FAA2533 wrote: Let $a$ and $b$ be real numbers such that$$\frac{a}{a^2-5} = \frac{b}{5-b^2} = \frac{ab}{a^2b^2-5}$$where $a+b \neq 0$. $a^4 + b^4 =$ ? Let $\frac 1c$ the common value. We have three equations : $a^2-ac-5=0$ $b^2+bc-5=0$ $(ab)^2-(ab)c-5=0$ First and third imply that $a$ and $ab$ both are roots of $X^2-cX-5=0$ and so : : Either $b=1$ (same root), either $a(ab)=-5$ and so $b=-\frac 5{a^2}$ Subtracting first and second, we get $(a+b)(a-b-c)=0$ and so, since $a+b\ne 0$ : $b=a-c=\frac 5a$ (from first) And so two cases : 1) $b=1$ and $b=\frac 5a$ and so $(a,b)=(5,1)$ which indeed fits 2) $b=-\frac 5{a^2}$ and $b=\frac 5a$ and so $(a,b)=(-1,-5)$, which indeed fits And in both cases : $\boxed{a^4+b^4=626}$

Replacing $b$ with $-b$ we got $\frac{a}{a^2-5}=\frac{b}{b^2-5}=-\frac{ab}{a^2b^2-5}$ ans easy to see, that $ab=-5$ $a+b=\frac{a^2-b^2}{a-b}=\frac{a^2-5}{a}=4$ $a^2+b^2=(a+b)^2-2ab=26$ $a^4+b^4=(a^2+b^2)-2a^2b^2=626$

Good problem.

Compute the real value of $a$ such that$$\sqrt{a(101b + 1)} - 1 = \sqrt{b(c - 1)}+ 10\sqrt{(a - c)b}.$$

By cross multiplication, we have \[5a-ab^2=a^2b-5b\implies 5(a+b)=ab(a+b)\implies ab=5\]Hence, \[\frac{a}{a^2-5}=\frac{b}{5-b^2}=\frac{1}{4}\]Again \[\frac{a}{a^2-5}=\frac{1}{4}\implies a^2-4a-5=0\implies a\in \{-1,5\}\]Similarly, $b\in \{1,-5\}$. Therefore, $(a,b)=(1,5)\text{ or }(-1,-5)\text{ or }(5,1)\text{ or }(-5,-1)$ as $ab=5$. Which implies \[\boxed{a^4+b^4=626}\]