rational number solutions to $\frac{a}{a^2+1} + \frac{b}{b^2+1} = \frac{c}{c^2+1}$ with $abc\ne 0$

Solution 1:

Let $C$ be the curve defined by the equation $$3(a^2 + 1)(b^2 + 1) = 10( a(b^2 + 1) + b(a^2 + 1)),$$ so rational points on $C$ correspond to solutions of your original equation with $c = 3$. Let $E$ be the elliptic curve with equation $$y^2 = x^3 + x^2 + 65x + 3458.$$

The point is that $E$ and $C$ are "birationally equivalent" over the rational numbers $\mathbf{Q}$ -- there are maps given by ratios of polynomials with $\mathbf{Q}$-coefficients from $E$ to $C$ and back again, which are well-defined away from a finite set of bad points. (I won't write down the polynomials explicitly; I got the computer program Magma to compute them for me).

Now, elliptic curves over $\mathbf{Q}$ are very well-studied objects, and we can do (or, rather, ask Magma to do) a "descent" computation to find that $E$ has infinitely many rational points and enumerate the first few. Then we just feed them into the map from $E$ to $C$ to get rational points on $C$. This gives the following solutions: $$ (a, b) = (-627/119, 160/123), (-1287933/7579360, 20097649/13679883), (176228877641067/9078545489689, 14356927982089/53923856957760), \dots $$

If $(a, b)$ is any such solution then clearly $(a, b, 3)$ is a solution to your original problem. So your equation has infinitely many rational solutions.

(I got to this by a little trial and error: I spotted that for a fixed $c$ the equation in $a, b$ gave an elliptic curve, and I tried a few small values of $c$ until I found an elliptic curve with nontrivial rational points. Here is the Magma session:

Magma V2.21-4     Thu Oct  8 2015 10:14:42 on fermat   [Seed = 3701259969]
Type ? for help.  Type <Ctrl>-D to quit.
> A<x, y> := AffineSpace(Rationals(), 2);
> S := Curve(A, 3/(3^2 + 1)*(x^2 + 1)*(y^2 + 1) - (x*(y^2 + 1) + y*(x^2 + 1)));
> Sp := ProjectiveClosure(S);
> E, phi := EllipticCurve(Sp, Sp ! [3, 0, 1]);
> Rank(E);
1 true
> _, nu := IsInvertible(phi);
> nu(2*Generators(E)[2]);
> (-627/119 : 123/160 : 1)
> nu(3*Generators(E)[2]);
(-1287933/7579360 : 13679883/20097649 : 1)

Try it yourself at the online Magma calculator. )

PS. One can even write down infinitely many parametric families of solutions: for instance, for any rational $c$ we can take $$ a = \frac{-c^6 + 3c^4 + c^2 + 1}{c^7 + c^5 + 3c^3 - c}, b = \frac{c^6 - c^4 - c^2 + 1}{2(c^5 + c)}. $$