2

I was looking for the integral solutions of equation of circle.

For a given integral radius,

All (x,y) such that both are positive integers.

I realised that it can't have more than 3 solutions (0,r),(r,0) and one Pythagorean triplet.

In fact I think we can have only one triplet for a given hypotenuse.

I am not able to prove it.

But not get any counter.

So please help me to know that I am correct or not.

super saiyan
  • 344
  • 3
    You can have more than $3$ solutions:

    the circle $${(x,y)|x^2+y^2=25}$$

    contains the points $(0,\pm 5), (\pm 5, 0), (\pm3,\pm4), (\pm4,\pm3)$ for a total of $12$ integer solutions.

     – 5xum Oct 06 '17 at 12:53

3 Answers3

3

Even, if you restrict to positive integers, there can be more than one solution, for example $$7^2+24^2=15^2+20^2=25^2$$

Peter
  • 84,454
2

And even if your restrict to primitive pythagorean triples, there can still be more than one solution in positive integers (counting reversed pairs only once), for example $$16^2 + 63^2 = 33^2 + 56^2 = 65^2$$

quasi
  • 58,772
  • This OEIS sequence lists all of these (a number is the hypotenuse of a primitive pythagorean triple iff it is the sum of two coprime squares, and for each way of writing it as such a sum of squares, there's another triple). – Arthur Oct 06 '17 at 13:05
1

In fact, for every positive integer $n$, there exists a positive integer $r$ such that the circle of radius $r$, centered at the origin, has $n$ distinct points $(a_1,b_1),...,(a_n,b_n)$, with positive integer coordinates.

This can be shown as follows, using what's called "rational trigonometry" . . .

Choose $n$ rational numbers $t_1,...,t_n$ with $0 < t_1 < \cdots < t_n < 1$.

For each $k \in \{1,...,n\}$, let \begin{align*} x_k &= \frac{1-t_k^2}{1+t_k^2}\\[4pt] y_k &= \frac{2t}{1+t_k^2} \end{align*} Then identically, \begin{align*} x_k^2 + y_k^2 &= \left({\small{\frac{1-t_k^2}{1+t_k^2}}}\right)^2 + \left({\small{\frac{2t_k^2}{1+t_k^2}}}\right)^2\\[4pt] &=\frac{(1-t_k^2)^2 + (2t_k)^2}{(1+t_k^2)^2}\\[4pt] &=\frac{(1-2t_k^2 + t_k^4) + (4t_k^2)}{(1+t_k^2)^2}\\[4pt] &=\frac{1+2t_k^2 + t_k^4}{(1+t_k^2)^2}\\[4pt] &=\frac{(1+t_k^2)^2}{(1+t_k^2)^2}\\[4pt] &=1\\[4pt] \end{align*} Since each $t_k$ is rational, with $0 < t_k < 1$, it follows that the point $(x_k,y_k)$ has positive rational coordinates.

From $0 < t_1 < \cdots < t_n < 1$, we get \begin{align*} &1 > 1 - t_1^2 > \cdots > 1-t_n^2 >0\\[4pt] &0 < 1 + t_1^2 < \cdots < 1 + t_n^2\\[4pt] \end{align*} Thus, the numerators of $x_1,...,x_n$ are positive and strictly decreasing, and the denominators of $x_1,...,x_n$ are positive and strictly increasing, hence $x_1 > \cdots > x_n$.

Let $r$ be a positive integer common denominator for all the elements of $\{x_1,...,x_n\} \cup \{y_1,...,y_n\}$.

Then for each $k \in \{1,...,n\}$, we can write \begin{align*} x_k &= \frac{a_k}{r}\\[4pt] y_k &= \frac{b_k}{r}\\[4pt] \end{align*} where $a_k,b_k$ are positive integers.

Since $x_1 > \cdots > x_n$, it follows that $a_1 > \cdots > a_n$, so the points $(a_1,b_1),...(a_n,b_n)$ are distinct.

Then for all $k \in \{1,...,n\}$, \begin{align*} &x_k^2 + y_k^2=1\\[4pt] \implies\;&\left(\frac{a_k}{r}\right)^2+\left(\frac{b_k}{r}\right)^2 = 1\\[4pt] \implies\;&a_k^2 + b_k^2 = r^2\\[4pt] \end{align*} Thus, as claimed, on the circle of radius $r$, centered at the origin, we have $n$ distinct points $(a_1,b_1),...,(a_n,b_n)$, with positive integer coordinates.

quasi
  • 58,772