Kvant Math Problem 103
The system is
Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m50s
Source on kvant.digital
Problem
Investigate how many solutions the system of equations $$\left{\begin{array}{l} x^2+y^2+xy=a,\ x^2-y^2=b, \end{array}\right.$$ has, where $a$ and $b$ are some real numbers.
E. A. Yasinovy̆ĭ
Exploration
The system is
$$\begin{cases} x^2+y^2+xy=a,\ x^2-y^2=b. \end{cases}$$
The first equation is a positive definite quadratic form, since
$$x^2+y^2+xy=\left(x+\frac y2\right)^2+\frac34y^2\ge 0,$$
with equality only at $(0,0)$. Hence $a\ge0$ is necessary for solvability.
The second equation factors:
$$(x-y)(x+y)=b.$$
Introducing
$$u=x+y,\qquad v=x-y$$
looks natural because the second equation becomes $uv=b$.
Expressing the first equation through $u,v$,
$$x=\frac{u+v}{2},\qquad y=\frac{u-v}{2},$$
gives
$$x^2+y^2+xy =\frac{3u^2+v^2}{4}.$$
Thus the system becomes
$$\begin{cases} 3u^2+v^2=4a,\ uv=b. \end{cases}$$
Eliminating $v$ by $v=b/u$ yields
$$3u^4-4au^2+b^2=0.$$
Setting $t=u^2\ge0$,
$$3t^2-4at+b^2=0.$$
The number of solutions should depend on the nonnegative roots of this quadratic.
The delicate point is translating the number of admissible roots $t$ into the number of pairs $(x,y)$. A positive root $t$ gives two values $u=\pm\sqrt t$, and then $v=b/u$ is determined. These two choices lead to two distinct pairs $(x,y)$ unless $(u,v)=(0,0)$, which can occur only when $a=b=0$.
Testing examples:
If $(a,b)=(1,0)$, then
$$3t^2-4t=t(3t-4),$$
so $t=0$ and $t=\frac43$. This gives three solutions:
$(0,0)$ and two nonzero ones.
If $(a,b)=\left(1,\frac12\right)$, then
$$3t^2-4t+\frac14=0$$
has two positive roots, hence four solutions.
This suggests that the maximum number of solutions is four.
Problem Understanding
We must determine, for arbitrary real parameters $a$ and $b$, how many real pairs $(x,y)$ satisfy the system
$$\begin{cases} x^2+y^2+xy=a,\ x^2-y^2=b. \end{cases}$$
This is a Type A problem. The task is to classify all parameter pairs $(a,b)$ according to the number of real solutions.
The core difficulty is converting the system into an equation whose roots can be counted and then relating those roots back to actual pairs $(x,y)$ without overlooking degenerate cases such as $u=0$ or $b=0$.
The answer will be expressed in terms of the discriminant of a quadratic equation in $t=(x+y)^2$.
Proof Architecture
Introduce $u=x+y$ and $v=x-y$, transforming the system into
$$3u^2+v^2=4a,\qquad uv=b.$$
Eliminating $v$ yields the quadratic equation
$$3t^2-4at+b^2=0,\qquad t=u^2\ge0.$$
The discriminant of this quadratic is
$$\Delta=16(a^2-3b^2).$$
If $a^2-3b^2<0$, there is no admissible $t$, hence no solutions.
If $a^2-3b^2=0$, there is one root $t=\frac{2a}{3}$; it gives two solutions when $b\ne0$ and one solution when $a=b=0$.
If $a^2-3b^2>0$, there are two distinct roots. Their signs determine the number of solutions. One proves that for $b\ne0$ both roots are positive, producing four solutions. For $b=0$ the roots are $0$ and $\frac{4a}{3}$, producing three solutions when $a>0$ and one solution when $a=0$.
The most delicate point is proving that when $b\ne0$ and $\Delta>0$, both roots of the quadratic in $t$ are positive.
Solution
Set
$$u=x+y,\qquad v=x-y.$$
Then
$$x=\frac{u+v}{2},\qquad y=\frac{u-v}{2}.$$
A direct computation gives
$$x^2-y^2=uv$$
and
$$x^2+y^2+xy =\frac{(u+v)^2+(u-v)^2+(u+v)(u-v)}4 =\frac{3u^2+v^2}{4}.$$
Hence the original system is equivalent to
$$\begin{cases} 3u^2+v^2=4a,\ uv=b. \end{cases}$$
First suppose $u\ne0$. From $uv=b$,
$$v=\frac bu.$$
Substituting into the first equation,
$$3u^2+\frac{b^2}{u^2}=4a.$$
Multiplying by $u^2$ and setting $t=u^2$, we obtain
$$3t^2-4at+b^2=0, \qquad t\ge0.$$
Conversely, every nonnegative root $t$ of this equation yields values
$$u=\pm\sqrt t, \qquad v=\frac bu,$$
which satisfy the transformed system.
The discriminant of the quadratic is
$$\Delta=16(a^2-3b^2).$$
Therefore solutions can exist only when
$$a^2-3b^2\ge0.$$
Since
$$a=x^2+y^2+xy =\left(x+\frac y2\right)^2+\frac34y^2\ge0,$$
the condition $a^2-3b^2\ge0$ already implies $a\ge0$ whenever solutions exist.
Consider the possible cases.
Assume first that
$$a^2-3b^2<0.$$
Then the quadratic in $t$ has no real roots, hence the system has no real solutions.
Now assume
$$a^2-3b^2=0.$$
Then the quadratic has the double root
$$t=\frac{2a}{3}.$$
If $a=b=0$, then $t=0$. The equations
$$3u^2+v^2=0,\qquad uv=0$$
give $u=v=0$, hence $(x,y)=(0,0)$. There is exactly one solution.
If $b\ne0$, then $a>0$ and $t=\frac{2a}{3}>0$. The two values
$$u=\pm\sqrt t$$
produce two corresponding values of $v=b/u$. These give two distinct pairs $(x,y)$. Thus there are exactly two solutions.
Finally assume
$$a^2-3b^2>0.$$
The quadratic has two distinct roots
$$t_{1,2} = \frac{2a\pm\sqrt{a^2-3b^2}}{3}.$$
If $b\ne0$, then
$$t_1t_2=\frac{b^2}{3}>0, \qquad t_1+t_2=\frac{4a}{3}>0.$$
Hence both roots are positive. Each root yields two choices of $u$, namely $\pm\sqrt t$, and for each such choice the value of $v=b/u$ is uniquely determined. Thus each root contributes exactly two solutions, for a total of four solutions.
If $b=0$, then
$$3t^2-4at=t(3t-4a).$$
Since $a^2-3b^2>0$, we have $a>0$. The roots are
$$t=0, \qquad t=\frac{4a}{3}.$$
The root $t=0$ gives $u=0$. From
$$3u^2+v^2=4a$$
we obtain
$$v=\pm2\sqrt a,$$
hence two solutions.
The root $t=\frac{4a}{3}$ gives
$$u=\pm2\sqrt{\frac a3},$$
and because $b=0$, the relation $uv=0$ forces $v=0$. This yields two more solutions.
Altogether there are four solutions when $b\ne0$ and $a^2-3b^2>0$, three solutions when $b=0$ and $a>0$.
Collecting all cases, the number $N(a,b)$ of real solutions is
$$N(a,b)= \begin{cases} 0,& a^2-3b^2<0,\[2mm] 1,& (a,b)=(0,0),\[2mm] 2,& a^2-3b^2=0,\ (a,b)\ne(0,0),\[2mm] 3,& b=0,\ a>0,\[2mm] 4,& a^2-3b^2>0,\ b\ne0. \end{cases}$$
Thus
$$\boxed{ N(a,b)= \begin{cases} 0,& a^2-3b^2<0,\ 1,& (a,b)=(0,0),\ 2,& a^2-3b^2=0,\ (a,b)\ne(0,0),\ 3,& b=0,\ a>0,\ 4,& a^2-3b^2>0,\ b\ne0. \end{cases}}$$
Verification of Key Steps
The first delicate step is the passage from $(u,v)$ to the quadratic in $t=u^2$. When $u=0$, division by $u$ is impossible. This case must be checked separately. If $u=0$, then $uv=b$ gives $b=0$, and the remaining equation becomes $v^2=4a$. These are exactly the solutions arising from the root $t=0$ when $b=0$. No solutions are lost.
The second delicate step is proving that both roots are positive when $a^2-3b^2>0$ and $b\ne0$. The product of the roots is
$$\frac{b^2}{3}>0,$$
so the roots have the same sign. Their sum is
$$\frac{4a}{3}.$$
Since $a^2-3b^2>0$, one has $a>\sqrt3,|b|>0$. Hence the sum is positive, forcing both roots to be positive.
The third delicate step is counting solutions produced by a positive root $t$. For a fixed positive root,
$$u=\sqrt t$$
and
$$u=-\sqrt t$$
lead to
$$v=\frac b{\sqrt t} \quad\text{and}\quad v=-\frac b{\sqrt t},$$
respectively. These correspond to distinct pairs $(x,y)$ because $(u,v)\mapsto(x,y)$ is an invertible linear transformation. Thus every positive root contributes exactly two solutions.
Alternative Approaches
Instead of introducing $u=x+y$ and $v=x-y$, one may solve directly for $x^2$ and $y^2$. Writing
$$x^2=\frac{a+b-xy}{2}, \qquad y^2=\frac{a-b-xy}{2},$$
and multiplying these expressions yields a quadratic equation for $xy$. After determining the possible values of $xy$, one reconstructs $x$ and $y$ from the sums and differences of their squares. This approach is longer because sign choices must be tracked carefully.
The substitution $u=x+y$, $v=x-y$ diagonalizes the second equation immediately and converts the first equation into a diagonal quadratic form. The resulting quadratic equation in $t=u^2$ makes both existence and counting transparent, which is why it is the most efficient method.