# If y forms Pythagorean triples with two different x, can some other y also form Pythagorean triples with those two x?

This question arises from an attempt to find whether a point P within an integer rectangle can lie at integer distances from the four vertices of the rectangle. It seems reducible to a question about Pythagorean triples: $x^2 + y^2 = z^2$.

I find that any odd integer $y$ which is not prime, or a power thereof, forms a PT with at least two different $x$. If some other $y$ can form triples with those same two $x$-es, then the rectangle problem has a solution.

The number of different $x$ forming a PT with a given $y$ is a function of the number of divisors of $y$. Every such $x$ is half the difference of the squares of two numbers whose product is $y$. For example, $y = 105$ forms a PT with $2x = 15^2 - 7^2,\ 21^2 - 5^2,\ 35^2 - 3^2$, and $105^2 - 1$.

For $y' = 1935,\ 45^2 - 43^2 = 2x' = 2x$. But the other five $x'$ forming a PT with $y' = 1935$ are different from the other three $x$-es forming a PT with $y = 105$. The five $2x'$-s are $129^2 - 15^2,\ 215^2 - 9^2,\ 387^2 - 5^2,\ 645^2 - 3^2$, and $1935^2 - 1$.

Must this be so in every case? My conjecture is, generally, if $(ab)^2 - c^2$ equals $(de)^2 - f^2$, then $(ac)^2 - b^2$ cannot equal $(df)^2 - e^2$, or $(ef)^2 - d^2$, or $(def)^2 - 1$. Similarly, $(bc)^2 - a^2$ cannot equal any of the last three, nor can $(abc)^2 - 1$.

The conjecture of course extends to y composed of more than three prime factors, where $2x$ accordingly will be the difference of the squares of two numbers each composed of any number of prime factors to any power.

Is this conjecture known to be true or false, specifically or in light of some more general theory? Has the rectangle question been answered, using this or some other approach?

Please advise if my question needs clarification, and excuse the inept mathematical notation. I am new to the site and need to learn how to use the resources available. Any hints will be welcome.

• I think the question of points at integer distances from all vertices of a square is discussed in Guy's Unsolved Problems In Number Theory. – Gerry Myerson Mar 29 '14 at 0:00
• The Guy reference is problem D19. The problem of finding a point at integer distance from each vertex of an integer rectangle was posed by Clayton Dodge, Math Mag 49 (1976) page 43. A solution was published in 50 (1977) 166-167. – Gerry Myerson Mar 30 '14 at 23:00

In the rectangle from $(0,0)$ to $(56,63)$, the point $(20,48)$ is at integer distances from all the corners. Going clockwise from the origin, the distances are $52, 25, 39$, and $60$.

Edit. Corresponding Pythagorean triples (in the same order) are $20^2+48^2=52^2$, $20^2+15^2=25^2$, $36^2+15^2=39^2$, and $36^2+48^2=60^2$.

• Thank you for the solution to the rectangle problem. Two of the four non-primitive right triangles have form 3-4-5, and two have form 5-12-13. My question was put too strictly, since it is not necessary that two different y's form triples with the same two x-es: in your solution, y forms triples with two x-es, and another x forms triples with two y's respectively equal to those two x-es. Hence the truth of my conjecture, if it is true, does not rule out a solution to the rectangle problem. – Edward Porcella Mar 30 '14 at 7:01

Mirko's answer can be generalized. Choose two pythagorean triples a,b,c and d,e,f

The four points $(1,0), (0,1), (-1,0), (0,-1)$ form a square all of them at unit distance from the centre, the integer point $(0,0)$. Now scale by an integer and translate by $(m, n,)$ both $m.n$ integers. The transformed centre is an integer point at integer distance from all the corners of the (transformed) square.