IMO 1973 Problem 3
The polynomial
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 11m52s
Problem
Let $a$ and $b$ be real numbers for which the equation $$ x^4 + ax^3 + bx^2 + ax + 1 = 0 $$ has at least one real solution. For all such pairs $(a, b)$, find the minimum value of $a^2 + b^2$.
Exploration
The polynomial
$$P(x)=x^4+ax^3+bx^2+ax+1$$
is palindromic. Such equations often simplify after dividing by $x^2$ and introducing
$$t=x+\frac1x.$$
If $x\neq 0$, then
$$x^2+\frac1{x^2}=t^2-2.$$
Since the constant term is $1$, the equation never has $x=0$ as a root, so this substitution is always legitimate for any real root.
Carrying this out gives
$$x^2+\frac1{x^2}+a\left(x+\frac1x\right)+b=0,$$
hence
$$t^2+at+(b-2)=0.$$
A real root $x$ produces a real number $t=x+1/x$, and for real $x$ one has
$$t\in(-\infty,-2]\cup[2,\infty).$$
Thus the existence condition becomes: there exists some real $t$ with $|t|\ge2$ such that
$$t^2+at+b-2=0.$$
Solving for $b$ yields
$$b=2-t^2-at.$$
Then
$$a^2+b^2=a^2+(2-t^2-at)^2.$$
For fixed $t$, this is a quadratic function of $a$. Minimizing with respect to $a$ should be straightforward.
Compute:
$$f_t(a)=a^2+(2-t^2-at)^2.$$
Expanding,
$$f_t(a)=a^2+a^2t^2-2at(2-t^2)+(2-t^2)^2.$$
Differentiate with respect to $a$:
$$f_t'(a)=2a(1+t^2)-2t(2-t^2).$$
Hence the minimizing value is
$$a=\frac{t(2-t^2)}{1+t^2}.$$
Substituting back should produce a function only of $t^2$. Let
$$c=2-t^2.$$
Then
$$a=\frac{tc}{1+t^2}, \qquad b=c-at =c-\frac{t^2c}{1+t^2} =\frac{c}{1+t^2}.$$
Thus
$$a^2+b^2 =\frac{t^2c^2+c^2}{(1+t^2)^2} =\frac{c^2}{1+t^2} =\frac{(2-t^2)^2}{1+t^2}.$$
Set
$$u=t^2.$$
Since $|t|\ge2$, we have $u\ge4$. Then the objective becomes
$$g(u)=\frac{(u-2)^2}{u+1}.$$
Evaluate several cases:
$$g(4)=\frac45,\qquad g(5)=\frac32,\qquad g(6)=\frac{16}7.$$
The function appears increasing for $u\ge4$. Differentiate:
$$g'(u)=\frac{2(u-2)(u+1)-(u-2)^2}{(u+1)^2} =\frac{(u-2)(u+4)}{(u+1)^2}.$$
For $u\ge4$, this is positive. Hence the minimum occurs at $u=4$, namely $t=\pm2$.
Then
$$a=\frac{2(2-4)}5=-\frac45$$
when $t=2$, and
$$a=\frac{(-2)(2-4)}5=\frac45$$
when $t=-2$.
Also
$$b=\frac{2-4}5=-\frac25.$$
The minimum value is
$$\frac45.$$
The delicate point is proving that every feasible pair $(a,b)$ indeed corresponds to some $t$ with $|t|\ge2$, and conversely that any such $t$ gives a real root. Another subtle point is avoiding minimization in the wrong variable before incorporating the constraint $|t|\ge2$.
Problem Understanding
We are given a reciprocal quartic polynomial
$$x^4+ax^3+bx^2+ax+1,$$
where $a,b\in\mathbb R$, and we assume that this polynomial has at least one real zero. Among all such pairs $(a,b)$, we must determine the minimum possible value of
$$a^2+b^2.$$
This is a Type C problem. We must prove two statements. First, we must exhibit coefficients $(a,b)$ for which the polynomial has a real root and for which $a^2+b^2$ attains the claimed value. Second, we must prove that no smaller value is possible.
The polynomial is symmetric under the transformation $x\mapsto1/x$. A direct attack on the quartic equation in terms of $a$ and $b$ is unwieldy because the existence of a real root imposes a nonlinear condition. The central difficulty is converting the root condition into a manageable algebraic constraint on $(a,b)$.
The substitution
$$t=x+\frac1x$$
reduces the quartic condition to a quadratic relation between $a,b,t$. Since real $x$ implies $|t|\ge2$, the problem becomes a constrained optimization problem in one variable. The expected extremum should occur at the boundary value $|t|=2$, corresponding to the repeated reciprocal root $x=\pm1$.
The answer will be
$$\boxed{\frac45},$$
attained for
$$(a,b)=\left(-\frac45,-\frac25\right) \quad\text{or}\quad \left(\frac45,-\frac25\right).$$
Proof Architecture
We first prove a lemma converting the quartic equation into a quadratic equation in
$$t=x+\frac1x.$$
The proof uses division by $x^2$ and the identity
$$x^2+\frac1{x^2}=t^2-2.$$
Next we prove that the existence of a real root of the quartic is equivalent to the existence of a real number $t$ with $|t|\ge2$ satisfying
$$t^2+at+b-2=0.$$
The forward implication comes from defining $t=x+1/x$. The reverse implication uses the fact that every real $t$ with $|t|\ge2$ can be written as $x+1/x$ for some real $x$.
We then parametrize all admissible pairs by
$$b=2-t^2-at$$
and minimize
$$a^2+b^2$$
for fixed $t$. This gives the sharp lower bound
$$a^2+b^2\ge\frac{(2-t^2)^2}{1+t^2}.$$
Finally we minimize the remaining one-variable function over the constraint $t^2\ge4$. The hardest direction is proving the global lower bound after incorporating the condition $|t|\ge2$. The most delicate lemma is the equivalence between the quartic root condition and the existence of such a parameter $t$.
Solution
Define
$$P(x)=x^4+ax^3+bx^2+ax+1.$$
Since the constant term of $P$ equals $1$, the equation $P(x)=0$ cannot have the root $x=0$.
Lemma 1
The equation $P(x)=0$ has a real solution if and only if there exists a real number $t$ satisfying
$$|t|\ge2$$
and
$$t^2+at+b-2=0.$$
Proof
Assume first that $P(x)=0$ for some real number $x$. Since $x\neq0$, divide the equation by $x^2$:
$$x^2+ax+b+\frac a x+\frac1{x^2}=0.$$
Set
$$t=x+\frac1x.$$
Then
$$x^2+\frac1{x^2} =\left(x+\frac1x\right)^2-2 =t^2-2.$$
Substituting into the previous equation gives
$$t^2-2+at+b=0,$$
hence
$$t^2+at+b-2=0.$$
Because $x$ is real,
$$\left(x-\frac1x\right)^2\ge0.$$
Expanding yields
$$x^2-2+\frac1{x^2}\ge0,$$
so
$$x^2+\frac1{x^2}\ge2.$$
Adding $2$ to both sides gives
$$\left(x+\frac1x\right)^2\ge4.$$
Therefore
$$t^2\ge4,$$
equivalently
$$|t|\ge2.$$
Conversely, assume there exists a real number $t$ such that
$$|t|\ge2 \quad\text{and}\quad t^2+at+b-2=0.$$
Consider the quadratic equation
$$x^2-tx+1=0.$$
Its discriminant equals
$$t^2-4\ge0,$$
so this equation has a real solution $x$. For such an $x$,
$$x+\frac1x=t.$$
Then
$$x^2+\frac1{x^2}=t^2-2.$$
Using the relation
$$t^2+at+b-2=0,$$
we obtain
$$x^2+\frac1{x^2}+a\left(x+\frac1x\right)+b=0.$$
Multiplying by $x^2$ yields
$$x^4+ax^3+bx^2+ax+1=0.$$
Hence $P(x)=0$ has a real solution.
This proves the equivalence. ∎
This lemma establishes an exact parametrization of all admissible pairs $(a,b)$, and omitting the converse direction would leave open the possibility that the parameter condition is insufficient.
We now use the parameter $t$ from Lemma 1. The relation
$$t^2+at+b-2=0$$
gives
$$b=2-t^2-at.$$
Hence
$$a^2+b^2 =a^2+(2-t^2-at)^2.$$
Fix a real number $t$ with $|t|\ge2$, and define
$$F_t(a)=a^2+(2-t^2-at)^2.$$
Expanding,
$$F_t(a) =a^2+a^2t^2-2at(2-t^2)+(2-t^2)^2.$$
Thus
$$F_t(a) =(1+t^2)a^2-2t(2-t^2)a+(2-t^2)^2.$$
Since
$$1+t^2>0,$$
this is a quadratic polynomial in $a$ with positive leading coefficient. Complete the square:
$$F_t(a) =(1+t^2)\left(a-\frac{t(2-t^2)}{1+t^2}\right)^2 +(2-t^2)^2-\frac{t^2(2-t^2)^2}{1+t^2}.$$
Simplifying the constant term,
$$F_t(a) =(1+t^2)\left(a-\frac{t(2-t^2)}{1+t^2}\right)^2 +\frac{(2-t^2)^2}{1+t^2}.$$
Therefore
$$a^2+b^2\ge\frac{(2-t^2)^2}{1+t^2},$$
with equality if and only if
$$a=\frac{t(2-t^2)}{1+t^2}.$$
Set
$$u=t^2.$$
Since $|t|\ge2$, we have
$$u\ge4.$$
The lower bound becomes
$$a^2+b^2\ge\frac{(u-2)^2}{u+1}.$$
Define
$$g(u)=\frac{(u-2)^2}{u+1}.$$
We evaluate the first relevant cases:
$$g(4)=\frac45, \qquad g(5)=\frac32, \qquad g(6)=\frac{16}7, \qquad g(7)=\frac{25}8, \qquad g(8)=4.$$
These values increase. To prove monotonicity for all $u\ge4$, differentiate:
$$g'(u) =\frac{2(u-2)(u+1)-(u-2)^2}{(u+1)^2}.$$
Factor the numerator:
$$2(u-2)(u+1)-(u-2)^2 =(u-2)(2u+2-u+2) =(u-2)(u+4).$$
Hence
$$g'(u)=\frac{(u-2)(u+4)}{(u+1)^2}.$$
For every $u\ge4$,
$$u-2>0, \qquad u+4>0, \qquad (u+1)^2>0,$$
so
$$g'(u)>0.$$
Thus $g$ is strictly increasing on $[4,\infty)$, and its minimum occurs at $u=4$.
Therefore
$$a^2+b^2\ge g(4)=\frac45.$$
It remains to prove that equality is attainable. Choose
$$t=2.$$
Then
$$a=\frac{2(2-4)}{1+4}=-\frac45,$$
and
$$b=2-4-2a =-2+\frac85 =-\frac25.$$
The corresponding polynomial is
$$x^4-\frac45x^3-\frac25x^2-\frac45x+1.$$
Since
$$t=2,$$
the equation
$$x+\frac1x=2$$
has the real solution
$$x=1.$$
Substituting directly,
$$1-\frac45-\frac25-\frac45+1=0.$$
Hence the polynomial has a real root. Moreover,
$$a^2+b^2 =\left(-\frac45\right)^2+\left(-\frac25\right)^2 =\frac{16}{25}+\frac4{25} =\frac{20}{25} =\frac45.$$
Thus the minimum value is
$$\boxed{\frac45},$$
attained for
$$(a,b)=\left(-\frac45,-\frac25\right)$$
and also for
$$(a,b)=\left(\frac45,-\frac25\right).$$
Verification of Key Steps
The first delicate step is the equivalence in Lemma 1. Starting from a real root $x$, define
$$t=x+\frac1x.$$
Then
$$x^2+\frac1{x^2}=t^2-2,$$
which converts the quartic into
$$t^2+at+b-2=0.$$
The hidden danger is forgetting that the converse requires the condition $|t|\ge2$. Without this restriction, the equation
$$x^2-tx+1=0$$
may have no real solution because its discriminant equals $t^2-4$.
The second delicate step is minimizing over $a$ for fixed $t$. The expression
$$a^2+(2-t^2-at)^2$$
contains mixed terms, and a careless manipulation can lose the dependence on $t$. Recomputing independently,
$$a^2+(2-t^2-at)^2 =(1+t^2)a^2-2t(2-t^2)a+(2-t^2)^2.$$
Completing the square gives
$$(1+t^2)\left(a-\frac{t(2-t^2)}{1+t^2}\right)^2 +\frac{(2-t^2)^2}{1+t^2}.$$
Since the squared term is nonnegative, the lower bound is exact.
The third delicate step is the monotonicity of
$$g(u)=\frac{(u-2)^2}{u+1}$$
for $u\ge4$. A common error is to minimize over all real $u$, which would incorrectly suggest the minimum occurs at $u=2$. The constraint $u=t^2\ge4$ is essential. Under this constraint,
$$g'(u)=\frac{(u-2)(u+4)}{(u+1)^2}>0,$$
so the minimum on the admissible region occurs at the endpoint $u=4$.
Alternative Approaches
A geometric approach is possible. From
$$t^2+at+b-2=0,$$
we obtain
$$at+b=2-t^2.$$
For fixed $t$, this describes a line in the $(a,b)$-plane. Minimizing
$$a^2+b^2$$
amounts to finding the squared distance from the origin to this line. The closest point to the origin on the line
$$ta+b=2-t^2$$
has squared distance
$$\frac{(2-t^2)^2}{1+t^2}.$$
The remainder of the argument is identical: minimize this quantity over $|t|\ge2$.
Another method avoids calculus entirely. After setting
$$u=t^2,$$
observe that
$$\frac{(u-2)^2}{u+1}-\frac45 =\frac{5(u-2)^2-4(u+1)}{5(u+1)} =\frac{u^2-8u+16}{5(u+1)} =\frac{(u-4)^2}{5(u+1)}.$$
Since $u\ge4$, the denominator is positive, and the numerator is nonnegative. Hence
$$\frac{(u-2)^2}{u+1}\ge\frac45,$$
with equality exactly when $u=4$. This approach is shorter than differentiating and isolates the equality condition immediately.