Kvant Math Problem 603
The denominators suggest introducing
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m39s
Source on kvant.digital
Problem
Solve the system of equations $$x+\dfrac{3x-y}{x^2+y^2}=3,\quad y-\dfrac{x+3y}{x^2+y^2}=0.$$
L. P. Kuptsov
Exploration
The denominators suggest introducing
$$r=x^2+y^2.$$
Since the system contains the expressions $\dfrac{3x-y}{r}$ and $\dfrac{x+3y}{r}$, multiplying by $r$ should convert the system into linear relations between $x,y$ and $r$.
Multiplying both equations by $r$ gives
$$xr+3x-y=3r,$$
$$yr-x-3y=0.$$
The second equation immediately yields
$$x=y(r-3).$$
Substituting into the first equation produces a relation involving only $y$ and $r$:
$$y\bigl((r-3)(r+3)-1\bigr)=3r,$$
hence
$$y(r^2-10)=3r.$$
Thus
$$y=\frac{3r}{r^2-10}, \qquad x=\frac{3r(r-3)}{r^2-10}.$$
Now impose the definition $r=x^2+y^2$. Substituting these expressions gives a single equation for $r$:
$$r=\frac{9r^2\bigl((r-3)^2+1\bigr)}{(r^2-10)^2}.$$
Since $r=x^2+y^2\ge0$, the case $r=0$ must be checked separately. Direct substitution into the original system is impossible because the denominators vanish, so $r>0$.
After canceling $r$:
$$(r^2-10)^2=9r\bigl((r-3)^2+1\bigr).$$
Expanding,
$$r^4-29r^2+54r+100=0.$$
Trying small integer values, $r=2$ gives
$$16-116+108+100=0.$$
Dividing by $r-2$ yields
$$r^3+2r^2-25r-50=(r+2)(r^2-25).$$
Hence
$$(r-2)(r+2)(r-5)(r+5)=0.$$
Because $r\ge0$, only $r=2$ and $r=5$ remain.
For $r=2$,
$$y=\frac{6}{-6}=-1,\qquad x=-1(-1)=1.$$
For $r=5$,
$$y=\frac{15}{15}=1,\qquad x=2.$$
Both satisfy $x^2+y^2=r$.
The potentially dangerous step is the cancellation of $r$; one must first exclude $r=0$ from the domain.
Problem Understanding
We must determine all real pairs $(x,y)$ satisfying
$$x+\frac{3x-y}{x^2+y^2}=3, \qquad y-\frac{x+3y}{x^2+y^2}=0.$$
This is a Type A problem, a classification problem. We must find every solution and prove that no others exist.
The core difficulty is eliminating the nonlinear denominators without losing solutions. Introducing
$$r=x^2+y^2$$
reduces the system to linear equations in $x$ and $y$ with parameter $r$, after which the condition $r=x^2+y^2$ yields an algebraic equation for $r$.
The expected answer is
$$(x,y)=(1,-1),\quad (2,1),$$
because the parameter reduction leads to only two admissible values of $r$.
Proof Architecture
First, set $r=x^2+y^2$ and observe that $r>0$ because the original denominators must be defined.
Second, derive the linear relation
$$x=y(r-3)$$
from the second equation after clearing denominators.
Third, substitute this relation into the first cleared equation to obtain
$$y(r^2-10)=3r,$$
and hence explicit formulas for $x$ and $y$ in terms of $r$.
Fourth, substitute those formulas into $r=x^2+y^2$ and derive the quartic equation
$$(r-2)(r+2)(r-5)(r+5)=0.$$
Fifth, use $r\ge0$ to obtain $r=2$ or $r=5$, then recover the corresponding pairs $(x,y)$.
The most delicate point is deriving and solving the equation for $r$ without dividing by a quantity that might vanish. The exclusion of $r=0$ must be handled before cancellation.
Solution
Let
$$r=x^2+y^2.$$
Since the denominators in the given system are $x^2+y^2$, every solution must satisfy
$$r>0.$$
Multiplying both equations by $r$ gives
$$xr+3x-y=3r, \tag{1}$$
$$yr-x-3y=0. \tag{2}$$
Equation (2) can be written as
$$x=y(r-3). \tag{3}$$
Substituting (3) into (1), we obtain
$$y(r-3)r+3y(r-3)-y=3r.$$
Factoring $y$,
$$y\bigl(r(r-3)+3(r-3)-1\bigr)=3r.$$
Since
$$r(r-3)+3(r-3)-1=r^2-10,$$
we get
$$y(r^2-10)=3r. \tag{4}$$
If $r^2-10=0$, then the left-hand side of (4) is $0$ while the right-hand side equals $3r>0$, impossible. Hence $r^2-10\ne0$, and
$$y=\frac{3r}{r^2-10}. \tag{5}$$
Using (3),
$$x=\frac{3r(r-3)}{r^2-10}. \tag{6}$$
Now impose the identity $r=x^2+y^2$. Substituting (5) and (6),
$$r = \frac{9r^2(r-3)^2}{(r^2-10)^2} + \frac{9r^2}{(r^2-10)^2}.$$
Thus
$$r= \frac{9r^2\bigl((r-3)^2+1\bigr)}{(r^2-10)^2}.$$
Because $r>0$, we may divide by $r$:
$$(r^2-10)^2 = 9r\bigl((r-3)^2+1\bigr).$$
Expanding both sides,
$$r^4-20r^2+100 = 9r^3-54r^2+90r.$$
Hence
$$r^4-9r^3+34r^2-90r+100=0.$$
Substituting $r=2$ gives zero, so $r-2$ is a factor. Dividing,
$$r^4-9r^3+34r^2-90r+100 = (r-2)(r^3-7r^2+20r-50).$$
The cubic factor vanishes at $r=5$, and division by $r-5$ yields
$$r^3-7r^2+20r-50=(r-5)(r^2-2r+10).$$
Therefore
$$(r-2)(r-5)(r^2-2r+10)=0.$$
Since
$$r^2-2r+10=(r-1)^2+9>0,$$
the only real roots are
$$r=2,\qquad r=5.$$
For $r=2$, formulas (5) and (6) give
$$y=\frac{6}{-6}=-1, \qquad x=\frac{6(-1)}{-6}=1.$$
For $r=5$,
$$y=\frac{15}{15}=1, \qquad x=\frac{15\cdot2}{15}=2.$$
Both pairs satisfy $x^2+y^2=r$, so they are genuine solutions.
Thus the complete set of solutions is
$$\boxed{{(1,-1),(2,1)}}.$$
Verification of Key Steps
The first delicate step is the cancellation of $r$. The original system contains the denominator $x^2+y^2$, so every solution must satisfy $r=x^2+y^2>0$. Only after establishing this may one divide by $r$.
The second delicate step is solving equation (4). If $r^2-10=0$, equation (4) becomes
$$0=3r,$$
which contradicts $r>0$. Thus division by $r^2-10$ is legitimate. Omitting this check would risk losing solutions.
The third delicate step is the factorization of the quartic. Expanding
$$(r-2)(r-5)(r^2-2r+10)$$
gives
$$r^4-9r^3+34r^2-90r+100,$$
which matches the derived polynomial exactly. Since
$$r^2-2r+10=(r-1)^2+9,$$
the quadratic factor has no real roots, leaving only $r=2$ and $r=5$.
Alternative Approaches
A more structural method treats the system as a linear transformation. Writing
$$\begin{pmatrix} 3x-y\ x+3y \end{pmatrix} = \begin{pmatrix} 3&-1\ 1&3 \end{pmatrix} \begin{pmatrix} x\ y \end{pmatrix},$$
the original equations become
$$\begin{pmatrix} 3-x\ y \end{pmatrix} = \frac1r \begin{pmatrix} 3&-1\ 1&3 \end{pmatrix} \begin{pmatrix} x\ y \end{pmatrix}.$$
Multiplying by the inverse matrix and expressing $(x,y)$ through $(3-x,y)$ yields formulas depending on $r$. Substituting into $r=x^2+y^2$ again produces the same polynomial equation for $r$.
The direct substitution method used in the main solution is preferable because the second equation immediately gives the simple relation $x=y(r-3)$, reducing the system to one parameter with minimal computation.