HP 15C: Pythagorean
Triples
This program calculates the Pythagorean triple (A, B, C)
such that A^2 + B^2 = C^2 by the formulas:
A = K * (M^2 – N^2)
B = K * (2 * M * N)
C = K * (M^2 + N^2)
The conditions are M, N, and K are all positive integers
where M > N.
Store M into memory 0, N into memory 1, and K into memory
2. A, B, and C are stored in memories 3,
4, and 5, respectively. If no such
combination can be found, a single zero (0) is returned.
|
Step
|
Key
|
Code
|
|
001
|
LBL A
|
42, 21, 11
|
|
002
|
RCL 1
|
45, 1
|
|
003
|
RCL 0
|
45, 0
|
|
004
|
X≤0
|
43, 10
|
|
005
|
GTO 0
|
22, 0
|
|
006
|
RCL 0
|
45, 0
|
|
007
|
X^2
|
43, 11
|
|
008
|
RCL 1
|
45, 1
|
|
009
|
X^2
|
43, 11
|
|
010
|
-
|
30
|
|
011
|
STO 3
|
44, 3
|
|
012
|
LST X
|
43, 36
|
|
013
|
2
|
2
|
|
014
|
*
|
20
|
|
015
|
+
|
40
|
|
016
|
STO 5
|
44, 5
|
|
017
|
RCL 0
|
45, 0
|
|
018
|
RCL* 1
|
45, 20, 1
|
|
019
|
2
|
2
|
|
020
|
*
|
20
|
|
021
|
STO 4
|
44, 4
|
|
022
|
RCL 2
|
45, 2
|
|
023
|
STO* 3
|
44, 20, 3
|
|
024
|
STO* 4
|
44, 20, 4
|
|
025
|
STO* 5
|
44, 20, 5
|
|
026
|
RCL 3
|
45, 3
|
|
027
|
X^2
|
43, 11
|
|
028
|
RCL 4
|
45, 4
|
|
029
|
X^2
|
43, 11
|
|
030
|
+
|
40
|
|
031
|
RCL 5
|
45, 5
|
|
032
|
X^2
|
43, 11
|
|
033
|
-
|
30
|
|
034
|
X=0
|
43, 20
|
|
035
|
GTO 1
|
22, 1
|
|
036
|
LBL 0
|
42, 22, 1
|
|
037
|
0
|
0
|
|
038
|
RTN
|
43, 32
|
|
039
|
LBL 1
|
42, 21, 1
|
|
040
|
RCL 3
|
45, 3
|
|
041
|
R/S
|
31
|
|
042
|
RCL 4
|
45, 4
|
|
043
|
R/S
|
31
|
|
044
|
RCL 5
|
45, 5
|
|
045
|
RTN
|
43, 32
|
Example: Input: R0 = M = 4,
R1 = N = 1, R2 = 2. Output: 30, 16, 34
This blog is property of Edward Shore, 2017.