On the divisibility of $(x+y)^k - 1$ by $xy$ For a fixed $k \geq 2$, are there infinitely many non-trivial coprime integer pairs $(x,y)$ for which $xy$ divides $(x+y)^k-1$? By trivial I mean parametrized pairs $(x,y)$ of the form $(-1,-1),(x,1),(1,y),(x,1-x),(x,1 - x^n),(1-y^n,y),(x,p(x)),(p(y),y)$, where $p(x)$ is a polynomial factor of $\frac{x^k-1}{x-1}$. When $k$ is even there are a few more trivial pairs, namely $(x,-1),(-1,y),(x,-1-x),(x,x^n-1),(y^n-1,y)$; where $n>0$ is a multiple of $k$.
 A: For fixed $k$ and $y=\frac{x^k-1}{x-1}$ we have $$x^k \equiv 1 \pmod y$$and $$y \equiv 1 \pmod x$$ so also $$y^k\equiv 1 \pmod x.$$
That isn't one of your trivial examples, but maybe it should be. 
Actually, $y$ could be any polynomial factor of $\frac{x^k-1}{x-1}.$ For example,
$\frac{x^{30}-1}{x-1}=\Phi_3\Phi_5\Phi_{15}\Phi_{30}=$
$ (x^2+x+1)(x^4+x^3+x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)$$(x^8+x^7-x^5-x^4-x^3+x^2+1)$
So for $k=30$ one has $x,y$ where $y$ is any of the $16$ products of the factors.
Maybe you want a parametric family that is not $y=p(x).$ 
Interesting Variant To have $xy$ divide $(x+y)^2+1$ is quite interesting. I saw it in a movie (in another form). It may have been a math olympiad problem.
A: If $(x,y)$ is an integer solution, then $((1-y^k)/x,y)$ also is, iterating this produces many series of solutions starting from 'trivial'.
A: After doing some research online I found out that there's a lot of theory behind this subject and it was studied by several authors (see the references below).
As it was pointed out by @Geoff Robinson, the question essentially reduces to finding all coprime integer pairs $(x,y)$ such that $x \mid 1-y^k$ and $y \mid 1-x^k$.
It turns out that the solutions have a recursive pattern, and this phenomenon was observed by @Aaron Meyerowitz in one of the answers given above. From a single solution one can produce an infinitude of solutions, which form a so-called 1 chain. A 1 chain is a recursive sequence $u_n = (1-u_{n-1}^k)/u_{n-2}$, with non-zero $u_0, u_1$ being given (note that this definition of a 1 chain differs from [4]).
The following construction is a slight variation of Mohanty's argument [4]:
Suppose that $x$ and $y$ are coprime integers which satisfy $x \mid 1-y^k$ and $y \mid 1-x^k$. Then there exists a unique $r \in \mathbb Z$ such that $xr=1-y^k$. But then
$$xr \equiv 1 \pmod{y},$$
$$(xr)^k \equiv 1 \pmod{y},$$
and since $x^k \equiv 1 \pmod{y}$ we have
$$(xr)^k-x^k\equiv 1 - 1 \pmod{y},$$
which results in $x^k(r^k - 1) \equiv 0 \pmod{y}$. Since $\gcd(x,y)=1$, we conclude that $y \mid r^k - 1$. Since $r \mid y^k - 1$, we conclude that $(y,r)$ is also a solution. By proceeding in the same fashion we can generate an infinitude of solutions (see [4] for more details).
Mohanty himself studied a slightly different problem, namely he looked for coprime solutions $(x,y)$ such that $x \mid y^3+1$ and $y \mid x^3 + 1$. One can see that $(x,y)=(1,1)$ is a solution, and the sequence $u_0=1, u_1=1, u_n=(u_{n-1}^3+1)/u_{n-2}$ looks as follows:
$$1,1,2,9,365,5403014, \ldots$$
This is a sequence A003818 in OEIS. According to Benoit Cloitre, this sequence "is asymptotic to $\exp(F_{2n}\cdot \log c)$ where $F_n$ is the $n$-th Fibonacci's number and $c\approx1.1137$".
In 1], Bier studies the polynomial system of congruences of the form $x \mid g(y)$ and $y \mid f(x)$, where $f(t), g(t) \in \mathbb Z[t]$. He fully classifies for which pairs $(\deg f, \deg g)$ it is possible to have only finitely many solutions. It turns out that there are only finitely many such pairs, and the pairs $(k,k)$ for $k \geq 2$ are not among them.
Any suggestions on how to obtain an asymptotic formula $A(M)$ for the number of solutions satisfying $\max(|x|,|y|)\leq M$ would be greatly appreciated.
References:
1] T. Bier, Finite or infinite number of solutions of polynomial congruences in two positive integer variables, Springer Proceedings in Mathematics & Statistics, pp. 11 - 26, 2014.
[2] J. Brzeziński, W. Holsztyński, P. Kurlberg, On the congruence $ax+by\equiv 1$ modulo $xy$, Experimental Mathematics 14 (4), pp. 391-401, 2005.
[3] E. Dofs, Solutions of $x^3+y^3+z^3=nxyz$, Acta Arithmetica 73 (3), pp. 201 - 213, 1995.
[4] S. P. Mohanty, A system of cubic Diophantine equations, Journal of Number Theory 9, pp. 153 - 159, 1977.
[5] L. J. Mordell, The congruence $ax^3+by^3+c\equiv 0 \pmod{xy}$, and integer solutions of cubic equations in three variables, Acta Mathematica 88, pp. 77 - 83, 1952.
A: Another family of answers which I don't think is of the other forms: For $k=3$, consider $x=t^2$, $y=1-t^3$. More generally, if $(t^j)^k-1$ has a factor of the form $\phi(t) = 1 + \sum_{m \geq j} c_m t^m$, even if $\phi(t)$ is not of the form $f(t^j)$, then $(t^j, \phi(t))$ works.
A: Felipe gave the "sporadic" example $x=3,y=5$ for $k=4.$ Other solutions for $k=4$ are $(x,y)=(2,3),(3,5),(5,8),(8,13),(13,21)\cdots$ and also $(x,y)=(2,5),(3,8),(5,13),(8,21),\cdots$
All these pairs have $x^2 \equiv \pm1 \pmod y$ and $y^2 \equiv \pm1 \pmod x$ which is of course related. 
The pairs $(3,8),(8,21),(21,55),\cdots$ are the ones with $x^2 \equiv 1 \pmod y$ and $y^2 \equiv 1 \pmod x$
I suggest computing to see what else turns up. I wouldn't be surprised to see other Fibonacci related families.
