1973 IMO Problems/Problem 3

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 $a^2 + b^2$ .

Solution

Substitute $z=x+1/x$ to change the original equation into $z^2+az+b-2=0$ . This equation has solutions $z=\frac{-a \pm \sqrt{a^2+8-4b}}{2}$ . We also know that

$|z|=|x+1/x| \geq 2.\ \ \ \ \ \ \ \ (1)$

So,

$\left | \frac{-a \pm \sqrt{a^2+8-4b}}{2} \right | \geq 2\ \ \ \ \ \ \ \ (2)$

$\frac{|a|+\sqrt{a^2+8-4b}}{2} \geq 2\ \ \ \ \ \ \ \ \ \ (3)$

$|a|+\sqrt{a^2+8-4b} \geq 4$

Rearranging and squaring both sides,

$a^2+8-4b \geq a^2-16|a|+16\ \ \ \ \ \ \ \ (4)$

$2|a|-b \geq 2\ \ \ \ \ \ \ \ \ \ \ \ (5)$

So,

$a^2+b^2 \geq a^2+(2-2|a|)^2\ \ \ \ \ \ \ \ (6)$

$= 5a^2-8|a|+4 = 5 \left( |a|-\frac{4}{5} \right )^2+\frac{4}{5}$ .

Therefore, the smallest possible value of $a^2+b^2$ is $\frac{4}{5}$ , when $a=\pm \frac{4}{5}$ and $b=\frac{-2}{5}$ .

Borrowed from [1]

Remarks (added by pf02, June 2025)

1. The solution above is incomplete and it has some serious errors. The result is correct, but the method for obtaining it is flawed. For the sake of reference to the steps in the solution I added labels, but I did not make any changes to the solution. I will highlight the errors and the missing steps:

A. (1) is true when we know that $x$ is real. By hypothesis the equation has at least one real solution, so (1) is true for this solution. A necessary condition for this is that $a^2 + 8 - 4b \ge 0$ . We have to use this in the proof, or if we obtain a result without having used it, we need to verify that the result satisfies this condition. This is missing from the solution.

B. (2) is equivalent to (3) when $\pm$ is interpreted as an $\mathbf{"or"}$ . (If the $\pm$ would be an $\mathbf{"and"}$ then (2) and (3) would not be equivalent.) In our problem the $\pm$ is an $\mathbf{"or"}$ so the proof can proceed. However, while easy, going from (2) to (3) is not obvious, and it is an important step in the proof, so it should be explained.

C. There is a small computational error in (4): we should have $8|a|$ , not $16|a|$ . This error is in writing, the computation proceeds as if it did not happen.

D. Going from (5) to (6) is a very serious error. Replacing $b^2$ by $(2 - 2|a|)^2$ can be done only when certain conditions are satisfied (for example if we know that $b \le 0$ and $2 - 2|a| \ge 0$ or if we know that $b = 2 - 2|a|$ ). So writing down (6) is completely unjustified.

2. I will not take the task of fixing this solution. Instead, I will give a different solution below.

Solution 2

The plan of this solution is the following: we will consider the plane of coordinates $a, b$ . We will write the condition(s) for the equation to have at least one real solution. These will be inequalities in $a, b$ , which delimit regions in the $a, b$ plane. $a^2 + b^2$ represents the square of the distance from the point of coordinates $(a, b)$ to the origin. We will find the point(s) in the region we delimited which are closest to the origin. These are minimizing $a^2 + b^2$ .

Dividing the given equation by $x^2$ and making the substitution $z = x + \frac{1}{x}$ , we can re-write the given equation as $z^2 + az + b - 2 = 0$ . The given equation has at least one real root when this equation in $z$ has real roots. This is true when $a^2 - 4b + 8 \ge 0$ . This represents the parabola $b = \frac{a^2}{4} + 2$ in the $a, b$ plane. Refer to this parabola as $P$ .

The two solutions of the equation in $z$ are $\frac{-a \pm \sqrt{a^2 -4b + 8}}{2}$ . We will take them one at a time, and write down the conditions for its solution $x$ to be real. Start with the $+$ sign, and consider the equation $2 x^2 - (-a + \sqrt{a^2 -4b + 8}) x + 2 = 0$

This equation has real solutions when $(a - \sqrt{a^2 - 4b + 8})^2 - 16 \ge 0$ . Work out the square, rearrange terms and simplify. We get $a \sqrt{a^2 - 4b + 8} \le a^2 - 2b - 4$ .

Note that at this point we could continue working with inequalities, being vary careful in considering all the cases as we do computations. Instead, we will consider the equation $a \sqrt{a^2 - 4b + 8} = a^2 - 2b - 4$ . This is a curve in the $a, b$ plane, delimiting several regions in the plane. We will study the curve, and decide which are the regions in the $a, b$ plane in which the inequality is true. This will give us a set of points $(a, b)$ for which the equation has at least one real root.

Square both sides and rearrange terms. We get $4a^2 = (b + 2)^2$ . As a curve, this is two lines $L_1, L_2$ with equations $b = 2a - 2$ and $b = -2a - 2$ . Before proceeding, we have to verify the validity of these solutions since we may have introduced "fake" solutions when we squared both sides.

Plug $b = 2a - 2$ into $a \sqrt{a^2 - 4b + 8} = a^2 - 2b - 4$ . We get $a \sqrt{(a - 4)^2} = a (a - 4)$ . This is true when $a = 0$ or $a \ge 4$ . Thus, $b = 2a - 2$ is valid when $a \ge 4, b \ge 6$ .

Similarly, plug $b = -2a - 2$ into $a \sqrt{a^2 - 4b + 8} = a^2 - 2b - 4$ , and get that $b = -2a - 2$ is valid when $a \ge -4, b \le 6$ .

The parabola $P$ , the lines $L_1, L_2$ and the lines $a = 4, a = -4, b = 6$ carve the $a, b$ plane into several regions. We now have to verify which are the regions in which $(a - \sqrt{a^2 - 4b + 8})^2 - 16 \ge 0$ . While mildly tedious, this is very easy: we just have to take a numerical sample in each region and verify the inequality. The results are shown in the image below (the one on the left): if the point $(a, b)$ is in the shaded area, then the inequality is satisfied, so equation in the problem has at least one real solution.

Now take the $-$ sign from $z = \frac{-a \pm \sqrt{a^2 -4b + 8}}{2}$ . We get the equation $2 x^2 + (a + \sqrt{a^2 -4b + 8}) x + 2 = 0$ . This has real solutions when $(a + \sqrt{a^2 - 4b + 8})^2 - 16 \ge 0$ . Consider the regions delimited by $(a + \sqrt{a^2 - 4b + 8})^2 - 16 = 0$ , and find the regions when the inequality is true. The computations are very similar to the ones in the preceding case. The regions for $(a, b)$ which guarantee that the equation from the problem has at least one real solution are shown in the image above (the one on the right hand side).

It is now clear that the points $(a, b)$ in these regions closest to $(0, 0)$ are the ones at the intersection of the line $b = 2a - 2$ with the perpendicular from $(0, 0)$ , and at the intersection of the line $b = -2a - 2$ with the perpendicular from $(0, 0)$ . A very simple computation shows that these points are $(\frac{4}{5}, -\frac{2}{5}), (-\frac{4}{5}, -\frac{2}{5})$ . Then the square of the distance from $(0, 0)$ to these points $= a^2 + b^2 = \frac{4}{5}$ .

[Solution by pf02, June 2025]

1973 IMO (Problems) • Resources
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1973 IMO Problems/Problem 3

Contents

Problem

Solution

Remarks (added by pf02, June 2025)

Solution 2

See Also