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Find the number of real solutions to the equation:

$$\left(16 x^{200}+1\right)\left(y^{200}+1\right)=16(x y)^{100}$$
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$$\left(16 x^{200}+1\right)\left(y^{200}+1\right)=16(x y)^{100}, x, y \in \mathbb{R}$$
Let $x^{100}=a$ and $y^{100}=b$, where $a, b$ are non-negative reals. So, we have.
\begin{aligned} \left(16 a^{2}+1\right)\left(b^{2}+1\right) &=16 a b \\ 16 a^{2} b^{2}+16 a^{2}+b^{2}+1 &=16 a b \end{aligned}
Applying AM-GM inequality on the terms $16 a^{2} b^{2}, 16 a^{2}, b^{2}, 1$ :
$$\frac{16 a^{2} b^{2}+16 a^{2}+b^{2}+1}{4} \geq \sqrt[4]{16 a^{2} b^{2} \cdot 16 a^{2} \cdot b^{2} \cdot 1}=4 a b$$
$$\Longrightarrow 16 a^{2} b^{2}+16 a^{2}+b^{2}+1 \geq 16 a b$$
Since $16 a^{2} b^{2}+16 a^{2}+b^{2}+1=16 a b$, from the equality condition of AM-GM inequality,
$$\begin{array}{c} 16 a^{2} b^{2}=16 a^{2}=b^{2}=1 \Longrightarrow(a, b)=\left(\frac{1}{4}, 1\right) \\ \therefore(x, y) \in\left\{\left(\sqrt[100] \frac{1}{4}, 1\right),\left(\sqrt[100]{\frac{1}{4}},-1\right),\left(-\sqrt[100]{\frac{1}{4}}, 1\right),\left(-\sqrt[100]{\frac{1}{4}},-1\right)\right\} \end{array}$$
Therefore, there are 4 pairs of reals $(x, y)$ that satisfy the given equation.
by Diamond (75,914 points)

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