Beyond Pythagoras
Pythagoras was a great mathematician who created theorems and one of his famous theorems was the "Pythagoras Theorem".
You start with a right-angled triangle. The hypotenuse is labeled "c". The bottom of the triangle is "b" and the side of the triangle is labeled "a".
Pythagoras Theorem says that in any right angled triangle, the lengths of the hypotenuse and the other two sides are related by a simple formula. So, if you know the lengths of any two sides of a right angled triangle, you can use Pythagoras Theorem to find the length of the third side:
Algebraically: a2 + b2 = c2
The numbers 3, 4 and 5 satisfy the condition
9 + 16 = 25
Because 3x3=9
4x4=16
5x5=25
And so 9 + 16 = 25
I now have to find out if the following sets of numbers satisfy a similar condition of:
(Shortest Side) 2 + (middle Side) 2 = (Longest side) 2
a) 5, 12, 13
a2 + b2 = c2
52 + 122 = 132
25 + 144 = 169
69 = 169
b) (7, 24, 25)
a2 + b2 = c2
72 + 242 = 252
49 + 576 = 625
625 = 625
(3, 4, 5), (5, 12, 13) and (7, 24, 25) are called Pythagorean triples because they satisfy the condition, (Shortest side)2 + (Middle side)2 = (Longest Side)2
We know from the Pythagorean triples the shortest side is always an odd number.
So far I have observed the following patterns:
The shortest side length advances by two each time.
Both the shortest and longest side lengths are always odd.
The longest length is always one unit more than the middle length.
I can immediately see the formula to get the shortest side length from the term number.
The terms multiplied by two add one equals the shortest side length.
Let Term number = Tn
Let Shortest Side length = SL
Tn ------ 2n + 1 = SL
If term number = 1; 2(1)+1=3
If term number = 2; 2(2)+1=5
If term number = 3; 2(3)+1=7
This works for all Pythagorean triples that have an odd numbered SL.
I have investigated another three Pythagorean triples and they are stated below.
Serial No.
Shortest side a
Middle Side b
Longest side c
3
4
5
2
5
2
3
3
7
24
25
4
9
40
41
5
1
60
61
6
3
84
85
I will now test the Pythagorean triples I have just found with the aid of Pythagoras theorem and a diagram.
a) (9, 40, 41)
a2 + b2 = c2
92 + 402 = 412
81 + 1600 = 1681
Scale: 1cm=2cm
b) (11, 60, 61)
a2 + b2 = c2
12 + 602 = 612
21 + 3600 = 3721
Scale: 1cm=2cm
c) (13, 84, 85)
a2 + b2 = c2
32 + 842 = 852
69 + 7056 = 7225
Scale: 1cm=4cm
I want to predict two other Pythagorean triples outside the table, and I will do this with the help of the observations I have made.
We know that the shortest side and longest side is an odd number and the longest side is one more than the middle side.
The shortest side increases by two each time, the middle side's difference increases by four each time, and the longest side is one more than the middle side.
. Since the shortest side increases by two each time, the last number we investigated was 13, therefore 13 + 2 = 15,
Shortest side = 15
Since the middle side's difference increases by four each time, the last
number was 84 and the value before that was 64, the difference between 84
and 64 is 24, so we add 4 to 24,
which is 28, therefore 84 + 28 = 112,
Middle Side = 112
The longest side is 1 more than the middle side, therefore 112 + 1 = 113
I would now like to check if the values found are accurate and would check it with the help of the Pythagoras theorem.
a2 + b2 = c2
52 + 1122 = 1132
2769 = 12769
The results show that (15,112,113) is a Pythagorean triple.
I would like to investigate another Pythagorean triple using the same method.
2. We add 2 to 15 which is 17.
Shortest Side = 17
112 - 84 = 28, I will add 4 to 28 which is 32, therefore 112 + 32 = 144
Middle Side = 144
The longest side is 1 more than the middle side, therefore 144 + 1 = 145
Longest side = 145
I would like to check if the values found are accurate with the aid of a Pythagoras theorem.
a2 + b2 = c2
72 + 1442 = 142
21025 = 21025
The results show that (17,144,145) are Pythagorean triples.
With the help of the site, http://www2.math.vic.edu/~fields/puzzle/triples.html I am able to generate around 70 more Pythagoras theorem.
Serial No.
Shorter Side a
Middle Side b
Longest side d
3
4
5
2
5
2
3
3
7
24
25
4
9
40
41
5
1
60
61
6
3
84
85
7
5
12
13
8
7
44
45
9
9
80
81
0
21
220
221
1
23
264
265
2
25
312
313
3
27
364
365
4
29
420
421
5
31
480
481
6
33
544
545
7
35
612
613
8
37
684
685
9
39
760
761
20
41
840
841
21
43
924
925
22
45
012
013
23
47
104
105
24
49
200
201
25
51
300
301
26
53
404
405
27
55
512
513
28
57
624
625
29
59
740
741
30
61
860
861
31
63
984
985
32
65
2112
2113
33
67
2244
2245
34
69
2380
2381
35
71
2520
2521
36
73
2664
2665
37
75
2812
2813
38
77
2964
2965
39
79
3120
3121
40
81
3280
3281
41
83
3444
3445
42
85
3612
3613
43
87
3784
3785
44
89
3960
3961
45
91
4140
4141
46
93
4324
4325
47
95
4512
4513
48
97
4704
4705
49
99
4900
4901
50
01
5100
...
This is a preview of the whole essay
2964
2965
39
79
3120
3121
40
81
3280
3281
41
83
3444
3445
42
85
3612
3613
43
87
3784
3785
44
89
3960
3961
45
91
4140
4141
46
93
4324
4325
47
95
4512
4513
48
97
4704
4705
49
99
4900
4901
50
01
5100
5101
51
03
5304
530
52
05
5512
5513
53
07
5724
5725
54
09
5940
5941
55
11
6160
6161
56
13
6384
6385
57
15
6612
6613
58
17
6844
684
59
19
7080
7081
60
21
7320
7321
61
23
7564
7565
62
25
7812
7813
63
27
8064
8065
64
29
8320
8321
65
31
8580
8581
66
33
8844
8845
67
35
9112
9113
68
37
9384
9385
69
39
9660
9661
70
41
9940
9941
I now have several different triangles, which I think is easily enough to find a relationship between each side. An easier and faster way to work out the sides would be by using the nth term. I will now try to work out the nth term for each side (shortest, middle & longest).
The formula I will work out first is for the shortest side.
3, 5, 7, 9, 11, 13
+2 +2 +2 +2 +2
The difference between the lengths of the shorter side is 2. This means the equation is something to do with 2n. So, I will write down all the answers for 2n.
2, 4, 6, 8, 10, 12
+1 +1 +1 +1 +1
There is only a difference of +1 between 2n and the shortest side, so this means the formula should be 2n+1. To see if I'm correct, I will now test this formula.
If n=1; Shorter side=2n+1
=2(1)+1
=3
If n=2; Shorter side=2(2)+1
=5
If n=3; Shorter Side=2(3)+1
=7
Therefore all these tests prove that: Shortest Side=2n+1
The next formula I need to work out is the formula for the middle side.
4, 12, 24, 40, 60, 84
+8 +12 +16 +20 +24
+4 +4 +4 +4
The second difference of the lengths of the middle sides is 4. This means that the formula has to have something to do with 4, most likely 4n. However, because 4 is the second difference, the formula must be n2. I now believe that the answer will have something to do with n2.
To see if I'm correct I'll test the formula n2 + 4.
If n=1; Middle side = n2 + 4
= 12 + 4
= 5
The formula is proved wrong for the first term itself. I will now try the formula 2n2.
If n=1; Middle Side =2n2
=2(1)2
=4
If n=2; Middle side =2(2)2
=16
The formula proved wrong in the second term. I will now try the formula 2n2 + 2n for the first three terms.
If n=1; Middle Side =2n2 + 2n
=2(1)2 + 2(1)
= 4
If n=2; Middle side =2(2)2 + 2(2)
=12
If n=3; Middle Side =2(3)2 + 2(3)
=24
I am now certain that 2n2 + 2n is the correct formula for finding the middle side.
Therefore all the tests prove that: Middle side = 2n2 + 2n
The next formula I need to work out is the formula for the longest side.
To start with, I am going to draw out a table containing the longest and middle side.
Serial No.
Shortest Side a
Middle Side b
Longest side c
3
4
5
2
5
2
3
From the table, I know that there is only a difference of 1 between the middle side and the long side. So:
(Middle Side) + 1= Longest Side
i.e. (2n2 + 2n) + 1 = Longest side
I will test it using the first two terms.
If n=1; Longest Side = (2n2 + 2n) + 1
= [2(1)2 + 2(1)] + 1
= 2 + 2 + 1
= 5
If n=2; Longest Side = [2(2)2 + 2(2)] + 1
= 8 + 4 + 1
= 13
Therefore all the tests prove that: Longest side = (2n2 + 2n) + 1
Since I have found all these formulas, I would like to test it by using the nth term and reconfirm with the Pythagoras theorem.
Shortest Side = 2n + 1
Middle Side = 2n2 + 2n
Longest Side = (2n2 + 2n) + 1
a) If n = 29;
Shortest Side = 2n + 1
= 2(29) + 1
= 58 + 1
= 59
Middle Side = 2n2 + 2n
= 2(29)2 + 2(29)
=1740
Longest Side = (2n2 + 2n) + 1
= [2(29)2 + 2(29)] + 1
= 1740 + 1
= 1741
Now I will reconfirm the Pythagorean triple with the Pythagoras theorem.
(59, 1740, 1741)
a2 + b2 = c2
592 + 17402 = 17412
3031081 = 3031081
b) If n = 31;
Shortest Side = 2n + 1
= 2(31) + 1
= 62 + 1
= 63
Middle Side = 2n2 + 2n
= 2(31)2 + 2(31)
= 1984
Longest side = (2n2 + 2n) +1
= [2(31)2 + 2(31)] + 1
= 1984 + 1
= 1985
(63, 1984, 1985)
a2 + b2 = c2
632 + 19842 = 19852
3940225 = 3940225
c) If n =109
Shortest Side = 2n + 1
= 2(109) + 1
= 218 + 1
= 219
Middle side = 2n2 + 2n
= 2(109)2 + 2(109)
= 23980
Longest Side = (2n2 + 2n) + 1
= [2(109)2 + 2(109)] + 1
= 23980 + 1
= 23981
(219, 23980, 23981)
a2 + b2 = c2
2192 + 239802 = 239812
575088361 = 575088361
From all the three testings, I am confirmed that the formulas are correct.
Now, I want to set up a relationship for, a (shortest side), b (middle side) and c (longest side), using the 1st and 2nd differences.
First I would like to analyze the shortest side.
n (Serial no)
a (shorter side)
3
2
5
3
7
4
9
5
1
The 1st difference remaining constant which is 2 ; Un = an + b
If n = 1 and If n = 2;
3 = a (1) + b (1)
5 = a (2) + b (2)
3 = a + b (1)
5 = 2a + b (2)
(2) - (1)
2 = a
Substituting a = 2 in (1);
3 = 2(1) + b
3 = 2 + b
b = 3 - 2
b = 1
Therefore a = 2 ; b = 1
nth term = 2n + 1
I would like to test if this sequence I established is exact.
If n = 20 ; nth term = 2(20) + 1
= 41
If n = 40 ; nth term = 2(40) + 1
= 81
I am sure that my formula is correct for the shortest side since I checked it from the table.
Now I would like to analyze the middle side.
n (serial number)
b (middle side)
4
2
2
3
24
4
40
5
60
The differences between the terms are increasing so you find second differences. The second difference in the sequence above is 4 ;
Un = an2 + bn + c
To find out the value of c, you need to work out the 0th term of the sequence. Counting back the 0th term is 0. So when n = 0, c = 0. The value of a is half the second difference.
If n = 1;
4 = 2 (1) +b (1)
4 = 2 + b
b = 4 - 2
b = 2
Therefore a =2 ; b = 2
nth term = 2n2 + 2n
I would like to test the formula with any two values and check from the table of the answers are right.
If n = 34 ; nth term = 2 (34)2 + 2 (34)
= 2380
If n = 56 ; nth term = 2 (56)2 + 2 (56)
= 6384
I checked from the table and the testings made were right.
Now I would like to analyze the longest side.
n (Serial number)
c (longest side)
5
2
3
3
25
4
41
5
61
Since we know that the longest side is one more than the middle side, therefore ; Un = (2n2 + 2n) + 1
I would like to check this formula by testing any two values.
If n = 1 ; nth term = [ 2 (1)2 + 2 (1) ] + 1
= 4 + 1
= 5
If n = 54 ; nth term = [ 2 (54)2 + 2 (54) ] + 1
= 5940 + 1
= 5941
Both the testings prove that the formula works for the longest side.
I would like to test the formulas I have found for the shortest, middle and longest sides by testing them algebraically using the Pythagoras theorem.
a2 + b2 = c2
(2n + 1)² + (2n² + 2n)² = (2n² + 2n + 1)²
(2n + 1) (2n + 1) + (2n² + 2n) (2n² + 2n) = (2n² + 2n + 1) (2n² + 2n + 1)
4n2 + 4n + 1 + 4n4 + 8n3 + 4n2 = 4n4 + 8n3 + 8n2 + 4n + 1
4n4 + 8n3 + 8n2 + 4n + 1 = 4n4 + 8n3 + 8n2 + 4n + 1
As you can see from the algebraic equation it balances out so it is equal on each side of the equation. This means that the short side and the middle side equal the longest side, so the equation works.
So far, my investigation on Pythagorean triples was, where 'a' was odd and c, the longest side was one more unit then the middle side, that is b + 1.
This was my first family of Pythagorean triples.
a = odd
b
c = b + 1
3
4
5
5
2
3
7
24
25
9
40
41
1
60
61
3
84
85
5
12
13
We know that the Pythagorean triples in the table will only work if a is odd and the longest side is 1 more than the middle side.
For instance if we choose (8,15,17), 6 is an even number and 10 is 2 more than the middle number, therefore we cannot say that (8,15,17) falls under this category. The formulas found out for this family will also not work for (8,15,17). We can use the formulas found to prove, that (8,15,17) does not work for.
If n = 1, the values are (8,15,17);
We will now substitute 1 for the formulas found:
Shorter side = 2n + 1
= 2 (1) + 1
= 3
Middle Side = 2n2 + 2n
= 2 (1)2 + 2 (1)
= 2 + 2
= 4
Longest Side = (2n2 + 2n) + 1
= [ 2 (1)2 + 2 (1) ] + 1
= 4 + 1
= 5
We see that the values found are not (8,15,17) and we realize that the formulas are not meant for values where a is even and c is 2 more than b.
If we choose another family where the n = 2, the values are (33,56,65), we see that a is an odd number and c is 9 more than b.
We can use the formulas found to prove that (33,56,65) are not applicable to the first family.
If n = 2, the values are (33,56,65);
Shorter Side = 2n + 1
= 2 (2) + 1
= 4 + 1
= 5
Middle Side = 2n2 + 2n
= 2 (2)2 + 2 (2)
= 8 + 4
= 12
Longer Side = (2n2 + 2n) + 1
= [ 2(2)2 + 2 (2) ] + 1
= 12 + 1
= 13
Here again we see that the values found are not (33,56,65), and we realize that the formulas found are not meant for values where a is odd and c is 9 more than b.
Now I would like to investigate other families of the Pythagorean triples, which I am able to do with the help of the site
www.oswego.edu/multi-campus-nsf/ David-Dennis-mathed.htm
My second family consists of a = even and c = b+2
a = even
b
c = b+2
6
8
0
8
5
7
0
24
26
2
35
37
4
48
50
6
63
65
8
80
82
I would like to find a, b and c in terms of the nth term by using the difference method.
First I would like to analyze the shortest side.
n ( serial number)
a (shortest side)
8
2
8
3
0
4
2
5
4
The 1st difference remaining constant which is 2 ; Un = an + b
If n = 3 and If n = 4 ;
0 = a (3) + b (1)
2 = a (4) + b (2)
0 = 3a + b (1)
2 = 4a + b (2)
(2) - (1) ;
2 = a
Substituting a = 2 in (1) ;
0 = 3 (2) + b
0 = 6 + b
b = 10 - 6
b = 4
Therefore a = 2 ; b = 4
nth term = 2n + 4
I would like to test this formula with any two values from the table.
If n = 2 ; nth term = 2 (2) + 4
= 8
If n = 5 ; nth term = 2 (5) + 4
= 14
Through the testings I have found the formula to be correct.
Now I will analyze the middle term.
n (serial number)
b (middle side)
8
2
5
3
24
4
35
5
48
The second difference in this sequence is 2 and therefore
Un = an2 + bn + c
We know that a = 1, since the half of the second difference is 1/2 x 2 = 1.
In order to find b and c we will substitute values with the 1st and 2nd values.
If n = 1 ; 8 = (1)2 + b (1) + c
8 = 1 + b + c
8 - 1 = b + c
7 = b + c
If n = 2 ; 15 = (2)2 + b (2) + c
15 = 4 + 2b + c
15 - 4 = 2b + c
11 = 2b + c
7 = b + c (1)
1 = 2b + c (2)
(2) - (1) ;
b = 4
Substitute b = 1 in (1);
7 = 4 + c
c = 7 - 4
c = 3
Therefore a = 1; b = 4; c = 3
nth term = n2 + 4n + 3
Now I would like to test if the formula works for any two values.
If n = 3 ; nth term = (3)2 + 4 (3) + 3
= 24
If n = 5 ; nth term = (5)2 + 4 (5) + 3
= 48
I have checked from the table and the values are right, therefore the formula is right.
Now I would like to analyze the longest side.
n ( serial number)
c (longest side)
5
2
0
3
7
4
26
5
37
Since we know that the longest side is b + 2, therefore ;
Un = n2 + 4n + 3 + 2
= n2 + 4n + 5
Now I will test this formula with any two values.
If n = 3 ; nth term = (3)2 + 4 (3) + 5
= 21 + 5
= 26
If n = 5 ; nth term = (5)2 + 4 (5) + 5
= 45 + 5
= 50
With the testings I am assured that the formula I found is correct.
Since I have found the formulas and they are working, I would like to use the values from the table and check using the Pythagoras theorem.
(6,8,10)
a2 + b2 = c2
62 + 82 = 102
00 = 100
(8,15,17)
a2 + b2 = c2
82 + 152 = 172
289 = 289
I have tested the values with Pythagoras theorem and am assure that my second family, where a is even and c = b + 2, is right.
Now I will construct my third family where it consists of a = odd and
c = b + 9.
a = odd
b
c = b + 9
27
36
45
33
56
65
39
80
89
45
08
17
51
40
49
57
76
85
63
216
225
Now I would like to investigate a, b and c in terms of the nth term by using the difference method.
First I will find the formula for the shortest side.
n (serial number)
a (shortest side)
27
2
33
3
39
4
45
5
51
The first difference remaining constant which is 6 ; Un = an + b
If n = 3 and If n = 4 ;
39 = a (3) + b (1)
45 = a (4) + b (2)
39 = 3a + b (1)
45 = 4a + b (2)
(2) - (1) ;
6 = a
Substituting a = 6 in (1) ;
39 = 3 (6) + b
39 = 18 + b
39 - 18 = b
21 = b
Therefore a = 6 ; b = 21
nth term = 6n + 21
I would like to test this formula with any two values.
If n = 2 ; nth term = 6 (2) + 21
= 33
If n = 5 ; nth term = 6 (5) + 21
= 51
With the testings done I am confirmed that the formula calculated is right.
Now I would like to analyze the middle side.
n (serial number)
b (middle side)
36
2
56
3
80
4
08
5
40
The second difference in this sequence is 4 and therefore ;
Un = an2 + bn + c
We know that a = 2 because 1/2 x 4 = 2
In order to find b and c we will substitute values with the 1st and 2nd values.
If n = 1 ; 36 = 2 (1)2 + b (1) + c
36 = 2 + b + c
36 - 2 = b + c
34 = b + c
If n = 2 ; 56 = 2 (2)2 + b (2) + c
56 = 8 + 2b + c
56 - 8 = 2b + c
48 = 2b + c
34 = b + c (1)
48 = 2b + c (2)
(2) - (1) ;
b = 14
Substitute b = 14 in (1);
34 = 14 + c
c = 34 - 14
c = 20
Therefore a = 2; b = 14; c = 20
nth term = 2n2 + 14n + 20
Now I would like to test if the formula found is applicable, by using two values.
If n = 1 ; nth term = 2 (1)2 + 14 (1) + 20
= 36
If n = 3 ; nth term = 2 (3)2 + 14 (3) + 20
= 80
With the results I am confirmed that the formula is right.
Now I would analyze the longest side.
n (serial number)
c (longest side)
45
2
65
3
89
4
17
5
49
Since we know that the longest side is 9 units more than the middle side, therefore ; Un = 2n2 + 14n + 20 + 9
= 2n2 + 14n + 29
I would like to test this formula using two values ;
If n = 3 ; nth term = 2 (3)2 + 14 (3) + 29
= 89
If n = 5 ; nth term = 2 (5)2 + 14 (5) + 29
= 149
With these testings I am sure that the formula calculated is correct.
Now I would like to use the values from the table and check using the Pythagoras theorem.
(39, 80, 89)
a2 + b2 = c2
392 + 802 = 892
7921 = 7921
(272, 362, 452)
a2 + b2 = c2
272 + 362 = 452
2025 = 2025
With the results I am assured that the values of my third family are correct.
Considering the first family where a = odd and c = b + 1 the formulas found are;
a = 2n + 1
b = 2n2 + 2n
c = 2n2 + 2n = 1
I would like to represent b and c in terms of a;
a = 2n + 1 n = a - 1
2
b = 2n2 + 2n
= 2 a - 1 2 + 2 a - 1
2 2
= 2 a - 1 2 + 2 a - 1
4 2
= (a - 1)2 + (a - 1)
2 1
=(a - 1 )2 + 2 (a - 1)
2
= a2 - 2a + 1 + 2a - 2
2
b = a2 - 1
2
c = b + 1
= a2 - 1 + 1
2
= a2 - 1 + 2
2
c = a2 + 1
2
Considering the 2nd family where a = even and c = b + 2 the formulas found are;
a = 2n + 4
b = n2 + 4n + 3
c = n2 + 4n + 5
I would like to represent b and c in terms of a;
a = 2n + 4 n = a - 4
2
b = n2 + 4n + 3
= a - 4 2 + 4 a - 4 + 3
2 2
= a - 4 2 + 2a - 8 + 3
4
= a2 - 8a + 16 + 8a - 32 + 12
4
= a2 - 4
4
c = b + 2
= a2 - 4 + 2
4
= a2 - 4 + 8
4
= a2 + 4
4
Considering the third family where a = odd and c = b + 9 the formulas found are;
a = 6n + 21
b = 2n2 + 14n + 20
c = 2n2 + 14n + 29
I would like to represent b and c in terms of a;
a = 6n + 21 n = a - 21
6
b = 2n2 + 14n + 20
= 2 a - 21 2 + 14 a - 21 + 20
6 6
= 2 a - 21 2 + 7a - 147 + 20
36 3
= 2 ( a2 - 42a + 441) + 84a - 1764 + 720
36
= 2a2 - 84a + 882 + 84a - 1764 + 720
36
= 2a2 - 162
36
= 2 ( a2 - 81 )
36
= a2 - 81
18
c = b + 9
= a2 - 81 + 9
18
= a2 - 81 + 162
18
= a2 + 81
18
I have put down my results in a tabulate form below.
a
b
c
st family
Odd
a2 - 1
2
a2 + 1
2
2nd family
Even
a2 - 4
4
a2 + 4
4
3rd family
Odd
a2 - 81
18
a2 + 81
18
The following are two other sets of families I have established.
a = odd and c = b + 25
a = odd
b
c = b + 25
85
32
57
95
68
93
05
208
233
15
252
277
25
300
325
35
352
377
45
506
531
The formulas found for this family are:
0n + 75
4n2 + 30n + 100
4n2 + 30n + 125
a = even and c = b + 10
a = even
b
c = b + 10
30
40
50
40
75
85
50
20
30
60
75
85
70
240
250
80
315
325
90
400
410
The formulas found for this family are:
0n + 20
5n2 + 20n + 15
5n2 + 20n + 25
In order to find a standard formula for b and c, I will generalize the formulas found.
b
c
a2 - 1
2
a2 + 1
2
a2 - 4
4
a + 4
4
a2 - 81
18
a2 + 81
18
a2 - 625
50
a2 + 625
50
a2 - 100
20
a2 + 100
20
a2 - z2
2z
a2 + z2
2z
All the families I have established are not any multiples of any family and therefore they are unique.
I have been able to establish three formulas for deriving Pythagorean triples. I have established this with the help of the website www.math.clemson.edu/~rsimms/neat/math/pyth
a = 2*s*t
b = t2 - s2
c = t2 + s2
The following shows how I have been able to establish these formulas:
First we rearrange a2 + b2 = c2 to get a2 equal to a product of two factors.
a2 = c2 - b2
Therefore we get a2 = (c - b ) ( c + b )
I gave two new variables to ( c - b ) and ( c + b ).
Let p = ( c - b )
Let q = ( c + b )
Therefore we have :
a = V (p * q )
b = ( q - p ) / 2
c = ( q + p ) / 2
To make selecting the factors u and v easier, replace them with expressions that are divisible by 2 and such that u * v is a perfect square, since we want a, b and c to be integers.
p -> 2 * s2
q -> 2 * s2
Then we finally have:
a = 2*s*t
b = t2 - s2
c = t2 + s2
Now I will show the primitive and imprimitive sets of Pythagorean triples.
s
t
Type of triple
Primitive
2
Primitive
3
Primitive
2
Primitive
2
2
Imprimitive
3
3
Imprimitive
3
Primitive
3
2
Primitive
I would now like to establish a generalized relationship for the various families.
I would like to establish a generalized formula for all the even families.
Serial no.
Family
b
c
b + 2
a2 - 22
2*2
a2 + 22
2*2
2
b + 4
a2 - 42
2*4
a2 + 42
2*4
3
b + 6
a2 - 62
2*6
a2 + 62
2*6
4
b + 8
a2 - 82
2*8
a2 + 82
2*8
5
b + 10
a2 - 102
2*10
a2 + 102
2*10
b + z
a2 - z2
2*z
a2 + z2
2*z
Next, I would like to establish a generalized formula for all the odd families.
Serial no.
Family
b
c
b + 5
a2 - 52
2*5
a2 + 5
2*5
2
b + 9
a2 - 92
2*9
a2 + 92
2*9
3
b + 11
a2 - 112
2*11
a2 + 11
2*11
4
b + 15
a2 - 152
2*15
a2 + 152
2*15
5
b + 17
a2 - 172
2*17
a2 + 172
2*17
b + z
a2 - z2
2*z
a2 + z2
2*z
Now, I would like to establish a generalized formula for all the square families.
Serial no.
Family
b
c
b + 4
a2 - 42
2*4
a2 + 42
2*4
2
b + 9
a2 - 92
2*9
a2 + 92
2*9
3
b + 25
a2 - 252
2*25
a2 + 252
2*25
4
b + 36
a2 - 36
2*36
a2 + 36
2*36
5
b + 49
a2 - 492
2*49
a2 + 49
2*49
b + z
a2 - z2
2*z
a2 + z2
2*z
Through all these investigations done, it has helped me to know the Pythagoras theorem in much more detail. It has even helped me in investigating on various families, which are odd, even, etc. It has even helped me find formulas which I could apply to find out the various Pythagorean triples.