IMO 1963 Problem 1
The equation is
Proposed by: -
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Verdicts: UNKNOWN + UNKNOWN
Solve time: 11m46s
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 is
$$\sqrt{x^2-p}+2\sqrt{x^2-1}=x,$$
with real parameter $p$. Since both square roots are real, the domain conditions are
$$x^2-p\ge 0,\qquad x^2-1\ge 0.$$
Hence $|x|\ge 1$.
The left-hand side is a sum of nonnegative quantities. Consequently the right-hand side must also be nonnegative, so
$$x\ge 0.$$
Combining this with $|x|\ge 1$ gives
$$x\ge 1.$$
Testing a few values is informative. If $x=1$, then
$$\sqrt{1-p}=1,$$
so $p=0$. Thus $(p,x)=(0,1)$ works.
The equation resembles situations where isolating one radical and squaring is effective, but repeated squaring may introduce extraneous roots. A cleaner approach is preferable.
Since
$$2\sqrt{x^2-1}\ge 0,$$
the equation implies
$$x\ge \sqrt{x^2-p}.$$
Squaring is legitimate because both sides are nonnegative:
$$x^2\ge x^2-p,$$
hence
$$p\ge 0.$$
Thus negative parameters are impossible.
Suppose now that $p\ge 0$. Rearranging gives
$$x-\sqrt{x^2-p}=2\sqrt{x^2-1}.$$
The left-hand side suggests rationalization:
$$x-\sqrt{x^2-p} = \frac{x^2-(x^2-p)}{x+\sqrt{x^2-p}} = \frac{p}{x+\sqrt{x^2-p}}.$$
Therefore
$$\frac{p}{x+\sqrt{x^2-p}} = 2\sqrt{x^2-1}.$$
The denominator is positive for $x\ge 1$, so the right-hand side is positive unless $x=1$. This suggests separating the case $x=1$.
Trying to eliminate radicals directly may become cumbersome. Another idea is to isolate one radical:
$$\sqrt{x^2-p}=x-2\sqrt{x^2-1}.$$
The right-hand side must be nonnegative:
$$x\ge 2\sqrt{x^2-1}.$$
Squaring gives
$$x^2\ge 4x^2-4,$$
so
$$3x^2\le 4.$$
Since $x\ge 1$,
$$1\le x^2\le \frac43.$$
This is very restrictive.
Now square the isolated equation:
$$x^2-p=x^2-4x\sqrt{x^2-1}+4x^2-4.$$
Hence
$$p=4x\sqrt{x^2-1}-4x^2+4.$$
It is natural to simplify this expression. Put
$$t=\sqrt{x^2-1}.$$
Then $x^2=t^2+1$, so
$$p=4xt-4t^2.$$
Factor:
$$p=4t(x-t).$$
Since
$$x-t=\frac{x^2-t^2}{x+t}=\frac1{x+t},$$
we obtain
$$p=\frac{4t}{x+t}.$$
Because $x\ge 1$ and $0\le t\le \frac1{\sqrt3}$, one gets
$$0\le p<2.$$
This suggests parametrization may exist.
A more decisive step is to square carefully after isolating:
$$\sqrt{x^2-p}=x-2\sqrt{x^2-1}.$$
Squaring:
$$x^2-p=x^2-4x\sqrt{x^2-1}+4x^2-4.$$
Thus
$$4x\sqrt{x^2-1}=4x^2+p-4.$$
Squaring again:
$$16x^2(x^2-1)=(4x^2+p-4)^2.$$
Set $y=x^2$. Then
$$16y(y-1)=(4y+p-4)^2.$$
Expanding:
$$16y^2-16y=16y^2+8y(p-4)+(p-4)^2.$$
Therefore
$$-16y=8y(p-4)+(p-4)^2.$$
Rearrange:
$$8y(2-p)=(p-4)^2.$$
Hence
$$x^2=y=\frac{(p-4)^2}{8(2-p)}.$$
This formula requires $p<2$ unless the numerator also vanishes. Since $(p-4)^2>0$ at $p=2$, there is no solution for $p=2$.
Simplify:
$$x^2=\frac{(4-p)^2}{8(2-p)}.$$
Try to rewrite:
$$(4-p)^2=(2-p+2)^2=(2-p)^2+4(2-p)+4.$$
No immediate simplification.
Now test whether substituting back yields a condition on $p$. Since $x^2\ge 1$,
$$\frac{(4-p)^2}{8(2-p)}\ge 1.$$
This becomes
$$(4-p)^2\ge 8(2-p).$$
Expanding:
$$p^2-8p+16\ge 16-8p,$$
so
$$p^2\ge 0,$$
always true. Equality occurs only at $p=0$, giving $x=1$.
This suggests that every $p$ with $0\le p<2$ gives a solution. The remaining issue is uniqueness and verification that the candidate indeed satisfies the original equation.
Compute
$$x^2-1 = \frac{(4-p)^2-8(2-p)}{8(2-p)} = \frac{p^2}{8(2-p)}.$$
Hence
$$\sqrt{x^2-1} = \frac{p}{2\sqrt{2(2-p)}},$$
because $p\ge 0$.
Also,
$$x^2-p = \frac{(4-p)^2-8p(2-p)}{8(2-p)} = \frac{(p-4)^2-16p+8p^2}{8(2-p)}.$$
Simplifying:
$$p^2-8p+16-16p+8p^2 = 9p^2-24p+16 = (3p-4)^2.$$
Therefore
$$\sqrt{x^2-p} = \frac{|3p-4|}{2\sqrt{2(2-p)}}.$$
This absolute value signals danger. The second squaring may have introduced extraneous candidates.
Substituting into the original equation:
$$\frac{|3p-4|}{2\sqrt{2(2-p)}} + \frac{p}{\sqrt{2(2-p)}} = \frac{4-p}{2\sqrt{2(2-p)}}.$$
Multiply by $2\sqrt{2(2-p)}$:
$$|3p-4|+2p=4-p.$$
Now solve.
If $3p-4\ge 0$, then
$$3p-4+2p=4-p,$$
hence
$$6p=8,$$
so
$$p=\frac43.$$
If $3p-4<0$, then
$$4-3p+2p=4-p,$$
which is an identity. Thus every
$$0\le p<\frac43$$
works.
At $p=\frac43$,
$$x^2=\frac{(8/3)^2}{8(2/3)}=\frac{64/9}{16/3}=\frac43,$$
so
$$x=\frac2{\sqrt3}.$$
Substitution works.
For $\frac43<p<2$, the candidates are extraneous.
Hence the complete parameter range is
$$0\le p\le \frac43,$$
and for each such $p$ there is exactly one positive root
$$x=\frac{4-p}{2\sqrt{2(2-p)}}.$$
Problem Understanding
The problem asks for all real numbers $x$ satisfying
$$\sqrt{x^2-p}+2\sqrt{x^2-1}=x,$$
where $p$ is a real parameter. The parameter is arbitrary, so the task is to determine exactly for which values of $p$ real solutions exist and to compute all corresponding roots.
This is a Type A problem. One must determine all real roots and prove both directions: every claimed root satisfies the equation, and no other real root exists.
The unknown appears both outside and inside square roots. Squaring is unavoidable, but each squaring step may introduce extraneous solutions. The main difficulty is controlling these false candidates and recovering the precise parameter restriction after the algebraic manipulations.
The final answer is that real roots exist exactly for
$$0\le p\le \frac43,$$
and in that case the unique real root is
$$x=\frac{4-p}{2\sqrt{2(2-p)}}.$$
The restriction arises because the second squaring introduces a family of formal candidates that satisfy a squared equation but fail the original equation when $p>\frac43$.
Proof Architecture
The proof begins by deriving necessary sign and domain conditions from the original equation.
The first lemma states that every real solution satisfies $x\ge 1$ and $p\ge 0$. This follows because all square roots are nonnegative and the equation forces the right-hand side to be nonnegative.
The second lemma derives an explicit formula for $x^2$ in terms of $p$:
$$x^2=\frac{(4-p)^2}{8(2-p)}.$$
This is obtained by isolating one radical and squaring twice.
The third lemma substitutes this candidate back into the original equation and determines exactly which parameters survive the verification step. The critical point is the appearance of an absolute value:
$$|3p-4|+2p=4-p.$$
Solving this equation yields
$$0\le p\le \frac43.$$
The hardest direction is excluding extraneous solutions after squaring. The most delicate step is simplifying $\sqrt{x^2-p}$ correctly, because the expression becomes
$$\sqrt{x^2-p}=\frac{|3p-4|}{2\sqrt{2(2-p)}},$$
and omitting the absolute value produces incorrect parameter ranges.
Solution
Assume that
$$\sqrt{x^2-p}+2\sqrt{x^2-1}=x$$
has a real solution.
Lemma 1
Every real solution satisfies
$$x\ge 1 \qquad\text{and}\qquad p\ge 0.$$
Proof
Since the square roots are real,
$$x^2-p\ge 0, \qquad x^2-1\ge 0.$$
Hence
$$|x|\ge 1.$$
The left-hand side of the equation is a sum of nonnegative numbers, so
$$x\ge 0.$$
Combining this with $|x|\ge 1$ gives
$$x\ge 1.$$
Furthermore,
$$x=\sqrt{x^2-p}+2\sqrt{x^2-1}\ge \sqrt{x^2-p}.$$
Both sides are nonnegative, so squaring preserves equivalence:
$$x^2\ge x^2-p.$$
Therefore
$$p\ge 0.$$
∎
This lemma establishes the fundamental sign restrictions, and without them the later squaring steps would not be logically reversible.
Lemma 2
Every real solution satisfies
$$x^2=\frac{(4-p)^2}{8(2-p)}.$$
Proof
From the original equation,
$$\sqrt{x^2-p}=x-2\sqrt{x^2-1}.$$
The right-hand side is nonnegative because the left-hand side is nonnegative.
Squaring both sides yields
$$x^2-p = x^2-4x\sqrt{x^2-1}+4(x^2-1).$$
Simplifying,
$$x^2-p = 5x^2-4-4x\sqrt{x^2-1}.$$
Hence
$$4x\sqrt{x^2-1}=4x^2+p-4.$$
Both sides are nonnegative, so squaring again gives
$$16x^2(x^2-1)=(4x^2+p-4)^2.$$
Set
$$y=x^2.$$
Then
$$16y(y-1)=(4y+p-4)^2.$$
Expanding both sides,
$$16y^2-16y = 16y^2+8y(p-4)+(p-4)^2.$$
Subtracting $16y^2$ from both sides,
$$-16y=8y(p-4)+(p-4)^2.$$
Hence
$$8y(2-p)=(p-4)^2.$$
Therefore
$$y=\frac{(p-4)^2}{8(2-p)}.$$
Since $(p-4)^2=(4-p)^2$,
$$x^2=\frac{(4-p)^2}{8(2-p)}.$$
∎
This lemma produces the only possible candidate for a solution, and any omission in the algebra would create false branches that are difficult to detect later.
Lemma 3
The equation has a real solution if and only if
$$0\le p\le \frac43.$$
For each such $p$, the unique real solution is
$$x=\frac{4-p}{2\sqrt{2(2-p)}}.$$
Proof
By Lemma 1,
$$p\ge 0.$$
By Lemma 2,
$$x^2=\frac{(4-p)^2}{8(2-p)}.$$
Since $x\ge 1$, we have $x>0$, so
$$x=\frac{4-p}{2\sqrt{2(2-p)}}.$$
We now verify which values of $p$ actually satisfy the original equation.
First compute
$$x^2-1 = \frac{(4-p)^2}{8(2-p)}-1.$$
Expanding,
$$x^2-1 = \frac{(4-p)^2-8(2-p)}{8(2-p)}.$$
Since
$$(4-p)^2-8(2-p) = 16-8p+p^2-16+8p = p^2,$$
we obtain
$$x^2-1=\frac{p^2}{8(2-p)}.$$
Because $p\ge 0$,
$$\sqrt{x^2-1} = \frac{p}{2\sqrt{2(2-p)}}.$$
Next compute
$$x^2-p = \frac{(4-p)^2}{8(2-p)}-p.$$
Hence
$$x^2-p = \frac{(4-p)^2-8p(2-p)}{8(2-p)}.$$
Expanding the numerator,
$$(4-p)^2-8p(2-p) = 16-8p+p^2-16p+8p^2 = 9p^2-24p+16.$$
Factorization gives
$$9p^2-24p+16=(3p-4)^2.$$
Therefore
$$x^2-p = \frac{(3p-4)^2}{8(2-p)},$$
so
$$\sqrt{x^2-p} = \frac{|3p-4|}{2\sqrt{2(2-p)}}.$$
Substituting into the original equation,
$$\frac{|3p-4|}{2\sqrt{2(2-p)}} + 2\cdot \frac{p}{2\sqrt{2(2-p)}} = \frac{4-p}{2\sqrt{2(2-p)}}.$$
Multiplying by the positive quantity $2\sqrt{2(2-p)}$ gives
$$|3p-4|+2p=4-p.$$
If
$$3p-4\ge 0,$$
then
$$3p-4+2p=4-p,$$
hence
$$6p=8,$$
so
$$p=\frac43.$$
If
$$3p-4<0,$$
then
$$|3p-4|=4-3p,$$
and the equation becomes
$$4-3p+2p=4-p,$$
which holds identically.
Thus all parameters satisfying
$$0\le p<\frac43$$
produce solutions, and additionally
$$p=\frac43$$
also produces a solution.
Hence the equation has a real solution exactly for
$$0\le p\le \frac43.$$
For every such $p$, the corresponding solution is
$$x=\frac{4-p}{2\sqrt{2(2-p)}}.$$
Finally, uniqueness follows from Lemma 2, because every solution must satisfy the uniquely determined value of $x^2$, and Lemma 1 gives $x>0$.
Therefore the complete set of real roots is
$$\boxed{ x=\frac{4-p}{2\sqrt{2(2-p)}} \quad\text{for}\quad 0\le p\le \frac43 }.$$
∎
Verification of Key Steps
The first delicate step is the deduction that $x\ge 1$. The domain condition alone gives
$$|x|\ge 1.$$
A careless argument may stop here and permit negative values of $x$. However, the left-hand side of the equation is nonnegative, so the right-hand side must also be nonnegative. Consequently
$$x\ge 0.$$
Combining the two conditions yields
$$x\ge 1.$$
Without this correction, the later passage from $x^2$ to $x$ would incorrectly introduce a negative branch.
The second delicate step is the second squaring. Starting from
$$4x\sqrt{x^2-1}=4x^2+p-4,$$
one must confirm that the right-hand side is nonnegative before squaring. The left-hand side is nonnegative because $x\ge 1$. Hence squaring preserves equivalence. Omitting this verification risks introducing additional false solutions.
The third delicate step is simplifying $\sqrt{x^2-p}$. Substituting the formula for $x^2$ gives
$$x^2-p=\frac{(3p-4)^2}{8(2-p)}.$$
Taking square roots yields
$$\sqrt{x^2-p} = \frac{|3p-4|}{2\sqrt{2(2-p)}}.$$
Replacing $|3p-4|$ by $3p-4$ would incorrectly eliminate all parameters satisfying
$$0\le p<\frac43.$$
The absolute value is the mechanism that separates genuine solutions from extraneous ones.
Alternative Approaches
A different strategy begins with the substitution
$$x=\cosh t, \qquad t\ge 0,$$
since $x\ge 1$. Then
$$\sqrt{x^2-1}=\sinh t,$$
and the equation becomes
$$\sqrt{\cosh^2 t-p}+2\sinh t=\cosh t.$$
Using
$$\cosh t-\sinh t=e^{-t},$$
one obtains
$$\sqrt{\cosh^2 t-p}=2e^{-t}-\cosh t.$$
This eventually leads to an algebraic condition on $e^t$. The method works, but the hyperbolic substitutions obscure the parameter dependence and create heavier algebra.
Another approach introduces
$$u=\sqrt{x^2-p}, \qquad v=\sqrt{x^2-1}.$$
Then
$$u+2v=x,$$
together with
$$u^2-v^2=1-p.$$
Solving the linear equation for $u$ and substituting into the quadratic relation yields a polynomial equation in $v$. This method also succeeds, but it requires more elimination steps than the direct squaring argument.
The chosen approach is preferable because it isolates the single obstruction introduced by squaring, namely the absolute value in
$$\sqrt{x^2-p},$$
and this obstruction immediately yields the exact parameter range.