$AM+GM+HM+RMS=66$ For two reals $a,b\in\mathbb{R}$ Arithmetic mean(AM), Geometric Mean(GM), Harmonic Mean(HM) and Root mean square (RMS) all are integers and 
$AM+GM+HM+RMS=66$ 
Find all such $a,b$
I have assumed $a+b=u$ and $ab=v$ and written all the means in the terms of $u,v$ and used trial and error after some analysis. But this is long method. Can there be a shorter method?
 A: $a+b=2u,ab=v^2 \implies AM=u,GM=v.HM=\dfrac{2v^2}{2u}=\dfrac{v^2}{u},RMS=\sqrt{2u^2-v^2}$
$HM$ is integer ,$\implies u=tq^2, t $ is not a perfect square number $  v=tqp \implies RMS=\sqrt{2t^2q^4-t^2q^2p^2}=tq\sqrt{2q^2-p^2} \implies tq^2+tqp+tp^2+tq\sqrt{2q^2-p^2}=66 \implies t(q^2+pq+p^2+q\sqrt{2q^2-p^2})=66 $
$u \ge v \implies q\ge p $,if $q=1 \implies p=1 ,t= \dfrac{33}{2}$, it is not possible . 
$q\ge2 \implies (q^2+pq+p^2+q\sqrt{2q^2-p^2})> 10, q^2+pq+p^2+q\sqrt{2q^2-p^2}$is mono increasing function for $q,p,\implies t=1,2,3,6$
look at $\sqrt{2q^2-p^2}$, there is a general possible solution $p=q \implies \sqrt{2q^2-p^2}=q$, but in this case, $4tq=66$ which is not possible also.
$t=1,(q^2+pq+p^2+q\sqrt{2q^2-p^2})=66, q_{max}=5$
$q=5,p=1$, there is no other possibility.
$q=4$,but $\sqrt{2q^2-p^2}$ can't be a integer when $p=3,2,1$
$q=3$,but $\sqrt{2q^2-p^2}$ can't be a integer when $p=2,1$
$q=2$,but $\sqrt{2q^2-p^2}$ can't be a integer when $p=1$
when $t \ge 2, \implies q \le 4$, but we already know there is no solutions for all the cases. 
the only solution is $u=25,v=5 \iff a+b=50,ab=25 \implies 25\pm 10\sqrt{6}$
