Solve $x^2+y^2=199^2$

Question

Question :

Find the number of integral points on the circle with centre $(199,0)$ and radius $199$.(Integral point is a point $(x,y)$ with both $x,y \in \mathbb{Z}$.

My Approach : Since no. of integral coordinates won't change on shifting of origin : I shifted origin to $(199,0)$. Now the equation of circle becomes $x^2+y^2=199^2$.

But how do I solve this equation.I can conclude that one of $x$ and $y$ is odd and other is even, since some of their squares is odd.

Trivial solutions are $(\pm 199,0)$ and $(0,\pm199)$ .

But how do I prove that the these are the only solutions, or others also exist, and if more points exist, how to find them.

Thanks.

Answer 1

By parametrization of Pythagorean triples,if the above equality is true, there must exist $u,v$ integers such that $x = u^2-v^2$, $y = 2uv$, $199 = u^2+v^2$. However, $199 \equiv 3 (4)$, so you can check that $199$ can't be written as the sum of two squares! (squares are congruent only to $0,1$ mod $4$, and hence a sum of two squares can only be congruent to $0,1$ or $2$ mod $4$).

ADDENDUM :

A proof Very long winded, but nevertheless worth mentioning:

The sum of squares function $r_2(n) = 4(d_1(n) -d_3(n))$, where $d_1$ is the number of divisors of $n$ congruent to $1$ modulo $4$, and $d_3$ is the number of divisors of $n$ congruent to $3$ modulo $4$. The proof of this is not easy by any means (at least at elementary level).

In our case, there are $3$ divisors: $1$, $199$ and $199^2$. You can check that $1$ and $199^2$ are equivalent to $1$ mod $4$, while $199$ is equivalent to $3$ mod $4$, so by the formula, $r_2(199^2) = 4(2-1) = 4$. These four correspond to the trivial solutions $(\pm 199,0)$ and $(0,\pm 199)$, and the formula tells you there aren't any more.

Answer 2

Without using any theory of Pythagorean triples here is an approach from first principle:

Note that $199$ is a prime number, hence gcd$(x,y)=1$. One among $x,y$ should be even and the other odd. Assume $x$ is odd. Now given condition means $ x^2= (199+y)(199-y)$. So for any prime $p$ dividing $x$ we have either $p^2$ divides $(199+y)$ or $p^2$ divides $(199-y)$, but not both. (otherwise, it will divide their difference $2y$, and being odd will divide $y$, contradicting the fact that $x,y$ are relatively prime).

SO we can conclude that $x^2=(199+y)(199-y)$ is a factorization of $x^2$ into two squares which are equidistant from 199. This can be verified to be false proceeding as below:

Look at all perfect squares less than 199 (e.g., 169, 144 etc, these are the candidates for $199-y$) Check that corresponding $199+y$ are not squares, e.g. $199+30, 199+55$ are not squares.

The advantage now is instead of looking at squares upto $199^2$, we reduced to checking with squares upto $199$.