Find all integers $x$ and $y$ for which $\frac{x^2}{x-1}+\frac{y^2}{y-1}$ is an integer.

Question

Find all integers $x$ and $y$ (not just positive) for which $\frac{x^2}{x-1}+\frac{y^2}{y-1}$ is an integer.

This is a generalization of Find integers $a$ in $[1,2014]$ to get solutions in integers of $x+y=a, \frac{x^2}{x-1}+\frac{y^2}{y-1}=4$

Here is what I have come up with:

The only solutions, with the value of $\frac{x^2}{x-1}+\frac{y^2}{y-1}$, are

$ (3, 3)\implies 9, \\ (2, 2)\implies 8,\\ (0, 0)\implies 0, \\ (0, 2)\implies 4,\\ (-1, -1)\implies -1,\\ (-1, 2)\implies -1,\\ (x, -x+2)\implies 4 $

I'll post my solution in a couple of days if there are no others.

Answer 1

Hint: Your term can be written in the form $$x+y+2+\frac{y-1+x-1}{(x-1)(y-1)}$$ so your last term must be an integer.

Answer 2

Hint:

Let $a=x-1$ and $b=y-1$. Then we have $${a^2+2a+1\over a}+{b^2+2b+1\over b} = a+2+b+2+{1\over a}+{1\over b}$$

So $ab\mid a+b$ so:

a) $a\mid a+b$ so $a\mid b$ and

b) $b\mid a+b$ so $b\mid a$ so $\boxed{a=\pm b}$.

If $a=b$ we get $a^2\mid 2a$ so $a\mid 2$ so $a\in \{-2,-1,1,2\}$

If $a=-b$ we get $y=2-x$.

Answer 3

For every prime $p$ we need the valuation $v_p\left( \frac{x^2}{x-1}+\frac{y^2}{y-1} \right)$ to be nonnegative. In particular, for $p$ dividing $x-1$ we have $v_p\left( \frac{x^2}{x-1}\right) < 0$ so we need $0>v_p\left( \frac{x^2}{x-1}\right) = v_p\left( \frac{y^2}{y-1}\right)$, which is equivalent to $v_p(x-1) = v_p(y-1)$. Similarly for $p \mid y-1$. We conclude that $x-1 = \pm(y-1)$ because they have the same valuation at every prime.

The problem is reduced to two problems in one variable $x$ by substituting $y$ in terms of $x$. We then have the following general result: Consider a rational function $S(x)/T(x) \in \mathbb Q(x)$ with $S,T \in \mathbb Z[x]$ and $T$ monic. Using Euclidean division we can write it as $Q(x)+R(x)/T(x)$ with $Q,R \in \mathbb Z[x]$ and $R(x)$ of degree less than $T(x)$. Then:

• If $R= 0$ then $S/T$ is integer for all integer arguments $x$.
• If $R \neq 0$ then $S/T$ is integer for only finitely many integer arguments $x$: It is noninteger as soon as $0 \neq |R(x)/T(x)| < 1$.

If $x-1 = -(y-1)$ we're in the first case. If $x-1=y+1$ we're in the second case with $R/T = 2/(x-1)$.

Thus: the solutions are all pairs $(x,2-x)$ (for $x \neq 1$) and $(x, x)$ for $x-1$ dividing $2$, i.e. for $x \in \{-1, 0, 2, 3\}$