thank you so much for all your help!
adityaguharoy
16.03.2017 06:19
MathPanda1 wrote: Dear angiland, I was just wondering, what is the motivation for your extraordinary proof? Thank you so much for all your help! His proofs are always wonderful and extraordinary.
angiland
16.03.2017 06:59
Thanks. My original idea was to use the fact that $\varphi(x)$ divides $\varphi(ax^2 + bx + c)$ and compare coefficients. The computation got pretty ugly, so I gave up and took a different route. The normalizations played an important role in the solution above, but one need not rely on $(b-u)(b-v)(b-w) = 1$. Instead, it is perhaps more natural to use $uvw = (au^2 + bu + c)(av^2 + bv + c)(aw^2 + bw + c)$, which also gives the final result.
R8450932
26.07.2018 15:17
angiland wrote: Since $\varphi(ax^2+bx+c)$ has $v$ as its root, $u, w$ must be the roots as well. Can anyone explain pls why this holds? I am really bad in polynomials ,so I hope I will get an elementary answer... Thanks!
R8450932
26.07.2018 15:36
rafayaashary1 wrote: Hence $\rho(v)=u$, $\rho(u)=w$, and $\rho(w)=v$. Any elementary explanation for this part? I really can't understand what rafayasharry wrote Pls help me to understand it ...
R8450932
29.07.2018 20:52
Any help?...
pavel kozlov
07.03.2019 23:51
Please read solution of angiland
cosinechicken
29.11.2019 00:49
R8450932 wrote: angiland wrote: Since $\varphi(ax^2+bx+c)$ has $v$ as its root, $u, w$ must be the roots as well. Can anyone explain pls why this holds? I am really bad in polynomials ,so I hope I will get an elementary answer... Thanks! Oops same question
CANBANKAN
28.06.2020 21:18
@cosinechicken we can show that the minimal polynomial is irreducible by taking gcds. Claim: If $P(x)$ is irreducible and $P(r)=0$, then $P(x)$ is the minimal polynomial of algebraic integer $r$. Proof. Assume not. Let $Q(r)$ be the minimal polynomial of $P$. Looking at $gcd(P,Q)$ we get it's either $P$ or $1$. It's clearly not $1$ because they share a root. Hence $P|Q$. Since $Q$ is the minimal polynomial, $P=Q$. If $P$ is minimal, it must be irreducible; if $P$ is reducible, then $P=QR$ and either $Q(r)=0$ or $R(r)=0$, contradicting $P$'s minimality. I'm confused why the $u+v+w=0$ transformation is valid. I tried to work out the bash with trying to show that $A(\frac{2u-v-w}{3})=\frac{2v-u-w}{3}$ where $A(x)=ax^2+bx+c$, but got this is equivalent to $4v^2+4w^2-4uv-4uw-2vw-2u^2+9c=0$, which seems hard to prove. If the transformation is not valid then angiland only solved the problem for $u+v+w=0$, in other words, Angiland fakesolved the problem.
bumjoooon
14.02.2021 11:24
Comments for angiland's(#4) assumption. When $(u,v,w)$ are the roots of cubic polynomial in $\mathbb{Z}[x]$, $(pu+q, pv+q, pw+q)$ are also the roots of cubic polynomial. The precise cubic would be $MP \left( \frac{x-q}{p} \right)$. ($q,p$ is any rational number and $M$ is just a natural number to guarantee that the coefficients are integer.) Precedure 1. Making $a=1$. There exists cubic with $(au, bu, cu)$ as root. Then $(au)=1\cdot(av)^2+b\cdot(av)+ac$. Note that this does not change the value $b^2-2b-4ac-7$. Precedure 2. Making $u+v+w=0$. There exists rational $q$ such that $(u+q)+(v+q)+(w+q)=0$. However, then $(u+q)=1\cdot(v+q)^2+(b-2q)\cdot(v+q)+(c-bq+q+q^2)$. So coefficient can be non-interger. However, this do not change $b^2-2b-4ac-7$ because \begin{align*} &(b-2q)^2-2(b-2q)-4(c-bq+q+q^2)-7\\ &=(b^2-4bq+4q^2)+(-2b+4q)+(-4c+4bq-4q-4q^2)\\ &=b^2-2b-4ac-7 \end{align*} So we can assume that $u+v+w=0$ with rational number $B,C$ such that $u=v^2+Bv+2C$. Here $B^2-2B-8C-7$ equals to original $b^2-2b-4ac-7$. Therefore, if we can prove that this is the square of some rational number, since $b^2-2b-4ac-7$ is integer, it is actually a square number. Note that we do not need $B, C$ to be integer, the proof works anyway.
bumjoooon
14.02.2021 11:49
However, I think we need to fix some parts of #4 because $b=0$ and thus dividing by $b$ may not be valid. (This must be the same condition for $u+v+w=3$ for my formulation. So it can be resolved. There is also other way do reformoluate the equation. Analog to #15, we can consider $(2au+b, 2av+b, 2aw+b)$. Then $$(2av+b)^2 = 2\cdot(2au+b) + (b^2-2b-4ac)$$This give following reformulation. Let $ \varphi(x)$ be a cubic polynomial with integer coefficients. Given that $ \varphi(x)$ has have 3 distinct real roots $u,v,w $ and $u,v,w $ are not rational number. Assume there are integer $d$ such that $v^2=2u+d$. Prove that $d-7$ is a square number. Here is my proof.
Similarly, we get $u^2=2w+d$, $w^2=2v+d$. Subtract each equation, we get
\begin{align*}
&(u-v)(u+v)=2(w-u)\\
&(w-u)(w+u)=2(v-w)\\
&(v-w)(v+w)=2(u-v)
\end{align*}Therefore, $\sum_{sym} u^2 v + 2uvw = (u+v)(v+w)(w+u) = 8$. Here we use the fact that $u,v,w$ is distinct. Therefore,
$$(u+v+w)^3 = u^3+v^3+w^3 + 24$$Next,
\begin{align*}
u^3+v^3+w^3 &= u\cdot u^2 + v\cdot v^2 + w \cdot w^2\\
&= u(2v+d)+v(2w+d)+w(2u+d)\\
&=(u+v+w)^2-(u^2+v^2+w^2)+d(u+v+w)\\
&=(u+v+w)^2-(2v+d+2w+d+2u+d)+d(u+v+w)\\
&=(u+v+w)^2-2(u+v+w)+d(u+v+w-3)
\end{align*}Let $u+v+w = M$. Mixing two equations we get $$d = \frac{M^3-M^2+2M-24}{M-3}=M^2+2M+8.$$Therefore $d-7=(M+1)^2$ and thus square of rational number. Since $d-7$ is integer, it is a square number.
Then $u^3+v^3+w^3=3$ as above and $u^2+v^2+w^2=6+3d$. Also
$$uv+vw+wu=2v^3+dv+2w^3+dw+2u^3+du=6+3d.$$Therefore, $u^2+v^2+w^2=uv+vw+wu$ and $u=v=w$ which contradicts that $u,v,w$ are distinct.
EthanWYX2009
09.04.2024 03:38
I feel like it is just a huge amount of calculating, but idea is quite straight. Let $f(x)=ax^2+bx+c.$ because $(x-u)(x-v)(x-w)\in\mathbb Q[x],$ $\sum u,\sum uv,uvw\in\mathbb Q.$ Therefore $(x-f(u))(x-f(v))(x-f(w))\in\mathbb Q[x],$ as $u=f(v),$ we know $v=f(w),w=f(u).$ So now we can represent $a,b,c$ by $u,v,w$ by solving a linear equation. And it is just calculating.