Mixing
Pythagorean quadruples
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
I am very interested in the problem that deals with the Pythagorean quadrruples, which are listed in rosettacode. Unfortunately, I have not been able to find any way to compute them with Mathematica. I thought that PowersRepresentations could help me, but I find I do not know how to use it for this problem. To test my ideas, I wrote this code:
simple = Range[50];
factors = PowersRepresentations[#^2, 3, 2]& /@ simple
I don't not know how to continue. I hope that someone can help me finish this task successfully.
programming algorithm recreational-mathematics
edited Sep 5 at 4:23
m_goldberg
81.8k869189
asked Sep 5 at 1:11
bullitohappy
600217
add a comment |Â
up vote
4
down vote
favorite
I am very interested in the problem that deals with the Pythagorean quadrruples, which are listed in rosettacode. Unfortunately, I have not been able to find any way to compute them with Mathematica. I thought that PowersRepresentations could help me, but I find I do not know how to use it for this problem. To test my ideas, I wrote this code:
simple = Range[50];
factors = PowersRepresentations[#^2, 3, 2]& /@ simple
I don't not know how to continue. I hope that someone can help me finish this task successfully.
programming algorithm recreational-mathematics
edited Sep 5 at 4:23
m_goldberg
81.8k869189
asked Sep 5 at 1:11
bullitohappy
600217
2
PowersRepresentations[#^2,3,2]&/@Range[50]seems like the right code, so what do you mean by "finish this task"?
– Jason B.
Sep 5 at 1:26
there is alsoReduce`SumOfSquaresReps[3, #^2] &but it is much slower thanPowersRepresentations[#^2,3,2]&
– kglr
Sep 5 at 1:27
@JasonB. Hi Jason, with "finish this task" I mean to get the values of d that are requested in the page that I shared. My problem is that I do not know how to debug $factors$ to get 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048, maybe you can suggest something to get them
– bullitohappy
Sep 5 at 2:34
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am very interested in the problem that deals with the Pythagorean quadrruples, which are listed in rosettacode. Unfortunately, I have not been able to find any way to compute them with Mathematica. I thought that PowersRepresentations could help me, but I find I do not know how to use it for this problem. To test my ideas, I wrote this code:
simple = Range[50];
factors = PowersRepresentations[#^2, 3, 2]& /@ simple
I don't not know how to continue. I hope that someone can help me finish this task successfully.
programming algorithm recreational-mathematics
edited Sep 5 at 4:23
m_goldberg
81.8k869189
asked Sep 5 at 1:11
bullitohappy
600217
I am very interested in the problem that deals with the Pythagorean quadrruples, which are listed in rosettacode. Unfortunately, I have not been able to find any way to compute them with Mathematica. I thought that PowersRepresentations could help me, but I find I do not know how to use it for this problem. To test my ideas, I wrote this code:
simple = Range[50];
factors = PowersRepresentations[#^2, 3, 2]& /@ simple
I don't not know how to continue. I hope that someone can help me finish this task successfully.
programming algorithm recreational-mathematics
edited Sep 5 at 4:23
m_goldberg
81.8k869189
asked Sep 5 at 1:11
bullitohappy
600217
edited Sep 5 at 4:23
m_goldberg
81.8k869189
edited Sep 5 at 4:23
m_goldberg
81.8k869189
edited Sep 5 at 4:23
m_goldberg
81.8k869189
81.8k869189
asked Sep 5 at 1:11
bullitohappy
600217
asked Sep 5 at 1:11
bullitohappy
600217
asked Sep 5 at 1:11
bullitohappy
600217
600217
2
PowersRepresentations[#^2,3,2]&/@Range[50]seems like the right code, so what do you mean by "finish this task"?
– Jason B.
Sep 5 at 1:26
there is alsoReduce`SumOfSquaresReps[3, #^2] &but it is much slower thanPowersRepresentations[#^2,3,2]&
– kglr
Sep 5 at 1:27
@JasonB. Hi Jason, with "finish this task" I mean to get the values of d that are requested in the page that I shared. My problem is that I do not know how to debug $factors$ to get 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048, maybe you can suggest something to get them
– bullitohappy
Sep 5 at 2:34
add a comment |Â
2
PowersRepresentations[#^2,3,2]&/@Range[50]seems like the right code, so what do you mean by "finish this task"?
– Jason B.
Sep 5 at 1:26
there is alsoReduce`SumOfSquaresReps[3, #^2] &but it is much slower thanPowersRepresentations[#^2,3,2]&
– kglr
Sep 5 at 1:27
@JasonB. Hi Jason, with "finish this task" I mean to get the values of d that are requested in the page that I shared. My problem is that I do not know how to debug $factors$ to get 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048, maybe you can suggest something to get them
– bullitohappy
Sep 5 at 2:34
2
2
PowersRepresentations[#^2,3,2]&/@Range[50] seems like the right code, so what do you mean by "finish this task"?– Jason B.
Sep 5 at 1:26
PowersRepresentations[#^2,3,2]&/@Range[50] seems like the right code, so what do you mean by "finish this task"?– Jason B.
Sep 5 at 1:26
there is also
Reduce`SumOfSquaresReps[3, #^2] & but it is much slower than PowersRepresentations[#^2,3,2]&– kglr
Sep 5 at 1:27
there is also
Reduce`SumOfSquaresReps[3, #^2] & but it is much slower than PowersRepresentations[#^2,3,2]&– kglr
Sep 5 at 1:27
@JasonB. Hi Jason, with "finish this task" I mean to get the values of d that are requested in the page that I shared. My problem is that I do not know how to debug $factors$ to get 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048, maybe you can suggest something to get them
– bullitohappy
Sep 5 at 2:34
@JasonB. Hi Jason, with "finish this task" I mean to get the values of d that are requested in the page that I shared. My problem is that I do not know how to debug $factors$ to get 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048, maybe you can suggest something to get them
– bullitohappy
Sep 5 at 2:34
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
simple = Range[50];
Since the elements of the quads must be positive, eliminate factors containing zero.
factors =
Select[Flatten[PowersRepresentations[#^2, 3, 2] & /@ simple, 1],
FreeQ[#, 0] &];
quads = Append[#, Sqrt[Total[#^2]]] & /@ factors
(* 1, 2, 2, 3, 2, 4, 4, 6, 2, 3, 6, 7, 1, 4, 8, 9, 3, 6, 6,
9, 4, 4, 7, 9, 2, 6, 9, 11, 6, 6, 7, 11, 4, 8, 8, 12, 3,
4, 12, 13, 4, 6, 12, 14, 2, 5, 14, 15, 2, 10, 11, 15, 5, 10,
10, 15, 1, 12, 12, 17, 8, 9, 12, 17, 2, 8, 16, 18, 6, 12,
12, 18, 8, 8, 14, 18, 1, 6, 18, 19, 6, 6, 17, 19, 6, 10, 15,
19, 4, 5, 20, 21, 4, 8, 19, 21, 4, 13, 16, 21, 6, 9, 18,
21, 7, 14, 14, 21, 8, 11, 16, 21, 4, 12, 18, 22, 12, 12, 14,
22, 3, 6, 22, 23, 3, 14, 18, 23, 6, 13, 18, 23, 8, 16, 16,
24, 9, 12, 20, 25, 12, 15, 16, 25, 6, 8, 24, 26, 2, 7, 26,
27, 2, 10, 25, 27, 2, 14, 23, 27, 3, 12, 24, 27, 7, 14, 22,
27, 9, 18, 18, 27, 10, 10, 23, 27, 12, 12, 21, 27, 8, 12,
24, 28, 3, 16, 24, 29, 11, 12, 24, 29, 12, 16, 21, 29, 4,
10, 28, 30, 4, 20, 22, 30, 10, 20, 20, 30, 5, 6, 30, 31, 6,
14, 27, 31, 6, 21, 22, 31, 14, 18, 21, 31, 1, 8, 32, 33, 4,
7, 32, 33, 4, 17, 28, 33, 6, 18, 27, 33, 7, 16, 28, 33, 8,
8, 31, 33, 8, 20, 25, 33, 11, 22, 22, 33, 17, 20, 20,
33, 18, 18, 21, 33, 2, 24, 24, 34, 16, 18, 24, 34, 1, 18,
30, 35, 6, 10, 33, 35, 6, 17, 30, 35, 10, 15, 30, 35, 15,
18, 26, 35, 4, 16, 32, 36, 12, 24, 24, 36, 16, 16, 28,
36, 3, 8, 36, 37, 3, 24, 28, 37, 8, 24, 27, 37, 12, 21, 28,
37, 2, 12, 36, 38, 12, 12, 34, 38, 12, 20, 30, 38, 2, 19,
34, 39, 2, 26, 29, 39, 9, 12, 36, 39, 10, 14, 35, 39, 13,
14, 34, 39, 13, 26, 26, 39, 14, 22, 29, 39, 19, 22, 26,
39, 4, 12, 39, 41, 4, 24, 33, 41, 9, 24, 32, 41, 12, 24, 31,
41, 23, 24, 24, 41, 8, 10, 40, 42, 8, 16, 38, 42, 8, 26,
32, 42, 12, 18, 36, 42, 14, 28, 28, 42, 16, 22, 32, 42, 2,
9, 42, 43, 2, 18, 39, 43, 6, 7, 42, 43, 7, 30, 30, 43, 9,
18, 38, 43, 18, 25, 30, 43, 8, 24, 36, 44, 24, 24, 28,
44, 4, 28, 35, 45, 5, 8, 44, 45, 5, 20, 40, 45, 6, 15, 42,
45, 6, 30, 33, 45, 8, 19, 40, 45, 13, 16, 40, 45, 15, 30,
30, 45, 16, 20, 37, 45, 20, 20, 35, 45, 20, 28, 29, 45, 6,
12, 44, 46, 6, 28, 36, 46, 12, 26, 36, 46, 2, 21, 42, 47, 6,
18, 43, 47, 6, 27, 38, 47, 11, 18, 42, 47, 18, 21, 38,
47, 18, 27, 34, 47, 16, 32, 32, 48, 4, 9, 48, 49, 4, 33, 36,
49, 9, 32, 36, 49, 12, 24, 41, 49, 12, 31, 36, 49, 14, 21,
42, 49, 15, 24, 40, 49, 23, 24, 36, 49, 18, 24, 40, 50, 24,
30, 32, 50 *)
Verifying,
And @@ (Total[Most[#]^2] == Last[#]^2 & /@ quads)
(* True *)
EDIT: If you are just looking for the d values not represented
simple = Range[600];
factors = Select[Flatten[PowersRepresentations[#^2, 3, 2] & /@ simple, 1],
FreeQ[#, 0] &];
notRepresented = Complement[simple, Union[Sqrt[Total[#^2]] & /@ factors]]
(* 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512 *)
answered Sep 5 at 2:24
Bob Hanlon
55.5k23589
Thank you very much for helping me solve my doubt, test your code and really return the results I expected, the only thing I notice is that to calculate $factors$ my computer used approximately 9 minutes or so, I do not know if you had to wait for that same time?. I should mention that it was using Range [600] no Range [2200] as requested by the exercise. Thanks again
– bullitohappy
Sep 5 at 23:33
The calculation offactorsforRange[600]took just under two minutes (114.561 sec) on my laptop. I agree that this does not scale well. For a quick result, evaluate2^Range[0, 11], 5 2^Range[0, 8] // Flatten // Sort
– Bob Hanlon
Sep 6 at 1:49
With the new form of evaluation that you suggested to me, I obtained a significant improvement in time, thank you very much for your help, sorry for the inconvenience that I generate to you.
– bullitohappy
Sep 6 at 17:31
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
simple = Range[50];
Since the elements of the quads must be positive, eliminate factors containing zero.
factors =
Select[Flatten[PowersRepresentations[#^2, 3, 2] & /@ simple, 1],
FreeQ[#, 0] &];
quads = Append[#, Sqrt[Total[#^2]]] & /@ factors
(* 1, 2, 2, 3, 2, 4, 4, 6, 2, 3, 6, 7, 1, 4, 8, 9, 3, 6, 6,
9, 4, 4, 7, 9, 2, 6, 9, 11, 6, 6, 7, 11, 4, 8, 8, 12, 3,
4, 12, 13, 4, 6, 12, 14, 2, 5, 14, 15, 2, 10, 11, 15, 5, 10,
10, 15, 1, 12, 12, 17, 8, 9, 12, 17, 2, 8, 16, 18, 6, 12,
12, 18, 8, 8, 14, 18, 1, 6, 18, 19, 6, 6, 17, 19, 6, 10, 15,
19, 4, 5, 20, 21, 4, 8, 19, 21, 4, 13, 16, 21, 6, 9, 18,
21, 7, 14, 14, 21, 8, 11, 16, 21, 4, 12, 18, 22, 12, 12, 14,
22, 3, 6, 22, 23, 3, 14, 18, 23, 6, 13, 18, 23, 8, 16, 16,
24, 9, 12, 20, 25, 12, 15, 16, 25, 6, 8, 24, 26, 2, 7, 26,
27, 2, 10, 25, 27, 2, 14, 23, 27, 3, 12, 24, 27, 7, 14, 22,
27, 9, 18, 18, 27, 10, 10, 23, 27, 12, 12, 21, 27, 8, 12,
24, 28, 3, 16, 24, 29, 11, 12, 24, 29, 12, 16, 21, 29, 4,
10, 28, 30, 4, 20, 22, 30, 10, 20, 20, 30, 5, 6, 30, 31, 6,
14, 27, 31, 6, 21, 22, 31, 14, 18, 21, 31, 1, 8, 32, 33, 4,
7, 32, 33, 4, 17, 28, 33, 6, 18, 27, 33, 7, 16, 28, 33, 8,
8, 31, 33, 8, 20, 25, 33, 11, 22, 22, 33, 17, 20, 20,
33, 18, 18, 21, 33, 2, 24, 24, 34, 16, 18, 24, 34, 1, 18,
30, 35, 6, 10, 33, 35, 6, 17, 30, 35, 10, 15, 30, 35, 15,
18, 26, 35, 4, 16, 32, 36, 12, 24, 24, 36, 16, 16, 28,
36, 3, 8, 36, 37, 3, 24, 28, 37, 8, 24, 27, 37, 12, 21, 28,
37, 2, 12, 36, 38, 12, 12, 34, 38, 12, 20, 30, 38, 2, 19,
34, 39, 2, 26, 29, 39, 9, 12, 36, 39, 10, 14, 35, 39, 13,
14, 34, 39, 13, 26, 26, 39, 14, 22, 29, 39, 19, 22, 26,
39, 4, 12, 39, 41, 4, 24, 33, 41, 9, 24, 32, 41, 12, 24, 31,
41, 23, 24, 24, 41, 8, 10, 40, 42, 8, 16, 38, 42, 8, 26,
32, 42, 12, 18, 36, 42, 14, 28, 28, 42, 16, 22, 32, 42, 2,
9, 42, 43, 2, 18, 39, 43, 6, 7, 42, 43, 7, 30, 30, 43, 9,
18, 38, 43, 18, 25, 30, 43, 8, 24, 36, 44, 24, 24, 28,
44, 4, 28, 35, 45, 5, 8, 44, 45, 5, 20, 40, 45, 6, 15, 42,
45, 6, 30, 33, 45, 8, 19, 40, 45, 13, 16, 40, 45, 15, 30,
30, 45, 16, 20, 37, 45, 20, 20, 35, 45, 20, 28, 29, 45, 6,
12, 44, 46, 6, 28, 36, 46, 12, 26, 36, 46, 2, 21, 42, 47, 6,
18, 43, 47, 6, 27, 38, 47, 11, 18, 42, 47, 18, 21, 38,
47, 18, 27, 34, 47, 16, 32, 32, 48, 4, 9, 48, 49, 4, 33, 36,
49, 9, 32, 36, 49, 12, 24, 41, 49, 12, 31, 36, 49, 14, 21,
42, 49, 15, 24, 40, 49, 23, 24, 36, 49, 18, 24, 40, 50, 24,
30, 32, 50 *)
Verifying,
And @@ (Total[Most[#]^2] == Last[#]^2 & /@ quads)
(* True *)
EDIT: If you are just looking for the d values not represented
simple = Range[600];
factors = Select[Flatten[PowersRepresentations[#^2, 3, 2] & /@ simple, 1],
FreeQ[#, 0] &];
notRepresented = Complement[simple, Union[Sqrt[Total[#^2]] & /@ factors]]
(* 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512 *)
answered Sep 5 at 2:24
Bob Hanlon
55.5k23589
Thank you very much for helping me solve my doubt, test your code and really return the results I expected, the only thing I notice is that to calculate $factors$ my computer used approximately 9 minutes or so, I do not know if you had to wait for that same time?. I should mention that it was using Range [600] no Range [2200] as requested by the exercise. Thanks again
– bullitohappy
Sep 5 at 23:33
The calculation offactorsforRange[600]took just under two minutes (114.561 sec) on my laptop. I agree that this does not scale well. For a quick result, evaluate2^Range[0, 11], 5 2^Range[0, 8] // Flatten // Sort
– Bob Hanlon
Sep 6 at 1:49
With the new form of evaluation that you suggested to me, I obtained a significant improvement in time, thank you very much for your help, sorry for the inconvenience that I generate to you.
– bullitohappy
Sep 6 at 17:31
add a comment |Â
up vote
7
down vote
accepted
simple = Range[50];
Since the elements of the quads must be positive, eliminate factors containing zero.
factors =
Select[Flatten[PowersRepresentations[#^2, 3, 2] & /@ simple, 1],
FreeQ[#, 0] &];
quads = Append[#, Sqrt[Total[#^2]]] & /@ factors
(* 1, 2, 2, 3, 2, 4, 4, 6, 2, 3, 6, 7, 1, 4, 8, 9, 3, 6, 6,
9, 4, 4, 7, 9, 2, 6, 9, 11, 6, 6, 7, 11, 4, 8, 8, 12, 3,
4, 12, 13, 4, 6, 12, 14, 2, 5, 14, 15, 2, 10, 11, 15, 5, 10,
10, 15, 1, 12, 12, 17, 8, 9, 12, 17, 2, 8, 16, 18, 6, 12,
12, 18, 8, 8, 14, 18, 1, 6, 18, 19, 6, 6, 17, 19, 6, 10, 15,
19, 4, 5, 20, 21, 4, 8, 19, 21, 4, 13, 16, 21, 6, 9, 18,
21, 7, 14, 14, 21, 8, 11, 16, 21, 4, 12, 18, 22, 12, 12, 14,
22, 3, 6, 22, 23, 3, 14, 18, 23, 6, 13, 18, 23, 8, 16, 16,
24, 9, 12, 20, 25, 12, 15, 16, 25, 6, 8, 24, 26, 2, 7, 26,
27, 2, 10, 25, 27, 2, 14, 23, 27, 3, 12, 24, 27, 7, 14, 22,
27, 9, 18, 18, 27, 10, 10, 23, 27, 12, 12, 21, 27, 8, 12,
24, 28, 3, 16, 24, 29, 11, 12, 24, 29, 12, 16, 21, 29, 4,
10, 28, 30, 4, 20, 22, 30, 10, 20, 20, 30, 5, 6, 30, 31, 6,
14, 27, 31, 6, 21, 22, 31, 14, 18, 21, 31, 1, 8, 32, 33, 4,
7, 32, 33, 4, 17, 28, 33, 6, 18, 27, 33, 7, 16, 28, 33, 8,
8, 31, 33, 8, 20, 25, 33, 11, 22, 22, 33, 17, 20, 20,
33, 18, 18, 21, 33, 2, 24, 24, 34, 16, 18, 24, 34, 1, 18,
30, 35, 6, 10, 33, 35, 6, 17, 30, 35, 10, 15, 30, 35, 15,
18, 26, 35, 4, 16, 32, 36, 12, 24, 24, 36, 16, 16, 28,
36, 3, 8, 36, 37, 3, 24, 28, 37, 8, 24, 27, 37, 12, 21, 28,
37, 2, 12, 36, 38, 12, 12, 34, 38, 12, 20, 30, 38, 2, 19,
34, 39, 2, 26, 29, 39, 9, 12, 36, 39, 10, 14, 35, 39, 13,
14, 34, 39, 13, 26, 26, 39, 14, 22, 29, 39, 19, 22, 26,
39, 4, 12, 39, 41, 4, 24, 33, 41, 9, 24, 32, 41, 12, 24, 31,
41, 23, 24, 24, 41, 8, 10, 40, 42, 8, 16, 38, 42, 8, 26,
32, 42, 12, 18, 36, 42, 14, 28, 28, 42, 16, 22, 32, 42, 2,
9, 42, 43, 2, 18, 39, 43, 6, 7, 42, 43, 7, 30, 30, 43, 9,
18, 38, 43, 18, 25, 30, 43, 8, 24, 36, 44, 24, 24, 28,
44, 4, 28, 35, 45, 5, 8, 44, 45, 5, 20, 40, 45, 6, 15, 42,
45, 6, 30, 33, 45, 8, 19, 40, 45, 13, 16, 40, 45, 15, 30,
30, 45, 16, 20, 37, 45, 20, 20, 35, 45, 20, 28, 29, 45, 6,
12, 44, 46, 6, 28, 36, 46, 12, 26, 36, 46, 2, 21, 42, 47, 6,
18, 43, 47, 6, 27, 38, 47, 11, 18, 42, 47, 18, 21, 38,
47, 18, 27, 34, 47, 16, 32, 32, 48, 4, 9, 48, 49, 4, 33, 36,
49, 9, 32, 36, 49, 12, 24, 41, 49, 12, 31, 36, 49, 14, 21,
42, 49, 15, 24, 40, 49, 23, 24, 36, 49, 18, 24, 40, 50, 24,
30, 32, 50 *)
Verifying,
And @@ (Total[Most[#]^2] == Last[#]^2 & /@ quads)
(* True *)
EDIT: If you are just looking for the d values not represented
simple = Range[600];
factors = Select[Flatten[PowersRepresentations[#^2, 3, 2] & /@ simple, 1],
FreeQ[#, 0] &];
notRepresented = Complement[simple, Union[Sqrt[Total[#^2]] & /@ factors]]
(* 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512 *)
answered Sep 5 at 2:24
Bob Hanlon
55.5k23589
Thank you very much for helping me solve my doubt, test your code and really return the results I expected, the only thing I notice is that to calculate $factors$ my computer used approximately 9 minutes or so, I do not know if you had to wait for that same time?. I should mention that it was using Range [600] no Range [2200] as requested by the exercise. Thanks again
– bullitohappy
Sep 5 at 23:33
The calculation offactorsforRange[600]took just under two minutes (114.561 sec) on my laptop. I agree that this does not scale well. For a quick result, evaluate2^Range[0, 11], 5 2^Range[0, 8] // Flatten // Sort
– Bob Hanlon
Sep 6 at 1:49
With the new form of evaluation that you suggested to me, I obtained a significant improvement in time, thank you very much for your help, sorry for the inconvenience that I generate to you.
– bullitohappy
Sep 6 at 17:31
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
simple = Range[50];
Since the elements of the quads must be positive, eliminate factors containing zero.
factors =
Select[Flatten[PowersRepresentations[#^2, 3, 2] & /@ simple, 1],
FreeQ[#, 0] &];
quads = Append[#, Sqrt[Total[#^2]]] & /@ factors
(* 1, 2, 2, 3, 2, 4, 4, 6, 2, 3, 6, 7, 1, 4, 8, 9, 3, 6, 6,
9, 4, 4, 7, 9, 2, 6, 9, 11, 6, 6, 7, 11, 4, 8, 8, 12, 3,
4, 12, 13, 4, 6, 12, 14, 2, 5, 14, 15, 2, 10, 11, 15, 5, 10,
10, 15, 1, 12, 12, 17, 8, 9, 12, 17, 2, 8, 16, 18, 6, 12,
12, 18, 8, 8, 14, 18, 1, 6, 18, 19, 6, 6, 17, 19, 6, 10, 15,
19, 4, 5, 20, 21, 4, 8, 19, 21, 4, 13, 16, 21, 6, 9, 18,
21, 7, 14, 14, 21, 8, 11, 16, 21, 4, 12, 18, 22, 12, 12, 14,
22, 3, 6, 22, 23, 3, 14, 18, 23, 6, 13, 18, 23, 8, 16, 16,
24, 9, 12, 20, 25, 12, 15, 16, 25, 6, 8, 24, 26, 2, 7, 26,
27, 2, 10, 25, 27, 2, 14, 23, 27, 3, 12, 24, 27, 7, 14, 22,
27, 9, 18, 18, 27, 10, 10, 23, 27, 12, 12, 21, 27, 8, 12,
24, 28, 3, 16, 24, 29, 11, 12, 24, 29, 12, 16, 21, 29, 4,
10, 28, 30, 4, 20, 22, 30, 10, 20, 20, 30, 5, 6, 30, 31, 6,
14, 27, 31, 6, 21, 22, 31, 14, 18, 21, 31, 1, 8, 32, 33, 4,
7, 32, 33, 4, 17, 28, 33, 6, 18, 27, 33, 7, 16, 28, 33, 8,
8, 31, 33, 8, 20, 25, 33, 11, 22, 22, 33, 17, 20, 20,
33, 18, 18, 21, 33, 2, 24, 24, 34, 16, 18, 24, 34, 1, 18,
30, 35, 6, 10, 33, 35, 6, 17, 30, 35, 10, 15, 30, 35, 15,
18, 26, 35, 4, 16, 32, 36, 12, 24, 24, 36, 16, 16, 28,
36, 3, 8, 36, 37, 3, 24, 28, 37, 8, 24, 27, 37, 12, 21, 28,
37, 2, 12, 36, 38, 12, 12, 34, 38, 12, 20, 30, 38, 2, 19,
34, 39, 2, 26, 29, 39, 9, 12, 36, 39, 10, 14, 35, 39, 13,
14, 34, 39, 13, 26, 26, 39, 14, 22, 29, 39, 19, 22, 26,
39, 4, 12, 39, 41, 4, 24, 33, 41, 9, 24, 32, 41, 12, 24, 31,
41, 23, 24, 24, 41, 8, 10, 40, 42, 8, 16, 38, 42, 8, 26,
32, 42, 12, 18, 36, 42, 14, 28, 28, 42, 16, 22, 32, 42, 2,
9, 42, 43, 2, 18, 39, 43, 6, 7, 42, 43, 7, 30, 30, 43, 9,
18, 38, 43, 18, 25, 30, 43, 8, 24, 36, 44, 24, 24, 28,
44, 4, 28, 35, 45, 5, 8, 44, 45, 5, 20, 40, 45, 6, 15, 42,
45, 6, 30, 33, 45, 8, 19, 40, 45, 13, 16, 40, 45, 15, 30,
30, 45, 16, 20, 37, 45, 20, 20, 35, 45, 20, 28, 29, 45, 6,
12, 44, 46, 6, 28, 36, 46, 12, 26, 36, 46, 2, 21, 42, 47, 6,
18, 43, 47, 6, 27, 38, 47, 11, 18, 42, 47, 18, 21, 38,
47, 18, 27, 34, 47, 16, 32, 32, 48, 4, 9, 48, 49, 4, 33, 36,
49, 9, 32, 36, 49, 12, 24, 41, 49, 12, 31, 36, 49, 14, 21,
42, 49, 15, 24, 40, 49, 23, 24, 36, 49, 18, 24, 40, 50, 24,
30, 32, 50 *)
Verifying,
And @@ (Total[Most[#]^2] == Last[#]^2 & /@ quads)
(* True *)
EDIT: If you are just looking for the d values not represented
simple = Range[600];
factors = Select[Flatten[PowersRepresentations[#^2, 3, 2] & /@ simple, 1],
FreeQ[#, 0] &];
notRepresented = Complement[simple, Union[Sqrt[Total[#^2]] & /@ factors]]
(* 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512 *)
answered Sep 5 at 2:24
Bob Hanlon
55.5k23589
simple = Range[50];
Since the elements of the quads must be positive, eliminate factors containing zero.
factors =
Select[Flatten[PowersRepresentations[#^2, 3, 2] & /@ simple, 1],
FreeQ[#, 0] &];
quads = Append[#, Sqrt[Total[#^2]]] & /@ factors
(* 1, 2, 2, 3, 2, 4, 4, 6, 2, 3, 6, 7, 1, 4, 8, 9, 3, 6, 6,
9, 4, 4, 7, 9, 2, 6, 9, 11, 6, 6, 7, 11, 4, 8, 8, 12, 3,
4, 12, 13, 4, 6, 12, 14, 2, 5, 14, 15, 2, 10, 11, 15, 5, 10,
10, 15, 1, 12, 12, 17, 8, 9, 12, 17, 2, 8, 16, 18, 6, 12,
12, 18, 8, 8, 14, 18, 1, 6, 18, 19, 6, 6, 17, 19, 6, 10, 15,
19, 4, 5, 20, 21, 4, 8, 19, 21, 4, 13, 16, 21, 6, 9, 18,
21, 7, 14, 14, 21, 8, 11, 16, 21, 4, 12, 18, 22, 12, 12, 14,
22, 3, 6, 22, 23, 3, 14, 18, 23, 6, 13, 18, 23, 8, 16, 16,
24, 9, 12, 20, 25, 12, 15, 16, 25, 6, 8, 24, 26, 2, 7, 26,
27, 2, 10, 25, 27, 2, 14, 23, 27, 3, 12, 24, 27, 7, 14, 22,
27, 9, 18, 18, 27, 10, 10, 23, 27, 12, 12, 21, 27, 8, 12,
24, 28, 3, 16, 24, 29, 11, 12, 24, 29, 12, 16, 21, 29, 4,
10, 28, 30, 4, 20, 22, 30, 10, 20, 20, 30, 5, 6, 30, 31, 6,
14, 27, 31, 6, 21, 22, 31, 14, 18, 21, 31, 1, 8, 32, 33, 4,
7, 32, 33, 4, 17, 28, 33, 6, 18, 27, 33, 7, 16, 28, 33, 8,
8, 31, 33, 8, 20, 25, 33, 11, 22, 22, 33, 17, 20, 20,
33, 18, 18, 21, 33, 2, 24, 24, 34, 16, 18, 24, 34, 1, 18,
30, 35, 6, 10, 33, 35, 6, 17, 30, 35, 10, 15, 30, 35, 15,
18, 26, 35, 4, 16, 32, 36, 12, 24, 24, 36, 16, 16, 28,
36, 3, 8, 36, 37, 3, 24, 28, 37, 8, 24, 27, 37, 12, 21, 28,
37, 2, 12, 36, 38, 12, 12, 34, 38, 12, 20, 30, 38, 2, 19,
34, 39, 2, 26, 29, 39, 9, 12, 36, 39, 10, 14, 35, 39, 13,
14, 34, 39, 13, 26, 26, 39, 14, 22, 29, 39, 19, 22, 26,
39, 4, 12, 39, 41, 4, 24, 33, 41, 9, 24, 32, 41, 12, 24, 31,
41, 23, 24, 24, 41, 8, 10, 40, 42, 8, 16, 38, 42, 8, 26,
32, 42, 12, 18, 36, 42, 14, 28, 28, 42, 16, 22, 32, 42, 2,
9, 42, 43, 2, 18, 39, 43, 6, 7, 42, 43, 7, 30, 30, 43, 9,
18, 38, 43, 18, 25, 30, 43, 8, 24, 36, 44, 24, 24, 28,
44, 4, 28, 35, 45, 5, 8, 44, 45, 5, 20, 40, 45, 6, 15, 42,
45, 6, 30, 33, 45, 8, 19, 40, 45, 13, 16, 40, 45, 15, 30,
30, 45, 16, 20, 37, 45, 20, 20, 35, 45, 20, 28, 29, 45, 6,
12, 44, 46, 6, 28, 36, 46, 12, 26, 36, 46, 2, 21, 42, 47, 6,
18, 43, 47, 6, 27, 38, 47, 11, 18, 42, 47, 18, 21, 38,
47, 18, 27, 34, 47, 16, 32, 32, 48, 4, 9, 48, 49, 4, 33, 36,
49, 9, 32, 36, 49, 12, 24, 41, 49, 12, 31, 36, 49, 14, 21,
42, 49, 15, 24, 40, 49, 23, 24, 36, 49, 18, 24, 40, 50, 24,
30, 32, 50 *)
Verifying,
And @@ (Total[Most[#]^2] == Last[#]^2 & /@ quads)
(* True *)
EDIT: If you are just looking for the d values not represented
simple = Range[600];
factors = Select[Flatten[PowersRepresentations[#^2, 3, 2] & /@ simple, 1],
FreeQ[#, 0] &];
notRepresented = Complement[simple, Union[Sqrt[Total[#^2]] & /@ factors]]
(* 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512 *)
answered Sep 5 at 2:24
Bob Hanlon
55.5k23589
edited Sep 5 at 3:47
answered Sep 5 at 2:24
Bob Hanlon
55.5k23589
answered Sep 5 at 2:24
Bob Hanlon
55.5k23589
answered Sep 5 at 2:24
Bob Hanlon
55.5k23589
55.5k23589
Thank you very much for helping me solve my doubt, test your code and really return the results I expected, the only thing I notice is that to calculate $factors$ my computer used approximately 9 minutes or so, I do not know if you had to wait for that same time?. I should mention that it was using Range [600] no Range [2200] as requested by the exercise. Thanks again
– bullitohappy
Sep 5 at 23:33
The calculation offactorsforRange[600]took just under two minutes (114.561 sec) on my laptop. I agree that this does not scale well. For a quick result, evaluate2^Range[0, 11], 5 2^Range[0, 8] // Flatten // Sort
– Bob Hanlon
Sep 6 at 1:49
With the new form of evaluation that you suggested to me, I obtained a significant improvement in time, thank you very much for your help, sorry for the inconvenience that I generate to you.
– bullitohappy
Sep 6 at 17:31
add a comment |Â
Thank you very much for helping me solve my doubt, test your code and really return the results I expected, the only thing I notice is that to calculate $factors$ my computer used approximately 9 minutes or so, I do not know if you had to wait for that same time?. I should mention that it was using Range [600] no Range [2200] as requested by the exercise. Thanks again
– bullitohappy
Sep 5 at 23:33
The calculation offactorsforRange[600]took just under two minutes (114.561 sec) on my laptop. I agree that this does not scale well. For a quick result, evaluate2^Range[0, 11], 5 2^Range[0, 8] // Flatten // Sort
– Bob Hanlon
Sep 6 at 1:49
With the new form of evaluation that you suggested to me, I obtained a significant improvement in time, thank you very much for your help, sorry for the inconvenience that I generate to you.
– bullitohappy
Sep 6 at 17:31
Thank you very much for helping me solve my doubt, test your code and really return the results I expected, the only thing I notice is that to calculate $factors$ my computer used approximately 9 minutes or so, I do not know if you had to wait for that same time?. I should mention that it was using Range [600] no Range [2200] as requested by the exercise. Thanks again
– bullitohappy
Sep 5 at 23:33
Thank you very much for helping me solve my doubt, test your code and really return the results I expected, the only thing I notice is that to calculate $factors$ my computer used approximately 9 minutes or so, I do not know if you had to wait for that same time?. I should mention that it was using Range [600] no Range [2200] as requested by the exercise. Thanks again
– bullitohappy
Sep 5 at 23:33
The calculation of
factors for Range[600] took just under two minutes (114.561 sec) on my laptop. I agree that this does not scale well. For a quick result, evaluate 2^Range[0, 11], 5 2^Range[0, 8] // Flatten // Sort– Bob Hanlon
Sep 6 at 1:49
The calculation of
factors for Range[600] took just under two minutes (114.561 sec) on my laptop. I agree that this does not scale well. For a quick result, evaluate 2^Range[0, 11], 5 2^Range[0, 8] // Flatten // Sort– Bob Hanlon
Sep 6 at 1:49
With the new form of evaluation that you suggested to me, I obtained a significant improvement in time, thank you very much for your help, sorry for the inconvenience that I generate to you.
– bullitohappy
Sep 6 at 17:31
With the new form of evaluation that you suggested to me, I obtained a significant improvement in time, thank you very much for your help, sorry for the inconvenience that I generate to you.
– bullitohappy
Sep 6 at 17:31
add a comment |Â
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2
PowersRepresentations[#^2,3,2]&/@Range[50]seems like the right code, so what do you mean by "finish this task"?– Jason B.
Sep 5 at 1:26
there is also
Reduce`SumOfSquaresReps[3, #^2] &but it is much slower thanPowersRepresentations[#^2,3,2]&– kglr
Sep 5 at 1:27
@JasonB. Hi Jason, with "finish this task" I mean to get the values of d that are requested in the page that I shared. My problem is that I do not know how to debug $factors$ to get 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 64, 80, 128, 160, 256, 320, 512, 640, 1024, 1280, 2048, maybe you can suggest something to get them
– bullitohappy
Sep 5 at 2:34