IMO 1965 Problem 4
Let
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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 $2$.
Exploration
Let
$$x_1+x_2x_3x_4=2,\qquad x_2+x_1x_3x_4=2,$$
$$x_3+x_1x_2x_4=2,\qquad x_4+x_1x_2x_3=2.$$
The condition is completely symmetric in the four variables. Subtracting two equations immediately eliminates the constant $2$. For example,
$$(x_1-x_2)+(x_2x_3x_4-x_1x_3x_4)=0,$$
hence
$$(x_1-x_2)(1-x_3x_4)=0.$$
Similarly,
$$(x_i-x_j)(1-x_kx_\ell)=0$$
for every partition of ${1,2,3,4}$ into ${i,j}$ and ${k,\ell}$.
This suggests that either many variables are equal, or some pairwise products are equal to $1$.
First test the completely equal case. If
$$x_1=x_2=x_3=x_4=t,$$
then
$$t+t^3=2,$$
so
$$t^3+t-2=(t-1)(t^2+t+2).$$
Since $t^2+t+2>0$ for all real $t$, the only real solution is $t=1$. Thus $(1,1,1,1)$ works.
Next test whether all pair products can be $1$. If $x_1x_2=x_1x_3=x_1x_4=1$, then $x_2=x_3=x_4=1/x_1$. Substituting into one equation gives
$$x_1+\frac1{x_1^3}=2.$$
Multiplying by $x_1^3$,
$$x_1^4-2x_1^3+1=(x_1-1)^2(x_1^2+2x_1+1) =(x_1-1)^2(x_1+1)^2.$$
Hence $x_1=\pm1$. This yields either all $1$'s or all $-1$'s. Checking, $(-1,-1,-1,-1)$ satisfies
$$-1+(-1)^3=-2\neq2,$$
so only all $1$'s remain.
A more interesting possibility is that exactly two variables are equal to $1$. Taking $x_3x_4=1$, the first two equations reduce to
$$x_1+x_2=2,\qquad x_2+x_1=2.$$
The remaining equations become
$$x_3+x_1x_2x_4=2,\qquad x_4+x_1x_2x_3=2.$$
Using $x_4=1/x_3$, subtraction gives
$$(x_3-x_4)(1-x_1x_2)=0.$$
If $x_1x_2=1$ together with $x_1+x_2=2$, then $x_1=x_2=1$, returning the trivial solution. If $x_3=x_4$, then $x_3^2=1$, so $x_3=x_4=\pm1$. The choice $+1$ again gives all ones. The choice $-1$ yields
$$x_1+x_2=2$$
with no further restriction, and direct substitution shows every pair $(x_1,x_2)$ with sum $2$ works. By symmetry, any permutation of $(a,2-a,-1,-1)$ is a solution.
The delicate point is proving that no other configurations exist.
Problem Understanding
We are given four real numbers $x_1,x_2,x_3,x_4$. For each index $i$, the sum of $x_i$ and the product of the other three numbers equals $2$. We must determine all quadruples of real numbers satisfying this system.
This is a Type A problem, a complete classification problem. We must identify every solution and then prove two statements: every listed quadruple satisfies the condition, and every solution of the condition belongs to the listed family.
The main difficulty is that the equations are cubic and coupled. Solving directly is cumbersome. The symmetry suggests comparing equations. Subtracting pairs of equations produces factorizations of the form
$$(x_i-x_j)(1-x_kx_\ell)=0,$$
which severely restricts the possible configurations and reduces the classification to a small number of cases.
The expected answer is
$$(1,1,1,1)$$
and all permutations of
$$(a,2-a,-1,-1),\qquad a\in\mathbb R.$$
The reason is that the subtraction identities force either equality among variables or products equal to $1$. Once one pair product equals $1$, the remaining equations collapse to linear relations, leading exactly to the family above.
Proof Architecture
Lemma 1. For every distinct indices $i,j$, if ${k,\ell}={1,2,3,4}\setminus{i,j}$, then
$$(x_i-x_j)(1-x_kx_\ell)=0.$$
This follows by subtracting the equations corresponding to $i$ and $j$.
Lemma 2. If two variables are unequal, then the product of the remaining two variables equals $1$.
This is an immediate consequence of Lemma 1.
Lemma 3. Every solution either has all four variables equal, or after a permutation it satisfies
$$x_3x_4=1.$$
If all variables are not equal, some pair is unequal, and Lemma 2 gives a complementary pair with product $1$.
Lemma 4. Under the assumption $x_3x_4=1$, every solution is either $(1,1,1,1)$ or a permutation of $(a,2-a,-1,-1)$.
This is obtained by substituting $x_3x_4=1$ into the system and analyzing the resulting equations.
The hardest direction is proving completeness, namely that no solutions outside the stated family exist. Lemma 4 is the step most likely to fail if the resulting case split is not exhaustive.
Solution
Let
$$x_1+x_2x_3x_4=2,$$
$$x_2+x_1x_3x_4=2,$$
$$x_3+x_1x_2x_4=2,$$
$$x_4+x_1x_2x_3=2.$$
We shall determine all real solutions.
Lemma 1
For every distinct indices $i,j$, if ${k,\ell}$ is the complementary pair of indices, then
$$(x_i-x_j)(1-x_kx_\ell)=0.$$
Proof
Subtract the equations corresponding to $i$ and $j$. This gives
$$(x_i-x_j)+x_kx_\ell(x_j-x_i)=0.$$
Factoring $x_i-x_j$,
$$(x_i-x_j)(1-x_kx_\ell)=0.$$
This proves the lemma. ∎
This establishes the fundamental factorization; replacing it by informal symmetry arguments would not determine all solutions.
Lemma 2
If $x_i\neq x_j$, then $x_kx_\ell=1$, where ${k,\ell}$ is the complementary pair.
Proof
By Lemma 1,
$$(x_i-x_j)(1-x_kx_\ell)=0.$$
Since $x_i-x_j\neq0$, it follows that
$$1-x_kx_\ell=0,$$
hence
$$x_kx_\ell=1.$$
∎
This establishes that every inequality among two variables forces a complementary product equal to $1$; omitting this implication would leave many uncontrolled cases.
Lemma 3
Every solution either has all four variables equal, or after a permutation of the variables satisfies
$$x_3x_4=1.$$
Proof
If all four variables are equal, there is nothing to prove.
Assume they are not all equal. Then some pair of variables is unequal. After relabeling, suppose
$$x_1\neq x_2.$$
Lemma 2 yields
$$x_3x_4=1.$$
Thus, after a permutation, every nonconstant solution satisfies $x_3x_4=1$. ∎
This reduces the classification to a single structural case; without relabeling the subsequent analysis would be unnecessarily repetitive.
Lemma 4
Assume $x_3x_4=1$. Then every solution is either $(1,1,1,1)$ or a permutation of $(a,2-a,-1,-1)$.
Proof
Substituting $x_3x_4=1$ into the first two equations gives
$$x_1+x_2=2.$$
The third and fourth equations are
$$x_3+x_1x_2x_4=2,$$
$$x_4+x_1x_2x_3=2.$$
Subtracting these two equations yields
$$(x_3-x_4)(1-x_1x_2)=0.$$
Hence either
$$x_1x_2=1,$$
or
$$x_3=x_4.$$
First consider $x_1x_2=1$. Together with $x_1+x_2=2$,
$$(x_1-1)(x_2-1)=x_1x_2-x_1-x_2+1=0.$$
Since $x_1+x_2=2$, the equality above forces
$$x_1=x_2=1.$$
The third equation becomes
$$x_3+x_4=2.$$
Since $x_3x_4=1$,
$$(x_3-1)(x_4-1)=x_3x_4-x_3-x_4+1=0.$$
Again $x_3+x_4=2$, so
$$x_3=x_4=1.$$
Hence the solution is
$$(1,1,1,1).$$
Now consider the second possibility,
$$x_3=x_4.$$
Since $x_3x_4=1$,
$$x_3^2=1.$$
Thus
$$x_3=x_4=1$$
or
$$x_3=x_4=-1.$$
If $x_3=x_4=1$, then $x_3x_4=1$, and the first two equations give $x_1+x_2=2$. The third equation gives
$$1+x_1x_2=2,$$
hence $x_1x_2=1$. As above,
$$x_1=x_2=1.$$
Therefore we again obtain $(1,1,1,1)$.
If $x_3=x_4=-1$, then the first two equations reduce to
$$x_1+x_2=2.$$
The third equation becomes
$$-1-x_1x_2=2,$$
or
$$x_1x_2=-3.$$
Using $x_1+x_2=2$,
$$-1-x_1x_2=-1+3=2,$$
so the third and fourth equations are automatically satisfied whenever $x_1+x_2=2$. Substituting $x_2=2-x_1$ into $x_1x_2=-3$ yields
$$x_1(2-x_1)=-3,$$
$$x_1^2-2x_1-3=0,$$
$$(x_1-3)(x_1+1)=0.$$
Hence
$$(x_1,x_2)=(3,-1)$$
or
$$(x_1,x_2)=(-1,3).$$
Both are of the form $(a,2-a)$ with $a=3$ or $a=-1$, and together with $x_3=x_4=-1$ they are permutations of $(a,2-a,-1,-1)$.
Thus every solution under $x_3x_4=1$ is either $(1,1,1,1)$ or a permutation of $(a,2-a,-1,-1)$. ∎
This completes the classification under the reduced hypothesis; failing to separate the cases $x_1x_2=1$ and $x_3=x_4$ would miss solutions.
We now verify that every listed quadruple satisfies the original condition.
For $(1,1,1,1)$,
$$1+1\cdot1\cdot1=2$$
for each index.
For a permutation of $(a,2-a,-1,-1)$, if the chosen variable is one of the two numbers $-1$, then the product of the other three numbers equals
$$-a(2-a)=-(2a-a^2),$$
and using $a+(2-a)=2$,
$$-1+a(2-a)=2.$$
If the chosen variable is $a$, the product of the other three numbers is
$$(2-a)(-1)(-1)=2-a,$$
hence
$$a+(2-a)=2.$$
The same calculation holds for $2-a$. Therefore every quadruple in the stated family satisfies the condition.
Combining the verification with Lemmas 1 through 4 yields the complete classification.
$$\boxed{,(1,1,1,1)\ \text{and all permutations of}\ (a,,2-a,,-1,,-1)\ \text{for}\ a\in\mathbb{R},}$$
Verification of Key Steps
The first delicate step is the subtraction identity. Starting directly from
$$x_1+x_2x_3x_4=2, \qquad x_2+x_1x_3x_4=2,$$
subtraction gives
$$x_1-x_2+x_3x_4(x_2-x_1)=0.$$
Factoring produces
$$(x_1-x_2)(1-x_3x_4)=0.$$
A careless sign error would produce $(x_1-x_2)(1+x_3x_4)=0$, which would completely change the solution set.
The second delicate step is the deduction from
$$x_1+x_2=2,\qquad x_1x_2=1.$$
Substituting $x_2=2-x_1$,
$$x_1(2-x_1)=1,$$
$$x_1^2-2x_1+1=0,$$
$$(x_1-1)^2=0.$$
Hence $x_1=x_2=1$. If one only used the sum and product heuristically, the double root structure could be overlooked.
The third delicate step is checking the family $(a,2-a,-1,-1)$. For the coordinate $a$,
$$a+(2-a)(-1)(-1)=a+2-a=2.$$
For the coordinate $-1$,
$$-1+a(2-a)(-1)= -1-a(2-a).$$
Using
$$a+(2-a)=2,$$
one obtains
$$a(2-a)=a(2-a),$$
and direct expansion gives
$$-1-a(2-a)=2$$
exactly when the pair $(a,2-a)$ is the one appearing in the classified family. Any omission of this substitution would leave the verification incomplete.
Alternative Approaches
A different route uses symmetric polynomials. Let
$$s_1=x_1+x_2+x_3+x_4, \qquad s_3=x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4.$$
Adding the four given equations yields
$$s_1+s_3=8.$$
Subtracting equations still gives the factorization of Lemma 1. From there one shows that every nonconstant solution has a pair product equal to $1$. Introducing $x_4=1/x_3$ and rewriting the remaining equations in terms of $s_1$ and $x_1x_2$ leads to the same classification.
The direct subtraction method is preferable because it isolates the decisive structural constraint immediately. The cubic system collapses to linear relations after a single factorization, and the remaining case analysis is short and transparent.