$$\exists a\in [1,n-1]\text{ such that }a(2n-a)\text{ is a square number }\iff \exists x,y\text{ such that }x^2+y^2=d\quad d\mid 2n$$$$x^2+y^2=d\iff d\text{'s prime factors which form }4k+3\text{ must occur to an even power}$$

If AM and GM are integers, then $a=tu^2,b=tv^2$ where $t $ is a square free and $ a\equiv b \pmod 2$. Let $u>v$. $AM=\frac{t(u^2+v^2)}{2}$ 1.) If $t $ is odd then $u \equiv v \pmod 2$ and $t_1=2t,u_1=\frac{u+v}{2},v_1=\frac{u-v}{2}$ is another solution with the same AM. $(a_1=t_1u_1^2 \ne tu^2$) 2.) If $t $ is even then another solution is $t_1=\frac t 2, u_1=u+v,v_1=u-v$ The answer is "yes".