Approaching a Divisibility Question

Question

Perhaps I'm a bit rusty on my number theory, but I have been struggling to figure out how to determine what positive integers $x$ and $y$ satisfy $$ x-y \text{ divides } x^2 + y^2 $$

Any thoughts? I have tried looking at a remainder after polynomial division, adding/subtracting multiples of the divisor, and factoring out a gcd term to see if that helped.

Any input would be appreciated.

Edit: If it is any help in answering the question, this is in order to help figure out how to prove the following:

Answer 1

To solve $x^2 + y^2 = k(x-y)$, you basically write it as $(2x-k)^2 + (2y+k)^2 = 2k^2$, and then use facts about the solutions to $a^2+b^2 = N$ alongside parity arguments to count the number of solutions.

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Observations towards a solution:

1. The condition $x+y > 4$ is somewhat forced / complicates things. For now, we remove that condition and add in the 4 solutions $(1, 2), (1, 3), (2, 1), (3, 1) $.

2. For now, we also remove the condition that $x, y$ are positive, and account for that later.

3. If $ \frac{ x^2 + y^2 } { x-y } = \frac{ 1995}{k}$, then $\frac{ (kx)^2 + (ky)^2 } { kx-ky} = 1995$.

   • So, it suffices to find all solutions to $\frac{x^2+y^2}{x-y} = 1995$, then account for the number of factors of $\gcd(x, y)$ that divide 1995.

4. To solve $x^2 + y^2 = 1995 (x-y)$, it is the same as solving $(2x-1995)^2 + (2y+1995)^2 = 2\times 1995^2 $.

5. Now we do number theory, especially using facts about sums of squares

   • Recall that if $ p = 4k+3$ is prime, and $ p ^2 \mid a^2 + b^2$, then $ p \mid a, p \mid b$.
   • Show that $ 399 \mid 2x-1995, 399 \mid 2y + 1995 $.
   • Conclude that the solutions over the integers are $(\frac{2x - 1995}{399}, \frac{2y + 1995}{399}) = ( \pm 1, \pm 7), (\pm 5, \pm 5), (\pm 7, \pm 1) $

6. Now, we add back the condition that $x, y> 0$.

   • Show that $x$ is positive iff $ \frac{ 2x-1995}{399} > -5$ and $y$ is positive iff $ \frac{2y+1995} {399} > 5$.
   • Hence, $(\frac{2x - 1995}{399}, \frac{2y + 1995}{399}) = ( \pm 1, 7) $
   • Hence $(x, y) = (798, 399), (1197, 399) $.

7. For each of the two solutions, $\gcd(x,y) = 399$ and the number of (positive) divisors of 399 is $\sigma_0(399) = 8$. As mentioned earlier, each divisor is a choice of $k$ that gives a unique solution. For the 8 additional negative choices of $k$, we get negative values for $x$ and $y$, which we turn into a valid solution by making both values positive and interchanging them ($(-x,-y)$ becomes $(y,x)$). These solutions are not counted for by the positive $k$ solutions since they give different values for the divisor of 1995 $\frac{x^2+y^2}{x-y}$.

8. Therefore, the final number of solutions is $16 + 16 - 4 = $ 28