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

Question

This question concerns the equation $$\frac{a}{a^2+1} + \frac{b}{b^2+1} = \frac{c}{c^2+1}$$ and the possibility of rational number solutions with $abc \ne 0$.

In comments arising from: Using Modularity Theorem and Ribet's Theorem to disprove existence of rational solutions
It is mentioned that this equation has an infinite number of such solutions.

How can we determine that there are an infinite number of such solutions?

And is there a way to generate these solutions (or at least find a few of them)?

Answer 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 $$ a = \frac{-c^6 + 3c^4 + c^2 + 1}{c^7 + c^5 + 3c^3 - c}, b = \frac{c^6 - c^4 - c^2 + 1}{2(c^5 + c)}. $$

Answer 2

-b/(bb+1)=a/(aa+1)-c/(cc+1). Put a=0. Then b/(bb+1)=c/(cc+1). Then b=c. Infinity number of rational solutions. We can replace the problem with -f(b)/f(b)f(b)+1=f(a)/(f(a)f(a)+1)-f(c)/(f(c)f(c)+1) such that f(0)=0 and f is 1:1.