## Task 3A - 341233A

Determine all pairs $(x, y)$ of real numbers $x, y$ that satisfy the following equations (1) and (2):

$$
\begin{aligned}
& \sin ^{4} x=y^{4}+x^{2} y^{2}-4 y^{2}+4 \\
& \cos ^{4} x=x^{4}+x^{2} y^{2}-4 x^{2}+1
\end{aligned}
$$
We substitute:

$$
y^{2}+x^{2}=a \quad y^{2}-x^{2}=b
$$

or

$$
x^{2}=\frac{1}{2}(a-b) \quad y^{2}=\frac{1}{2}(a+b)
$$

Then it follows from (1):

$$
\sin ^{4} x=\frac{1}{2}(a+b)(a-4)+4
$$

and from (2):

$$
\cos ^{4} x=\frac{1}{2}(a-b)(a-4)+1
$$

We subtract the two equations (3) and (4) and use $\sin ^{2} x+\cos ^{2} x=1$:

$$
\begin{gathered}
\sin ^{4} x-\cos ^{4} x=\left(\sin ^{2} x+\cos ^{2} x\right)\left(\sin ^{2} x-\cos ^{2} x\right)=\sin ^{2} x-\cos ^{2} x \\
\sin ^{2} x-\cos ^{2} x=b(a-4)+3 \\
2 \sin ^{2} x=b(a-4)+4
\end{gathered}
$$

$$
2 \cos ^{2} x=-b(a-4)-2
$$

We substitute (5a) into (3):

$$
\begin{gathered}
4\left(\frac{1}{2}(a+b)(a-4)+4\right)=(b(a-4)+4)^{2} \\
2(a+b)(a-4)+16=b^{2}(a-4)^{2}+8 b(a-4)+16 \\
2(a+b)=b^{2}(a-4)+8 b
\end{gathered}
$$

In the same way, we get, if we substitute (5b) into (4):

$$
\begin{gathered}
4\left(\frac{1}{2}(a-b)(a-4)+1\right)=(b(a-4)+2)^{2} \\
2(a-b)=b^{2}(a-4)+4 b
\end{gathered}
$$

We add (6) and (7):

$$
\begin{gathered}
4 a=2 b^{2}(a-4)+12 b \\
2 a=b^{2} a-4 b^{2}+6 b \\
\left(2-b^{2}\right) a=b(6-4 b) \\
a=\frac{b(6-4 b)}{2-b^{2}}
\end{gathered}
$$

Substituting (8) into (6):

$$
2\left(\frac{b(6-4 b)}{2-b^{2}}+b\right)=b^{2}\left(\frac{b(6-4 b)}{2-b^{2}}-4\right)+8 b
$$

Dividing by $2 b$:

$$
\begin{gathered}
\frac{6-4 b}{2-b^{2}}+1=b\left(\frac{b(3-2 b)}{2-b^{2}}-2\right)+4 \\
6-4 b+2-b^{2}=b^{2}(3-2 b)+(4-2 b)(3-2 b) \\
8-4 b-b^{2}=3 b^{2}-2 b^{3}+12-6 b-8 b+4 b^{2} \\
0=b^{3}-4 b^{2}+5 b-2
\end{gathered}
$$

One solution that is easily guessed is $b=1$. Polynomial division by $(b-1)$ yields

$$
b^{2}-3 b+2=0
$$

with solutions $b=1$ (double root) and $b=2$, so we have the following solutions overall, using (8):

$$
b_{1}=1 \quad a_{1}=2 \quad \text { and } \quad b_{2}=2 \quad a_{2}=2
$$

Solution a): Substituting into the substitution, we get

$$
y^{2}=\frac{3}{2} \quad x^{2}=\frac{1}{2}
$$

But according to (5a), it must also hold that

$$
2 \sin ^{2} x=2
$$

which is incompatible with

$$
x^{2}=\frac{1}{2}
$$

This is therefore only a spurious solution.

Solution b): From the substitution, we get

$$
x^{2}=0 \quad y^{2}=2
$$

$x=0$ is also consistent with (5a) and (5b).

We therefore have the only solutions $(x ; y) \in\{(0 ; \sqrt{2}),(0 ;-\sqrt{2})\}$.