Pythagorean Triangle Search

[Eddie W. Shore]

Three Integers Make a Right Triangle

A Pythagorean triple is a trio of positive integers a, b, and c, that describe the lengths of a right triangle, where a and b are the lengths of the sides while c is the length of the hypotenuse.

The variables a, b, and c follow the Pythagorean Theorem:

a^2 + b^2 = c^2

Where the following measurements perimeter and area are calculated as:

p = perimeter = a + b + c
r = area = a * b / 2

With two positive integers m and n where m > n, Euclid gives a formula where a, b, and c are generated:

a = 2 * m * n
b = m^2 - n^2
c = m^2 + n^2

This can easily verified to satisfy the Pythagorean Theorem:

a^2 + b^2 = c^2
(2*m*n)^2 + (m^2 - n^2)^2 = (m^2 + n^2)^2
4*m^2*n^2 + m^4 - 2*m^2*m^2 + n^4 = m^4 + 2*m^2*m^2 + n^4
4*m^2*n^2 + m^4 - 4*m^2*m^2 + n^4 = m^4 + n^4
m^4 + n^4 = m^4 + n^4

The following program PYTHRI asks you for m and n and generates a Pythagorean triple. Make sure that m > n.

Code:

HP Prime Program PYTHRI

EXPORT PYTHTRI()
BEGIN
// 2020-06-13 EWS
// r: area
LOCAL a,b,c,p,r,m,n;
INPUT({m,n},"Pythagorean Triple
 Generator",{"m = ","n = "},
{"m > n, m,n ∈ Z+","m > n, m,n ∈ Z+"});
a:=2*m*n; b:=m^2-n^2; c:=m^2+n^2;
p:=a+b+c; r:=a*b/2;
PRINT();
PRINT(a+"^2+"+b+"^2="+c+"^2");
PRINT("a = "+a);
PRINT("b = "+b);
PRINT("c = "+c);
PRINT("perimeter = "+p);
PRINT("area = "+r);
END;

Can We Go the Other Way?

Let's say we have the area and the perimeter of a right triangle. Can we find a Pythagorean triple? In order to do so, we need to solve for m and n, and make sure that m and n are positive integers.

Recall that:

a = 2 * m * n
b = m^2 - n^2
c = m^2 + n^2

Perimeter:

p = a + b + c
p = 2 * m * n + m^2 - n^2 + m^2 + n^2
p = 2 * m^2 + 2 * m * n

Area:

r = a * b / 2
r = m * n * (m^2 - n^2)
r = m^3 * n - m * n^3

Let's solve for n in the perimeter equation:

p = 2 * m^2 + 2 * m * n
p - 2 * m^2 = 2 * m * n
Since m is a positive integer, m ≠ 0 and by dividing by 2 * m:
p / (2* m) - m = n

Substitute in the area equation:

r = m^3 * n - m * n^3
r = m^3 * (p / (2* m) - m) - m * (p / (2* m) - m)^3

The program IPYTHTRI attempts to find a Pythagorean triple by solving for m in the above equation. A first initial guess of 0 is used, but the initial guess uses powers of 10 for any further iterations that are needed.

Should a triple not be found, the program will indicate the finding. Perfect search is not guaranteed.

If a suitable solution is found, then the program calculates and displays a, b, and c.

HP Prime Program IPYTHTRI
(inverse PYTHTRI)

Code:

EXPORT IPYTHTRI()
BEGIN
// 2020-06-13 EWS
// r: area
LOCAL a,b,c,p,r,m,n,k;
INPUT({p,a},"Pythagorean Triple
 Search",{"p = ","r = "},
{"perimeter","area"});
// search for integers
FOR k FROM 0 TO 7 DO
m:=fsolve(X^3*(p/(2*X)-X)
-X*(p/(2*X)-X)^3-a,X,10*k);
IF (FP(m)==0) AND (m>0) THEN
BREAK;
END;
END;
n:=p/(2*m)-m;
PRINT();
PRINT("m = "+m);
PRINT("n = "+n);
IF (FP(m)==0) AND (FP(n)==0) THEN
PRINT("Integer Solutions Found");
a:=2*m*n; b:=m^2-n^2; c:=m^2+n^2;
PRINT(a+"^2+"+b+"^2="+c+"^2");
PRINT("a = "+a);
PRINT("b = "+b);
PRINT("c = "+c);
ELSE
PRINT("No integer solutions found");
END;
END;

Examples
(p = perimeter, r = area)

m = 5, n = 3
a = 30, b = 16, c = 34
p = 80, r = 240

m = 11, n = 6
a = 132, b = 85, c = 157
p = 374, r = 5610

m = 18, n = 14
a = 504, b = 128, c = 520
p = 1152, r = 32256

m = 164, n = 133
a = 43624, b = 9207, c = 44585
p = 97416, r = 200,823,084

Source:
Pythagorean triple. Wikipedia. Last Edited June 13, 2020. https://en.wikipedia.org/wiki/Pythagorean_triple Accessed June 13, 2020

Eddie

http://edspi31415.blogspot.com/2020/07/h...earch.html