System of irrational equations

Question

Given that $a$, $b$ and $c$ are positive real numbers that satisfy $b = \dfrac{64a}{a^2 - 64}= \dfrac{81c}{2c^2 - 81}= \sqrt{a^2 + c^2}$, find $b$.

Answer 1

Squaring yields two polynomial equations in $a$ and $c$, $$ a^6 + a^4c^2 - 128a^4 - 128a^2c^2 + 4096c^2=0, $$ and $$ - 81a^2c + 128ac^2 - 5184a + 5184c=0. $$ Over the complex numbers all solutions can be computed by using Groebner bases. Among them the positive real solutions are $(a,b,c)=(0,0,0)$, $(a,b,c)=(24/\sqrt{5},30/\sqrt{5}, 18/\sqrt{5})$. All other solutions are either real with one of the values $a,b,c$ negative, or non-real solutions. We have $b^2=a^2+c^2$, which is, up to a factor here $30^2=24^2+18^2$, i.e., $5^2=4^2+3^2$.