No Way Out Of It

$(16x^{200}+1)(y^{200}+1)=16(xy)^{100}\;,\;\; x,y\in \mathbb{R}$ Let $x^{100}=a$ and $y^{100}=b$ , where $a,b$ are non-negative reals. So, we have, $\begin{aligned} (16a^2+1)(b^2+1)&=16ab \\ 16a^2b^2+16a^2+b^2+1&=16ab \end{aligned}$ Applying AM-GM inequality on the terms $16a^2b^2,\;16a^2,\;b^2,\;1$ : $\frac{16a^2b^2+16a^2+b^2+1}{4}\ge \sqrt[{4}]{16a^2b^2\cdot 16a^2\cdot b^2\cdot1}=4ab$ $\implies 16a^2b^2+16a^2+b^2+1\ge 16ab\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;$ Since $\; 16a^2b^2+16a^2+b^2+1=16ab,$ from the equality condition of AM-GM inequality, $16a^2b^2=16a^2=b^2=1 \; \implies (a,b)=\left(\frac{1}{4},1\right)$ $\therefore \; (x,y)\in \left \{\left(\sqrt[100]{\frac{1}{4}},1\right),\; \left(\sqrt[100]{\frac{1}{4}},-1\right),\; \left(-\sqrt[100]{\frac{1}{4}},1\right),\; \left(-\sqrt[100]{\frac{1}{4}},-1\right)\right\}$ Therefore, there are $\boxed{4}$ pairs of reals $(x,y)$ that satisfy the given equation.

$\begin{aligned} (16x^{200}+1)(y^{200}+1) & = 16(xy)^{100} \\ (16x^{200}+1)y^{200} - 16x^{100}y^{100} + 16x^{200}+1 & = 0 \\ \implies y^{100} & = \frac {8x^{100} \pm \sqrt{-(16x^{200}-1)^2}}{16x^{200}+1} & \small \blue{\text{For real }x, (16x^{200}-1)^2 \ge 0} \\ & = \frac {8x^{100} \pm (16x^{200}-1)i}{16x^{200}+1} \end{aligned}$

For $y$ to be real,

$\begin{aligned} \implies 16x^{200} - 1 & = 0 \\ (4x^{100}-1)(4x^{100} + 1) & = 0 \\ (2x^{50}-1)(2x^{50}+1)(4x^{100} + 1) & = 0 \\ \implies x & = \pm \frac 1{\sqrt[50] 2} & \small \blue{\text{For real }x} \end{aligned} \\ \implies \begin{cases} x = \dfrac 1{\sqrt[50]2} & y = 1 \\ x = \dfrac 1{\sqrt[50]2} & y = - 1 \\ x = - \dfrac 1{\sqrt[50]2} & y = 1 \\ x = - \dfrac 1{\sqrt[50]2} & y = - 1 \end{cases}$

Therefore there are $\boxed 4$ real solutions.

For simplicity, let $X:=4x^{100}$ and $Y:=y^{100}$ with $X,\:Y\geq 0$ . Then the equation simplifies to $\begin{aligned} 4XY&=(1+X^2)(1+Y^2) \end{aligned}$ We check the cases $X=0$ and $Y=0$ and notice they can never be a solution, so we have $X,\:Y>0$ and may divide by $XY$ : $\begin{aligned} X,\:Y&>0:&&& 4&=(X + X^{-1})(Y+Y^{-1})=\left( 2+\Bigl(X^{\frac{1}{2}}-X^{-\frac{1}{2}}\Bigr)^2 \right)\left( 2+\Bigl(Y^{\frac{1}{2}}-Y^{-\frac{1}{2}}\Bigr)^2 \right) \end{aligned}$ The cases $X\neq 1$ or $Y\neq 1$ can never lead to a solution, as the product on the right will be greater than $4$ . The remaining case $X=Y=1$ leads to solutions $(x,\:y)\in \left\{(\pm 2^{-50},\:\pm 1)\right\}=:M \quad\Rightarrow\quad |M|=\boxed{4}$

Let a,b be real numbers. Then we can have, (ab-1)²+(a-b)²>=0 => a²b²+a²+b²+1>=4ab => (a²+1)(b²+1)>=4ab. Equality holds if a-b=0 and ab=1 => a=b=1. Put a=4x¹⁰⁰,b=y¹⁰⁰ to get given inequality. Since x,y have even powers they can also have negative values, thus 4 total solutions.