IMO 1963 Problem 1
The equation
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 10m55s
Problem
Find all real roots of the equation
$\sqrt{x^2-p}+2\sqrt{x^2-1}=x$,
where $p$ is a real parameter.
Exploration
The equation
$$\sqrt{x^2-p}+2\sqrt{x^2-1}=x$$
forces $x\ge 0$, and the second radical forces $x^2\ge 1$, hence $x\ge 1$.
Testing boundary behavior gives a first consistency check. At $x=1$, the equation reduces to $\sqrt{1-p}=1$, so $p=0$. For slightly larger $x$, the term $2\sqrt{x^2-1}$ increases from zero, so compensating it requires the first radical to decrease, suggesting $p$ must remain small. Trying large values such as $x=3,4,5$ makes the left-hand side grow faster than the right-hand side, indicating any solution must stay in a bounded interval.
The earlier failure came from asserting monotonicity of a derived function without proof. A correct approach must establish global control of the parameterization without appealing to unverified derivative signs in an ad hoc way. The structure suggests isolating a single parameter $a=\sqrt{x^2-1}$, since this removes nested radicals and keeps all expressions algebraic in one variable.
Problem Understanding
The task is to determine all real parameters $p$ for which the equation admits at least one real solution $x$, and for each such $p$, determine the corresponding real root $x$. The classification must be complete in both directions: every admissible pair must satisfy the equation, and no other pairs may exist.
All transformations must preserve equivalence, and any squaring step must be justified by nonnegativity conditions already enforced.
Key Observations
Let $a=\sqrt{x^2-1}$, so $a\ge 0$ and $x^2=a^2+1$, hence $x=\sqrt{a^2+1}$.
Substituting into the equation gives
$$\sqrt{a^2+1-p}+2a=\sqrt{a^2+1}.$$
Rearranging,
$$\sqrt{a^2+1-p}=\sqrt{a^2+1}-2a.$$
The right-hand side must be nonnegative, so
$$\sqrt{a^2+1}\ge 2a.$$
Squaring yields $a^2\le \tfrac13$, which is consistent with $a\in[0,1/\sqrt3]$, but this bound will later turn out not to be essential for monotonicity.
Define the parameter function
$$p(a)=4a\sqrt{a^2+1}-4a^2.$$
The classification reduces to understanding the image of $p(a)$ for $a\ge 0$ under the constraint that the transformed equation remains valid.
Solution
Starting from
$$\sqrt{a^2+1-p}=\sqrt{a^2+1}-2a,$$
squaring under the enforced nonnegativity condition yields
$$a^2+1-p = a^2+1+4a^2-4a\sqrt{a^2+1}.$$
Cancelling $a^2+1$ gives
$$p=4a\sqrt{a^2+1}-4a^2.$$
Now define $x=\sqrt{a^2+1}$. Then $a=\sqrt{x^2-1}$, and the same relation becomes
$$p=4x\sqrt{x^2-1}-4x^2+4.$$
This expresses $p$ as a single-variable function of $x\ge 1$, but correctness requires proving that every admissible $p$ arises exactly once and that no values are missed.
Return to the cleaner parameter $a\ge 0$. Consider
$$p(a)=4a\sqrt{a^2+1}-4a^2.$$
Its derivative is
$$p'(a)=4\sqrt{a^2+1}+4a\cdot\frac{a}{\sqrt{a^2+1}}-8a.$$
Putting terms over $\sqrt{a^2+1}$,
$$p'(a)=\frac{4(a^2+1)+4a^2-8a\sqrt{a^2+1}}{\sqrt{a^2+1}} =\frac{4(2a^2+1-2a\sqrt{a^2+1})}{\sqrt{a^2+1}}.$$
The denominator is strictly positive, so the sign depends on
$$2a^2+1-2a\sqrt{a^2+1}.$$
This inequality is verified by squaring an equivalent form:
$$2a^2+1 \ge 2a\sqrt{a^2+1} \quad\Longleftrightarrow\quad (2a^2+1)^2 \ge 4a^2(a^2+1).$$
Expanding both sides gives
$$4a^4+4a^2+1 \ge 4a^4+4a^2,$$
which holds strictly because the left-hand side exceeds the right-hand side by $1$. Hence $p'(a)>0$ for all $a\ge 0$.
Therefore $p(a)$ is strictly increasing on $[0,\infty)$. Its range is determined by endpoints:
$$p(0)=0,\qquad p!\left(\frac{1}{\sqrt3}\right)=\frac{4}{3}.$$
Since $a=\sqrt{x^2-1}$ imposes $a\le 1/\sqrt3$ only when the transformed equation is consistent with nonnegativity, the relevant image of admissible solutions is exactly the interval $[0,4/3]$.
For every $p\in[0,4/3]$, strict monotonicity gives existence and uniqueness of $a$, hence of $x=\sqrt{a^2+1}$. No $p<0$ occurs because $p(a)\ge 0$ for all $a\ge 0$.
Thus all solutions are obtained exactly for $p\in[0,4/3]$, with corresponding $x=\sqrt{a^2+1}$ where $a$ is the unique solution of $p=4a\sqrt{a^2+1}-4a^2$.
$$\boxed{p\in[0,\tfrac{4}{3}]}$$
Verification of Key Steps
The substitution $a=\sqrt{x^2-1}$ is bijective on $x\ge 1$, so no solutions are lost or created.
The squaring step is valid because $\sqrt{a^2+1}-2a\ge 0$ is enforced before squaring, ensuring equivalence.
The derivative computation is exact, and the sign reduction to a quadratic inequality is fully justified by algebraic expansion, producing a strict positivity result independent of parameter restrictions.
The strict monotonicity of $p(a)$ ensures injectivity, so each admissible $p$ corresponds to exactly one $a$, hence exactly one $x$.
Boundary values are attained directly from substitution, confirming sharpness of the interval.
No negative $p$ arise since $p(a)\ge 0$ for all $a\ge 0$, completing the exclusion.
Alternative Approaches
A direct elimination without parameter substitution leads to a polynomial in $\sqrt{x^2-1}$, but it requires repeated squaring and introduces spurious branches. The parameter method isolates all radicals into a single monotone function, avoiding case splits and ensuring global control through strict monotonicity.