How many pairs of positive integers $(x,y)$ are there such that $x<y$ and $\dfrac{x^2+y^2}{x+y}$ is an integer which is a divisor of $2835$? BIMC 2018
Solution: Based on the result of Number Theory Challenge – 1, if $x^2+y^2=pq$, where $p=4k+3$ is a prime number, we have $x\equiv y\equiv 0\pmod{p}$. Therefore, if $(x,y)$ is a solution for $\dfrac{x^2+y^2}{x+y}$ as an integer, then $(\dfrac{x}{p},\dfrac{y}{p})$ is also a solution.
Because $2835=3^{4}\cdot 5\cdot 7$, and $3$ and $7$ are prime numbers in the format of $4k+3$, therefore, if we find all solutions for $$\dfrac{x^2+y^2}{x+y}=q$$ where $q$ is not divisible by $3$ nor $7$, we can find all solutions. Therefore, we only need to consider two cases, where $q=1$ or $q=5$.
Case 1: if $q=1$, we have $\dfrac{x^2+y^2}{x+y}=1$, i.e. $$(2x-1)^2+(2y-1)^2=2$$ The above equation have no integer solution with $0<x<y$.
Case 2: if $q=5$, we have $\dfrac{x^2+y^2}{x+y}=5$, i.e. $$(2x-5)^2+(2y-5)^2=50$$ The two perfect square numbers on the left side of the above equation must be $1$ and $49$.
Because $0<x<y$, we have $$2x-5=\pm 1$$ $$2y-5=\ \ \ 7$$
which leads two solutions of $(x,y)$ as $(2, 6)$ or $(3,6)$.
Therefore, the total number of solutions is $\boxed{2\cdot(4+1)\cdot(1+1)=20}$.