1965 IMO Problems/Problem 4

Problem

Find all sets of four real numbers $x_1$ , $x_2$ , $x_3$ , $x_4$ such that the sum of any one and the product of the other three is equal to $2$ .

Solution

Let $P = x_1x_2x_3x_4$ be the product of the four real numbers.

Then, for $i = 1,2,3,4$ we have: $x_i + \prod_{j \neq i}x_j = 2$ .

Multiplying by $x_i$ yields:

$x^2_i + P = 2x_i \Longleftrightarrow x^2_i-2x_i+1 = (x_i-1)^2 = 1-P \Longleftrightarrow x_i = 1 \pm t$ where $t = \pm \sqrt{1-P} \in \mathbb{R}$ .

If $t=0$ , then we have $(x_1,x_2,x_3,x_4)=(1,1,1,1)$ which is a solution.

So assume that $t \neq 0$ . WLOG, let at least two of $x_i$ equal $1+t$ , and $x_1 \ge x_2 \ge x_3 \ge x_4$ OR $x_1 \le x_2 \le x_3 \le x_4$ .

Case I: $x_1 = x_2 = x_3 = x_4 = 1+t$

Then we have:

$(1+t)+(1+t)^3 = 2 \Longleftrightarrow t^3+3t^2+4t = 0 \Longleftrightarrow t(t^2+3t+4) = 0$

Which has no non-zero solutions for $t$ .

Case II: $x_1 = x_2 = x_3 = 1+t$ AND $x_4 = 1-t$

Then we have:

$(1-t)+(1+t)^3 = 2 \Longleftrightarrow t^3+3t^2+2t = 0$ $\Longleftrightarrow t(t+1)(t+2) = 0 \Longleftrightarrow t \in \{0,-1,-2\}$

AND

$(1+t)+(1-t)(1+t)^2 = 2 (1+t)+(1-t)(1+t)^2 = 2 -t^3-t^2+2t = 0$ $\Longleftrightarrow -t(t-1)(t+2) = 0 \Longleftrightarrow t \in \{0,1,-2\}$

So, we have $t = -2$ as the only non-zero solution, and thus, $(x_1,x_2,x_3,x_4) = (-1,-1,-1,3)$ and all permutations are solutions.

Case III: $x_1 = x_2 = 1+t$ AND $x_3 = x_4 = 1-t$

Then we have:

$(1-t)+(1-t)(1+t)^2 = 2 \Longleftrightarrow -t^3-t^2 = 0$ $\Longleftrightarrow -t^2(t+1) = 0 \Longleftrightarrow t \in \{0,-1\}$

AND

$(1+t)+(1+t)(1-t)^2 = 2 \Longleftrightarrow t^3-t^2 = 0$ $\Longleftrightarrow t^2(t-1) = 0 \Longleftrightarrow t \in \{0,1\}$

Thus, there are no non-zero solutions for $t$ in this case.

Therefore, the solutions are: $(1,1,1,1)$ ; $(3,-1,-1,-1)$ ; $(-1,3,-1,-1)$ ; $(-1,-1,3,-1)$ ; $(-1,-1,-1,3)$ .

Solution 2

We have to solve the system of equations

$x_1 + x_2x_3x_4 = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \\ x_2 + x_1x_3x_4 = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2) \\ x_3 + x_1x_2x_4 = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) \\ x_4 + x_1x_2x_3 = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

Subtract (2) from (1) and factor. We get

$(x_1 - x_2)(1 - x_3x_4) = 0$ ,

which implies $x_1 - x_2 = 0$ or $1 - x_3x_4 = 0$ .

Similarly, subtracting (3) and then (4) from (1) and factoring, we get

$(x_1 - x_3)(1 - x_2x_4) = 0 \\ (x_1 - x_4)(1 - x_2x_3) = 0$

They imply $x_1 - x_3 = 0$ or $1 - x_2x_4 = 0$ , and $x_1 - x_4 = 0$ or $1 - x_2x_3 = 0$ .

We will consider four possibilities:

1. $x_2 = x_1, x_3 = x_1, x_4 = x_1$

2. $x_2 = x_1, x_3 = x_1$ and $x_2x_3 = 1$

3. $x_2 = x_1$ and $x_2x_3 = 1, x_2x_4 = 1$

4. $x_2x_3 = 1, x_2x_4 = 1, x_3x_4 = 1$

Note that in fact, there are four more possibilities, but they just correspond to permutations in $x_1, x_2, x_3, x_4$ of cases 2. and 3., so there is no harm in not dealing with them explicitly.

Case 1. Plug $x_2, x_3, x_4$ in equation (1). We get $x_1 + x_1^3 = 2$ . This is an equation of degree $3$ whose only real root is $x_1 = 1$ . We get the solution $x_1 = x_2 = x_3 = x_4 = 1$ .

Case 2. Plug $x_2, x_3$ into $x_2x_3 = 1$ , and get $x_1^2 = 1$ . We get $x_1 = 1$ or $x_1 = -1$ . The first solution, $x_1 = 1$ , yields $x_2 = x_3 = 1$ , and using (4), $x_4 = 1$ . The second solution, $x_1 = -1$ , yields $x_2 = x_3 = -1$ , and using (4), $x_4 = 3$ . We get the solution $x_1 = x_2 = x_3 = -1, x_4 = 3.$ Because of the permutations of $x_1, x_2, x_3, x_4$ , we also get the solutions $(-1, -1, 3, -1), (-1, 3, -1, -1), (3, -1, -1, -1)$ .

Case 3.

(Solution by pf02, November 2024)

TO BE CONTINUED. I AM SAVING MID WAY SO I DON'T LOSE WORK DONE SO FAR.

1965 IMO (Problems) • Resources
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1965 IMO Problems/Problem 4

Contents

Problem

Solution

Solution 2

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