Wednesday, March 22, 2017

HP 15C: Pythagorean Triples



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.

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