Question:
What are the solutions to $$a^2+b^2+c^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$$ where $a,b,c\in \mathbb{Q}$ are rational numbers. There is the easy solution $a=b=c=\pm 1$. And in fact, if one variable is $\pm1$, say $c=\pm1$, then we get a family of solutions with $b=\pm\frac{1}{a}$. So I'd like to know: Are there other solutions where $a,b,c\not=\pm 1$?
My Three (Edit: Four) Attempts:
(1) Firstly, if $c=\pm1$, then we get a simpler equation. $$a^2+b^2=\frac{1}{a^2}+\frac{1}{b^2}$$ We can multiply this through by $a^2b^2$ to get $$a^2b^2(a^2+b^2)=a^2b^2\Big(\frac{1}{a^2}+\frac{1}{b^2}\Big)=a^2+b^2$$ Thus either $a^2+b^2=0$ or $a^2b^2=1$. The first case would imply $a=b=0$, creating a divide-by-zero error in the original equation. The second case gives us the $b=\pm \frac{1}{a}$ family of solutions.
I tried to recreate this line of logic with all three variables. That is, we can multiply the original equation through by $a^2b^2c^2$ to get $$a^2b^2c^2(a^2+b^2+c^2)=a^2b^2c^2\Big(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\Big)=b^2c^2+a^2c^2+a^2b^2$$ I tried to manipulate this further with some symmetric polynomial identities like $$2(b^2c^2+a^2c^2+a^2b^2)=(a^2+b^2+c^2)^2-(a^4+b^4+c^4)$$ which gets us $$2a^2b^2c^2(a^2+b^2+c^2)=(a^2+b^2+c^2)^2-(a^4+b^4+c^4)$$ and this can be rearranged to $$a^4+b^4+c^4=(a^2+b^2+c^2)(a^2+b^2+c^2-2a^2b^2c^2)$$ which looks useful but I didn't see where to go from here.
(2) I noticed that if $\frac{a^2+1}{a^2-1}$ is a perfect square, say $d^2=\frac{a^2+1}{a^2-1}$ then we have a strange identity $$d^2-\frac{1}{d^2}=\frac{(a^2+1)^2-(a^2-1)^2}{(a^2+1)(a^2-1)}=\Big(\frac{a^4-1}{4a^2}\Big)^{-1}=4\Big( a^2-\frac{1}{a^2}\Big)^{-1}$$ Or in other words, we have $$\Big( a^2-\frac{1}{a^2}\Big)\Big( d^2-\frac{1}{d^2}\Big)=4$$ Again, I wasn't sure where exactly to go with this but it seemed useful since the original equation can be rewritten as $$\Big( a^2-\frac{1}{a^2}\Big)+\Big( b^2-\frac{1}{b^2}\Big)+\Big( c^2-\frac{1}{c^2}\Big)=0$$
(3) My last attempt was to work with integers instead of fractions by defining $a,b,c=\frac{u_a}{v_a}, \frac{u_b}{v_b}, \frac{u_c}{v_c}$. Plugging this into the original equation, collecting terms, clearing denominators, etc... we get $$(u_a^4-v_a^4)u_b^2v_b^2u_c^2v_c^2+u_a^2v_a^2(u_b^4-v_b^4)u_c^2v_c^2+u_a^2v_a^2u_b^2v_b^2(u_c^4-v_c^4)=0$$ To cut down on the clutter here, we can define $P_j=u_j^2v_j^2$ and $Q_j=u_j^4-v_j^4$ with $j=a,b,c$. To get $$Q_aP_bP_c+P_aQ_bP_c+P_aP_bQ_c=0$$ Where our original constraint that $a,b,c\not = \pm1$ I think now becomes $Q_a, Q_b, Q_c\not= 0$ since, to take $a$ for example, $$a^2-\frac{1}{a^2}=\frac{u_a^2}{v_a^2}-\frac{v_a^2}{u_a^2}=\frac{u_a^4-v_a^4}{u_a^2v_a^2}=\frac{Q_a}{P_a}$$ And here I was able to make a significant insight. If a non-trivial solution exists, we can pick each $\frac{u_j}{v_j}$ to be in reduced form and still have a solution. That is, we can always pick $u_j$ and $v_j$ coprime. This in turn implies each $P_j$ and $Q_j$ are coprime since if not, we would have that $u_a^4-v_a^4$ and $u_a$ (or something similar of this form) are both multiples of some prime $p$, implying $v_a^4$ must also be a multiple of $p$, implying $u_a$ and $v_a$ weren't actually coprime.
This has the side-effect that no prime can divide just one of $P_a, P_b, P_c$. If, for instance, a prime $p$ only divided $P_a$ then clearly $P_aQ_bP_c+P_aP_bQ_c$ would be a multiple of $p$ but $Q_aP_bP_c$ would not (by assumption and by coprimality of the $P$'s and $Q$'s). And we would have an equation $$Q_aP_bP_c=-P_aQ_bP_c-P_aP_bQ_c$$ where the left is not a multiple of $p$ and the right is. More strictly, no $P_j$ may be more divisible by any one prime than the other two. That is, if you think of divisibility by a given prime as a race between three competitors $P_a, P_b, P_c$, there must always be a tie for first place.
But again, I didn't see any way to show that this means there are no non-trivial solutions; only that solutions must satisfy some rather tricky prime factorization constraints.
Edit: (4) I forgot to include when first posting that quaternions offer another opportunity of attack. The sum of three squares itself alway factors as perfect square in the quaternions since $$a^2+b^2+c^2=-(ai+bj+ck)^2$$ Thus the original problem may be rephrased as $$(ai+bj+ck)^2=\Big(\frac{i}{a}+\frac{j}{b}+\frac{k}{c}\Big)^2$$ At first glance, I thought this solved the problem since in $\mathbb{Z}$ (and even in $\mathbb{Z}[i]$) having $x^2=y^2$ implies $x=\pm y$. Thus in our case, we would have $$\{a, b, c\}=\pm\Big\{\frac{1}{a},\frac{1}{b},\frac{1}{c}\Big\}$$ which yields only trivial solutions. However, this isn't necesarily the case because the quaternions are not commutative and one can have solutions to $x^2=y^2$ where $x\not=\pm y$. For example $$(i+5j+6k)^2=(2i+3j+7k)^2=-62$$ but obviously $\{1,5,6\}\not=\pm\{2,3,7\}$. And I wasn't sure where to go from here.