$(x^2+1)(y^2+1)=z^2+1$

Prove that $$(x^2+1)(y^2+1)=z^2+1$$ has infinitely many solutions when $$x,y,z \in \Bbb N$$

I couldn't even find one solution for this equation.

Any ideas on how to prove it?

$$(n^2+1)((n+1)^2+1)=(n^2+n+1)^2+1$$

• (+1) Category: "proof without words" Dec 7, 2019 at 21:00
• So elegant, yep “proof without words”. Dec 9, 2019 at 12:03

$$\boxed{\color{red}{(x^2+1)(4x^4+1)= (2x^3+x)^2+1}}$$

Some motivation for that "proof without words" $$z^2 = x^2+y^2+x^2y^2$$

Let $$y=xt$$ then we have $$z^2 = x^2(1+t^2+x^2t^2)$$

so if we put $$t=2x$$ we get $$z^2 = x^2(1+4x^2+4x^4)=x^2(2x^2+1)^2$$

So we have triples $$(x,2x^2,2x^3+x)$$ where $$x$$ is arbitrary integer.

Actually we can find all the solutions $$(x,y,z) \in {\mathbb Z}^3$$ of the given equation $$(x^2+1)(y^2+1)=z^2+1$$. Note right from the start that if $$(x,y,z)$$ is a solution, so is any $$(x', y', z')$$ obtained by changing independently the signs of $$x,y,z$$. It will be convenient to introduce the Gaussian ring $$\mathbb Z[i]= \{a+bi\mid a,b \in \mathbb Z\}$$, and the norm map $$N:\mathbb Q(i)\to \mathbb Q$$, defined by $$x+yi \to (x+yi)(x-yi)=x^2+y^2$$, with $$x,y \in \mathbb Q$$. This norm is obviously a multiplicative function.

A solution $$(x,y,z)\in {\mathbb Z}^3$$ of our equation is s.t. $$N(x+i)N(y+i)=N(z+i)$$, or equivalently, $$N((x+i)(y+i)/(z+i))=1$$. In a Galois extension with cyclic group generated by $$\sigma$$, an element $$\alpha$$ has norm $$1$$ iff it is of the form $$\beta/\sigma(\beta)$$ : this is Hilbert's thm.90, which can be shown "by hand" in the quadratic case. So $$(x+i)(y+i)/(z+i)=(m+ni)/(m-ni)$$ here, with $$(0,0)\neq (m,n) \in \mathbb Q$$ a priori (we could take $$m,n\in \mathbb Z$$ because of homogeneity). Clearing denominators and identifying the real and imaginary parts, we get a "parametrization" system consisting of two equations, $$ma+nb=0, mb-na=0$$, with $$a=z+xy-1$$ and $$b=x+y-1$$. Since $$(m,n)\neq (0,0)$$, the determinant $$a^2+b^2$$ is null, i.e. $$a=b=0$$, so $$m,n$$ are arbitrary and play no role.

Summarizing : all the solutions in $$\mathbb Z^3$$ of our diophantine equation consist of all the triples $$(x,y,z)$$ with $$y=-x+1, z=-x^2+x-1$$, and all the triples obtained by changing independently the signs of $$x,y,z$$. In particular, we recover the triples $$(a,a+1,a^2+a+1)$$ given by @Don Thousand.

• Great answer! I think some spacing could make the middle paragraph a bit more digestible. Also, for your norm map, I don't see why the function's codomain is $\mathbb N$, as $x^2+y^2$ isn't always in $\mathbb N$. The argument works just as well if you expand the codomain. Dec 9, 2019 at 13:57
• True. The misprint comes from the fact that originally I had defined the norm map only for the Gaussian integers. I edit that. Dec 9, 2019 at 14:06