Integral Coordinates on a Circle

The problem essentially asks the number of integral solutions to the equation $x^2+y^2={85}^2$ . Then, $x^2={85}^2-y^2=(85+y)(85-y)$ .

This means that there is a number $\frac{a}{b}$ where $a<b$ are coprime positive integers such that $\frac{b}{a}x=(85+y)$ and $\frac{a}{b}x=(85-y)$ . Adding these two equations show that $170=x(\frac{b}{a}+\frac{a}{b})=x(\frac{a^2+b^2}{ab})$ , or $x=\frac{170ab}{a^2+b^2}$ . Since $a^2+b^2$ and $ab$ are coprime and $x$ is an integer, $a^2+b^2|170$ . The prime factors of $170$ are $2\cdot5\cdot17$ .

If $a^2+b^2=1$ , the possible solution is $a=0,b=1$ . This gives $x=0,y=±85$ (remembering that both $x$ and $y$ can be either positive or negative).

If $a^2+b^2=2$ , the possible solution is $a=1,b=1$ . This gives $x=±85,y=0$ .

If $a^2+b^2=5$ , the possible solution is $a=1,b=2$ . This gives $x=±68,y=±51$ .

If $a^2+b^2=10$ , the possible solution is $a=1,b=3$ . This gives $x=±51,y=±68$ .

If $a^2+b^2=17$ , the possible solution is $a=1,b=4$ . This gives $x=±40,y=±75$ .

If $a^2+b^2=34$ , the possible solution is $a=3,b=5$ . This gives $x=±75,y=±40$ .

If $a^2+b^2=85$ , the possible solutions are $a=2,b=9$ and $a=6,b=7$ . This gives $x=±36,y=±77$ and $x=±84,y=±13$ .

If $a^2+b^2=170$ , the possible solutions are $a=1,b=13$ and $a=7,b=11$ . This gives $x=±13,y=±84$ and $x=±77,y=±36$ .

This shows that there are $36$ integral coordinates.