Pythagorean like Diophantine Equation
$begingroup$
I am trying to solve this problem.
http://www.javaist.com/rosecode/problem-527-1-2-3-type-Pythagorean-triangles-askyear-2018
Here we have to find all positive integral solution of $a^2+2b^2=3c^2$ where $a+b+cle N$. In the question, $N=25000000$.
My first approach was to loop over $a$ and for fixed value of $a$ find the solutions of $3c^2-2b^2=a^2$ using LMM algorithm (https://thilinaatsympy.wordpress.com/2013/07/06/solving-the-generalized-pell-equation/). But this is very slow. I am unable to find a faster method. Any help will be appreciated.
number-theory diophantine-equations pythagorean-triples
asked Dec 10 '18 at 11:37
AsifAsif
1069
$endgroup$
|
show 1 more comment
$begingroup$
I am trying to solve this problem.
http://www.javaist.com/rosecode/problem-527-1-2-3-type-Pythagorean-triangles-askyear-2018
Here we have to find all positive integral solution of $a^2+2b^2=3c^2$ where $a+b+cle N$. In the question, $N=25000000$.
My first approach was to loop over $a$ and for fixed value of $a$ find the solutions of $3c^2-2b^2=a^2$ using LMM algorithm (https://thilinaatsympy.wordpress.com/2013/07/06/solving-the-generalized-pell-equation/). But this is very slow. I am unable to find a faster method. Any help will be appreciated.
number-theory diophantine-equations pythagorean-triples
asked Dec 10 '18 at 11:37
AsifAsif
1069
$endgroup$
$begingroup$
math.stackexchange.com/questions/2773097/…
$endgroup$
– individ
Dec 10 '18 at 13:14
$begingroup$
This doesn't give all the solutions.
$endgroup$
– Asif
Dec 11 '18 at 6:02
$begingroup$
Gives all solutions. It is necessary to accurately and attentively all to write down. Although you can write such a simple solution. $$a^2+2b^2=3c^2$$ $$a=p^2+6ps+3s^2$$ $$b=p^2-3s^2$$ $$c=p^2+2ps+3s^2$$
$endgroup$
– individ
Dec 11 '18 at 6:41
$begingroup$
I am not sure how you arrived at this parametric solution, but it certainly doesn't give all the solutions and also produces lots of negative integer solutions, which aren't necessary.
$endgroup$
– Asif
Dec 11 '18 at 16:22
$begingroup$
Rewrite the negative into a positive solution. You want to get a solution what you like, not a solution to the problem. That's not how math works. We may not like the answer, but the solution is the solution.
$endgroup$
– individ
Dec 11 '18 at 16:37
|
show 1 more comment
$begingroup$
I am trying to solve this problem.
http://www.javaist.com/rosecode/problem-527-1-2-3-type-Pythagorean-triangles-askyear-2018
Here we have to find all positive integral solution of $a^2+2b^2=3c^2$ where $a+b+cle N$. In the question, $N=25000000$.
My first approach was to loop over $a$ and for fixed value of $a$ find the solutions of $3c^2-2b^2=a^2$ using LMM algorithm (https://thilinaatsympy.wordpress.com/2013/07/06/solving-the-generalized-pell-equation/). But this is very slow. I am unable to find a faster method. Any help will be appreciated.
number-theory diophantine-equations pythagorean-triples
asked Dec 10 '18 at 11:37
AsifAsif
1069
$endgroup$
I am trying to solve this problem.
http://www.javaist.com/rosecode/problem-527-1-2-3-type-Pythagorean-triangles-askyear-2018
Here we have to find all positive integral solution of $a^2+2b^2=3c^2$ where $a+b+cle N$. In the question, $N=25000000$.
My first approach was to loop over $a$ and for fixed value of $a$ find the solutions of $3c^2-2b^2=a^2$ using LMM algorithm (https://thilinaatsympy.wordpress.com/2013/07/06/solving-the-generalized-pell-equation/). But this is very slow. I am unable to find a faster method. Any help will be appreciated.
number-theory diophantine-equations pythagorean-triples
number-theory diophantine-equations pythagorean-triples
asked Dec 10 '18 at 11:37
AsifAsif
1069
asked Dec 10 '18 at 11:37
AsifAsif
1069
asked Dec 10 '18 at 11:37
AsifAsif
1069
asked Dec 10 '18 at 11:37
AsifAsif
1069
asked Dec 10 '18 at 11:37
AsifAsif
1069
1069
$begingroup$
math.stackexchange.com/questions/2773097/…
$endgroup$
– individ
Dec 10 '18 at 13:14
$begingroup$
This doesn't give all the solutions.
$endgroup$
– Asif
Dec 11 '18 at 6:02
$begingroup$
Gives all solutions. It is necessary to accurately and attentively all to write down. Although you can write such a simple solution. $$a^2+2b^2=3c^2$$ $$a=p^2+6ps+3s^2$$ $$b=p^2-3s^2$$ $$c=p^2+2ps+3s^2$$
$endgroup$
– individ
Dec 11 '18 at 6:41
$begingroup$
I am not sure how you arrived at this parametric solution, but it certainly doesn't give all the solutions and also produces lots of negative integer solutions, which aren't necessary.
$endgroup$
– Asif
Dec 11 '18 at 16:22
$begingroup$
Rewrite the negative into a positive solution. You want to get a solution what you like, not a solution to the problem. That's not how math works. We may not like the answer, but the solution is the solution.
$endgroup$
– individ
Dec 11 '18 at 16:37
|
show 1 more comment
$begingroup$
math.stackexchange.com/questions/2773097/…
$endgroup$
– individ
Dec 10 '18 at 13:14
$begingroup$
This doesn't give all the solutions.
$endgroup$
– Asif
Dec 11 '18 at 6:02
$begingroup$
Gives all solutions. It is necessary to accurately and attentively all to write down. Although you can write such a simple solution. $$a^2+2b^2=3c^2$$ $$a=p^2+6ps+3s^2$$ $$b=p^2-3s^2$$ $$c=p^2+2ps+3s^2$$
$endgroup$
– individ
Dec 11 '18 at 6:41
$begingroup$
I am not sure how you arrived at this parametric solution, but it certainly doesn't give all the solutions and also produces lots of negative integer solutions, which aren't necessary.
$endgroup$
– Asif
Dec 11 '18 at 16:22
$begingroup$
Rewrite the negative into a positive solution. You want to get a solution what you like, not a solution to the problem. That's not how math works. We may not like the answer, but the solution is the solution.
$endgroup$
– individ
Dec 11 '18 at 16:37
$begingroup$
math.stackexchange.com/questions/2773097/…
$endgroup$
– individ
Dec 10 '18 at 13:14
$begingroup$
math.stackexchange.com/questions/2773097/…
$endgroup$
– individ
Dec 10 '18 at 13:14
$begingroup$
This doesn't give all the solutions.
$endgroup$
– Asif
Dec 11 '18 at 6:02
$begingroup$
This doesn't give all the solutions.
$endgroup$
– Asif
Dec 11 '18 at 6:02
$begingroup$
Gives all solutions. It is necessary to accurately and attentively all to write down. Although you can write such a simple solution. $$a^2+2b^2=3c^2$$ $$a=p^2+6ps+3s^2$$ $$b=p^2-3s^2$$ $$c=p^2+2ps+3s^2$$
$endgroup$
– individ
Dec 11 '18 at 6:41
$begingroup$
Gives all solutions. It is necessary to accurately and attentively all to write down. Although you can write such a simple solution. $$a^2+2b^2=3c^2$$ $$a=p^2+6ps+3s^2$$ $$b=p^2-3s^2$$ $$c=p^2+2ps+3s^2$$
$endgroup$
– individ
Dec 11 '18 at 6:41
$begingroup$
I am not sure how you arrived at this parametric solution, but it certainly doesn't give all the solutions and also produces lots of negative integer solutions, which aren't necessary.
$endgroup$
– Asif
Dec 11 '18 at 16:22
$begingroup$
I am not sure how you arrived at this parametric solution, but it certainly doesn't give all the solutions and also produces lots of negative integer solutions, which aren't necessary.
$endgroup$
– Asif
Dec 11 '18 at 16:22
$begingroup$
Rewrite the negative into a positive solution. You want to get a solution what you like, not a solution to the problem. That's not how math works. We may not like the answer, but the solution is the solution.
$endgroup$
– individ
Dec 11 '18 at 16:37
$begingroup$
Rewrite the negative into a positive solution. You want to get a solution what you like, not a solution to the problem. That's not how math works. We may not like the answer, but the solution is the solution.
$endgroup$
– individ
Dec 11 '18 at 16:37
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
Above equation shown below has parametric solution:
$a^2+2b^2=3c^2$
$(a,b,c)=((3k^2-6k-5),(3k^2+6k-1),(3k^2+2k+3))$
For $k=4$ we get:
$(a,b,c)=(5,13,11)$
answered Dec 10 '18 at 13:11
SamSam
1
$endgroup$
add a comment |
$begingroup$
If we use Pythagorean triples we have:
$(m^2-n^2)^2+(2 mn)^2=(m^2-n^2)^2$
$(m^2+n^2)^2-(m^2-n^2)^2=(2mn)^2=2(2m^2n^2)$
Here $a=m^2+n^2$, $c=m^2-n^2$
Now we rearrange the equation as:
$a^2-c^2=2(c^2-b^2)$
Comparing these relations means:
$c^2-b^2=2m^2n^2$
$2m^2n^2=(m^2-n^2)^2-b^2$
$b^2=m^4+n^4-4m^2n^2$
For example with $m=2$ and $n=1$ we have:
$a=2^2+1^2=5$, $c=2^2-1^2=3$ and:
$b^2=2^4+1^4-4times2^2times1^2=1$ or $b=1$
and we have:
$5^2+2times1^2=3times3^2$
This is primitive solution and subsequent solutions can be found by multiplying both sides of this relation by a number like $n^2$:
$(5n)^2+2(n)^2=(3n)^2$
$a+b+c<=25000000$
$5n+n+3n=9n<=25000000$
$n<=2777777$
This gives a set of solutions. Also $a=b=c=1$ is a primitives solution which gives another set of solutions. It is not known if the parametric formula for $b$ gives more integer solutions for b; if it does there will be more primitive solutions and thereby more solutions.
answered Dec 10 '18 at 14:55
siroussirous
1,6391513
$endgroup$
add a comment |
$begingroup$
$$(2 a^2 - 3 b^2)^2 + 2 (2 a^2 + 3 b^2 - 6 a b)^2 = 3 (2 a^2 + 3 b^2 - 4 a b)^2$$
$$(a^2 - 6 b^2)^2 + 2 (a^2 + 6 b^2 - 6 a b)^2 = 3 (a^2 + 6 b^2 - 4 a b)^2$$
answered Dec 11 '18 at 10:07
S. I.S. I.
112
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Above equation shown below has parametric solution:
$a^2+2b^2=3c^2$
$(a,b,c)=((3k^2-6k-5),(3k^2+6k-1),(3k^2+2k+3))$
For $k=4$ we get:
$(a,b,c)=(5,13,11)$
answered Dec 10 '18 at 13:11
SamSam
1
$endgroup$
add a comment |
$begingroup$
Above equation shown below has parametric solution:
$a^2+2b^2=3c^2$
$(a,b,c)=((3k^2-6k-5),(3k^2+6k-1),(3k^2+2k+3))$
For $k=4$ we get:
$(a,b,c)=(5,13,11)$
answered Dec 10 '18 at 13:11
SamSam
1
$endgroup$
add a comment |
$begingroup$
Above equation shown below has parametric solution:
$a^2+2b^2=3c^2$
$(a,b,c)=((3k^2-6k-5),(3k^2+6k-1),(3k^2+2k+3))$
For $k=4$ we get:
$(a,b,c)=(5,13,11)$
answered Dec 10 '18 at 13:11
SamSam
1
$endgroup$
Above equation shown below has parametric solution:
$a^2+2b^2=3c^2$
$(a,b,c)=((3k^2-6k-5),(3k^2+6k-1),(3k^2+2k+3))$
For $k=4$ we get:
$(a,b,c)=(5,13,11)$
answered Dec 10 '18 at 13:11
SamSam
1
answered Dec 10 '18 at 13:11
SamSam
1
answered Dec 10 '18 at 13:11
SamSam
1
answered Dec 10 '18 at 13:11
SamSam
1
1
add a comment |
add a comment |
$begingroup$
If we use Pythagorean triples we have:
$(m^2-n^2)^2+(2 mn)^2=(m^2-n^2)^2$
$(m^2+n^2)^2-(m^2-n^2)^2=(2mn)^2=2(2m^2n^2)$
Here $a=m^2+n^2$, $c=m^2-n^2$
Now we rearrange the equation as:
$a^2-c^2=2(c^2-b^2)$
Comparing these relations means:
$c^2-b^2=2m^2n^2$
$2m^2n^2=(m^2-n^2)^2-b^2$
$b^2=m^4+n^4-4m^2n^2$
For example with $m=2$ and $n=1$ we have:
$a=2^2+1^2=5$, $c=2^2-1^2=3$ and:
$b^2=2^4+1^4-4times2^2times1^2=1$ or $b=1$
and we have:
$5^2+2times1^2=3times3^2$
This is primitive solution and subsequent solutions can be found by multiplying both sides of this relation by a number like $n^2$:
$(5n)^2+2(n)^2=(3n)^2$
$a+b+c<=25000000$
$5n+n+3n=9n<=25000000$
$n<=2777777$
This gives a set of solutions. Also $a=b=c=1$ is a primitives solution which gives another set of solutions. It is not known if the parametric formula for $b$ gives more integer solutions for b; if it does there will be more primitive solutions and thereby more solutions.
answered Dec 10 '18 at 14:55
siroussirous
1,6391513
$endgroup$
add a comment |
$begingroup$
If we use Pythagorean triples we have:
$(m^2-n^2)^2+(2 mn)^2=(m^2-n^2)^2$
$(m^2+n^2)^2-(m^2-n^2)^2=(2mn)^2=2(2m^2n^2)$
Here $a=m^2+n^2$, $c=m^2-n^2$
Now we rearrange the equation as:
$a^2-c^2=2(c^2-b^2)$
Comparing these relations means:
$c^2-b^2=2m^2n^2$
$2m^2n^2=(m^2-n^2)^2-b^2$
$b^2=m^4+n^4-4m^2n^2$
For example with $m=2$ and $n=1$ we have:
$a=2^2+1^2=5$, $c=2^2-1^2=3$ and:
$b^2=2^4+1^4-4times2^2times1^2=1$ or $b=1$
and we have:
$5^2+2times1^2=3times3^2$
This is primitive solution and subsequent solutions can be found by multiplying both sides of this relation by a number like $n^2$:
$(5n)^2+2(n)^2=(3n)^2$
$a+b+c<=25000000$
$5n+n+3n=9n<=25000000$
$n<=2777777$
This gives a set of solutions. Also $a=b=c=1$ is a primitives solution which gives another set of solutions. It is not known if the parametric formula for $b$ gives more integer solutions for b; if it does there will be more primitive solutions and thereby more solutions.
answered Dec 10 '18 at 14:55
siroussirous
1,6391513
$endgroup$
add a comment |
$begingroup$
If we use Pythagorean triples we have:
$(m^2-n^2)^2+(2 mn)^2=(m^2-n^2)^2$
$(m^2+n^2)^2-(m^2-n^2)^2=(2mn)^2=2(2m^2n^2)$
Here $a=m^2+n^2$, $c=m^2-n^2$
Now we rearrange the equation as:
$a^2-c^2=2(c^2-b^2)$
Comparing these relations means:
$c^2-b^2=2m^2n^2$
$2m^2n^2=(m^2-n^2)^2-b^2$
$b^2=m^4+n^4-4m^2n^2$
For example with $m=2$ and $n=1$ we have:
$a=2^2+1^2=5$, $c=2^2-1^2=3$ and:
$b^2=2^4+1^4-4times2^2times1^2=1$ or $b=1$
and we have:
$5^2+2times1^2=3times3^2$
This is primitive solution and subsequent solutions can be found by multiplying both sides of this relation by a number like $n^2$:
$(5n)^2+2(n)^2=(3n)^2$
$a+b+c<=25000000$
$5n+n+3n=9n<=25000000$
$n<=2777777$
This gives a set of solutions. Also $a=b=c=1$ is a primitives solution which gives another set of solutions. It is not known if the parametric formula for $b$ gives more integer solutions for b; if it does there will be more primitive solutions and thereby more solutions.
answered Dec 10 '18 at 14:55
siroussirous
1,6391513
$endgroup$
If we use Pythagorean triples we have:
$(m^2-n^2)^2+(2 mn)^2=(m^2-n^2)^2$
$(m^2+n^2)^2-(m^2-n^2)^2=(2mn)^2=2(2m^2n^2)$
Here $a=m^2+n^2$, $c=m^2-n^2$
Now we rearrange the equation as:
$a^2-c^2=2(c^2-b^2)$
Comparing these relations means:
$c^2-b^2=2m^2n^2$
$2m^2n^2=(m^2-n^2)^2-b^2$
$b^2=m^4+n^4-4m^2n^2$
For example with $m=2$ and $n=1$ we have:
$a=2^2+1^2=5$, $c=2^2-1^2=3$ and:
$b^2=2^4+1^4-4times2^2times1^2=1$ or $b=1$
and we have:
$5^2+2times1^2=3times3^2$
This is primitive solution and subsequent solutions can be found by multiplying both sides of this relation by a number like $n^2$:
$(5n)^2+2(n)^2=(3n)^2$
$a+b+c<=25000000$
$5n+n+3n=9n<=25000000$
$n<=2777777$
This gives a set of solutions. Also $a=b=c=1$ is a primitives solution which gives another set of solutions. It is not known if the parametric formula for $b$ gives more integer solutions for b; if it does there will be more primitive solutions and thereby more solutions.
answered Dec 10 '18 at 14:55
siroussirous
1,6391513
answered Dec 10 '18 at 14:55
siroussirous
1,6391513
answered Dec 10 '18 at 14:55
siroussirous
1,6391513
answered Dec 10 '18 at 14:55
siroussirous
1,6391513
1,6391513
add a comment |
add a comment |
$begingroup$
$$(2 a^2 - 3 b^2)^2 + 2 (2 a^2 + 3 b^2 - 6 a b)^2 = 3 (2 a^2 + 3 b^2 - 4 a b)^2$$
$$(a^2 - 6 b^2)^2 + 2 (a^2 + 6 b^2 - 6 a b)^2 = 3 (a^2 + 6 b^2 - 4 a b)^2$$
answered Dec 11 '18 at 10:07
S. I.S. I.
112
$endgroup$
add a comment |
$begingroup$
$$(2 a^2 - 3 b^2)^2 + 2 (2 a^2 + 3 b^2 - 6 a b)^2 = 3 (2 a^2 + 3 b^2 - 4 a b)^2$$
$$(a^2 - 6 b^2)^2 + 2 (a^2 + 6 b^2 - 6 a b)^2 = 3 (a^2 + 6 b^2 - 4 a b)^2$$
answered Dec 11 '18 at 10:07
S. I.S. I.
112
$endgroup$
add a comment |
$begingroup$
$$(2 a^2 - 3 b^2)^2 + 2 (2 a^2 + 3 b^2 - 6 a b)^2 = 3 (2 a^2 + 3 b^2 - 4 a b)^2$$
$$(a^2 - 6 b^2)^2 + 2 (a^2 + 6 b^2 - 6 a b)^2 = 3 (a^2 + 6 b^2 - 4 a b)^2$$
answered Dec 11 '18 at 10:07
S. I.S. I.
112
$endgroup$
$$(2 a^2 - 3 b^2)^2 + 2 (2 a^2 + 3 b^2 - 6 a b)^2 = 3 (2 a^2 + 3 b^2 - 4 a b)^2$$
$$(a^2 - 6 b^2)^2 + 2 (a^2 + 6 b^2 - 6 a b)^2 = 3 (a^2 + 6 b^2 - 4 a b)^2$$
answered Dec 11 '18 at 10:07
S. I.S. I.
112
answered Dec 11 '18 at 10:07
S. I.S. I.
112
answered Dec 11 '18 at 10:07
S. I.S. I.
112
answered Dec 11 '18 at 10:07
S. I.S. I.
112
112
add a comment |
add a comment |
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math.stackexchange.com/questions/2773097/…
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– individ
Dec 10 '18 at 13:14
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This doesn't give all the solutions.
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– Asif
Dec 11 '18 at 6:02
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Gives all solutions. It is necessary to accurately and attentively all to write down. Although you can write such a simple solution. $$a^2+2b^2=3c^2$$ $$a=p^2+6ps+3s^2$$ $$b=p^2-3s^2$$ $$c=p^2+2ps+3s^2$$
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– individ
Dec 11 '18 at 6:41
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I am not sure how you arrived at this parametric solution, but it certainly doesn't give all the solutions and also produces lots of negative integer solutions, which aren't necessary.
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– Asif
Dec 11 '18 at 16:22
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Rewrite the negative into a positive solution. You want to get a solution what you like, not a solution to the problem. That's not how math works. We may not like the answer, but the solution is the solution.
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– individ
Dec 11 '18 at 16:37