Number theory on square numbers [duplicate]
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This question already has an answer here:
Number of integer solutions of $x^2 + y^2 = k$
3 answers
Can a square number be split as the sum of two other squares in two different ways?
The Pythagorean numbers help us to identify numbers which can be split as sum of two squares. Is this unique?
number-theory
asked Dec 17 '18 at 1:39
Narayanan RamanNarayanan Raman
1
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marked as duplicate by Saad, Jyrki Lahtonen, Ben, Lord Shark the Unknown, apnorton Dec 17 '18 at 7:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Number of integer solutions of $x^2 + y^2 = k$
3 answers
Can a square number be split as the sum of two other squares in two different ways?
The Pythagorean numbers help us to identify numbers which can be split as sum of two squares. Is this unique?
number-theory
asked Dec 17 '18 at 1:39
Narayanan RamanNarayanan Raman
1
$endgroup$
marked as duplicate by Saad, Jyrki Lahtonen, Ben, Lord Shark the Unknown, apnorton Dec 17 '18 at 7:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
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Number of integer solutions of $x^2 + y^2 = k$ gives all the information you need. And then some. Voting to close as a duplicate.
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– Jyrki Lahtonen
Dec 17 '18 at 5:24
add a comment |
$begingroup$
This question already has an answer here:
Number of integer solutions of $x^2 + y^2 = k$
3 answers
Can a square number be split as the sum of two other squares in two different ways?
The Pythagorean numbers help us to identify numbers which can be split as sum of two squares. Is this unique?
number-theory
asked Dec 17 '18 at 1:39
Narayanan RamanNarayanan Raman
1
$endgroup$
This question already has an answer here:
Number of integer solutions of $x^2 + y^2 = k$
3 answers
Can a square number be split as the sum of two other squares in two different ways?
The Pythagorean numbers help us to identify numbers which can be split as sum of two squares. Is this unique?
This question already has an answer here:
Number of integer solutions of $x^2 + y^2 = k$
3 answers
number-theory
number-theory
asked Dec 17 '18 at 1:39
Narayanan RamanNarayanan Raman
1
asked Dec 17 '18 at 1:39
Narayanan RamanNarayanan Raman
1
asked Dec 17 '18 at 1:39
Narayanan RamanNarayanan Raman
1
asked Dec 17 '18 at 1:39
Narayanan RamanNarayanan Raman
1
asked Dec 17 '18 at 1:39
Narayanan RamanNarayanan Raman
1
1
marked as duplicate by Saad, Jyrki Lahtonen, Ben, Lord Shark the Unknown, apnorton Dec 17 '18 at 7:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Saad, Jyrki Lahtonen, Ben, Lord Shark the Unknown, apnorton Dec 17 '18 at 7:07
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
$begingroup$
Number of integer solutions of $x^2 + y^2 = k$ gives all the information you need. And then some. Voting to close as a duplicate.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 5:24
add a comment |
3
$begingroup$
Number of integer solutions of $x^2 + y^2 = k$ gives all the information you need. And then some. Voting to close as a duplicate.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 5:24
3
3
$begingroup$
Number of integer solutions of $x^2 + y^2 = k$ gives all the information you need. And then some. Voting to close as a duplicate.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 5:24
$begingroup$
Number of integer solutions of $x^2 + y^2 = k$ gives all the information you need. And then some. Voting to close as a duplicate.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 5:24
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
By the Brahmagupta–Fibonacci identity, which states that $$(a^2 + b^2)(c^2 + d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$ you can get a number which is a sum of two squares two different ways by multiplying two sums of two squares.
So from two Pythagorean triples $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ , the left hand side simplifies to $(a_1a_2)^2$, you can get a number that can be written as a sum of two squares.
answered Dec 17 '18 at 1:59
twnlytwnly
1,1221213
$endgroup$
$begingroup$
Love whipping out the Brahmagupta-Fibonacci identity. math.stackexchange.com/questions/1012733/…
$endgroup$
– Matt Samuel
Dec 17 '18 at 2:40
1
$begingroup$
Of course if you have $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ then you also have $(a_2a)^2 + (a_2b)^2 = (a_1a_2)^2$ and $(a_1c)^2 + (a_1d)^2 = (a_1a_2)^2.$ This is how $65^2$ comes to be the sum of two squares in four different ways: take $a=4,b=3,c=12,d=5.$
$endgroup$
– David K
Dec 17 '18 at 12:00
add a comment |
$begingroup$
HINT
$3,4,5$ is a PPT, meaning that any multiple (like $6,8,10$) is a PT
$5,12,13$ is a PPT as well ...
answered Dec 17 '18 at 1:51
Bram28Bram28
63.7k44793
$endgroup$
1
$begingroup$
Why the downvote? Not a good hint?
$endgroup$
– Bram28
Dec 17 '18 at 2:02
add a comment |
$begingroup$
For example
begin{eqnarray*}
65^2=63^2+16^2=33^2+56^2.
end{eqnarray*}
answered Dec 17 '18 at 1:51
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
$begingroup$
In fact this happens infinitely often, see this sequence of the Online Encyclopedia of Integer Sequences.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 2:26
$begingroup$
In addition, there are the obvious $65^2=26^2+52^2=25^2+60^2$ (multiples of $3,4,5$ and $5,12,13$). So that's not just two, but four ways for $65^2.$
$endgroup$
– David K
Dec 17 '18 at 2:43
$begingroup$
@BrianHopkins Most of the numbers in that list are not squares, but of course some are.
$endgroup$
– David K
Dec 17 '18 at 2:46
1
$begingroup$
@DavidK Oops, yes: of the many related sequences, I think A009177 is more relevant.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 3:15
$begingroup$
Thanks for the solution and illuminating discussion
$endgroup$
– Narayanan Raman
Dec 30 '18 at 1:03
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By the Brahmagupta–Fibonacci identity, which states that $$(a^2 + b^2)(c^2 + d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$ you can get a number which is a sum of two squares two different ways by multiplying two sums of two squares.
So from two Pythagorean triples $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ , the left hand side simplifies to $(a_1a_2)^2$, you can get a number that can be written as a sum of two squares.
answered Dec 17 '18 at 1:59
twnlytwnly
1,1221213
$endgroup$
$begingroup$
Love whipping out the Brahmagupta-Fibonacci identity. math.stackexchange.com/questions/1012733/…
$endgroup$
– Matt Samuel
Dec 17 '18 at 2:40
1
$begingroup$
Of course if you have $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ then you also have $(a_2a)^2 + (a_2b)^2 = (a_1a_2)^2$ and $(a_1c)^2 + (a_1d)^2 = (a_1a_2)^2.$ This is how $65^2$ comes to be the sum of two squares in four different ways: take $a=4,b=3,c=12,d=5.$
$endgroup$
– David K
Dec 17 '18 at 12:00
add a comment |
$begingroup$
By the Brahmagupta–Fibonacci identity, which states that $$(a^2 + b^2)(c^2 + d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$ you can get a number which is a sum of two squares two different ways by multiplying two sums of two squares.
So from two Pythagorean triples $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ , the left hand side simplifies to $(a_1a_2)^2$, you can get a number that can be written as a sum of two squares.
answered Dec 17 '18 at 1:59
twnlytwnly
1,1221213
$endgroup$
$begingroup$
Love whipping out the Brahmagupta-Fibonacci identity. math.stackexchange.com/questions/1012733/…
$endgroup$
– Matt Samuel
Dec 17 '18 at 2:40
1
$begingroup$
Of course if you have $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ then you also have $(a_2a)^2 + (a_2b)^2 = (a_1a_2)^2$ and $(a_1c)^2 + (a_1d)^2 = (a_1a_2)^2.$ This is how $65^2$ comes to be the sum of two squares in four different ways: take $a=4,b=3,c=12,d=5.$
$endgroup$
– David K
Dec 17 '18 at 12:00
add a comment |
$begingroup$
By the Brahmagupta–Fibonacci identity, which states that $$(a^2 + b^2)(c^2 + d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$ you can get a number which is a sum of two squares two different ways by multiplying two sums of two squares.
So from two Pythagorean triples $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ , the left hand side simplifies to $(a_1a_2)^2$, you can get a number that can be written as a sum of two squares.
answered Dec 17 '18 at 1:59
twnlytwnly
1,1221213
$endgroup$
By the Brahmagupta–Fibonacci identity, which states that $$(a^2 + b^2)(c^2 + d^2) = (ac-bd)^2 + (ad+bc)^2 = (ac+bd)^2 + (ad-bc)^2$$ you can get a number which is a sum of two squares two different ways by multiplying two sums of two squares.
So from two Pythagorean triples $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ , the left hand side simplifies to $(a_1a_2)^2$, you can get a number that can be written as a sum of two squares.
answered Dec 17 '18 at 1:59
twnlytwnly
1,1221213
answered Dec 17 '18 at 1:59
twnlytwnly
1,1221213
answered Dec 17 '18 at 1:59
twnlytwnly
1,1221213
answered Dec 17 '18 at 1:59
twnlytwnly
1,1221213
1,1221213
$begingroup$
Love whipping out the Brahmagupta-Fibonacci identity. math.stackexchange.com/questions/1012733/…
$endgroup$
– Matt Samuel
Dec 17 '18 at 2:40
1
$begingroup$
Of course if you have $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ then you also have $(a_2a)^2 + (a_2b)^2 = (a_1a_2)^2$ and $(a_1c)^2 + (a_1d)^2 = (a_1a_2)^2.$ This is how $65^2$ comes to be the sum of two squares in four different ways: take $a=4,b=3,c=12,d=5.$
$endgroup$
– David K
Dec 17 '18 at 12:00
add a comment |
$begingroup$
Love whipping out the Brahmagupta-Fibonacci identity. math.stackexchange.com/questions/1012733/…
$endgroup$
– Matt Samuel
Dec 17 '18 at 2:40
1
$begingroup$
Of course if you have $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ then you also have $(a_2a)^2 + (a_2b)^2 = (a_1a_2)^2$ and $(a_1c)^2 + (a_1d)^2 = (a_1a_2)^2.$ This is how $65^2$ comes to be the sum of two squares in four different ways: take $a=4,b=3,c=12,d=5.$
$endgroup$
– David K
Dec 17 '18 at 12:00
$begingroup$
Love whipping out the Brahmagupta-Fibonacci identity. math.stackexchange.com/questions/1012733/…
$endgroup$
– Matt Samuel
Dec 17 '18 at 2:40
$begingroup$
Love whipping out the Brahmagupta-Fibonacci identity. math.stackexchange.com/questions/1012733/…
$endgroup$
– Matt Samuel
Dec 17 '18 at 2:40
1
1
$begingroup$
Of course if you have $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ then you also have $(a_2a)^2 + (a_2b)^2 = (a_1a_2)^2$ and $(a_1c)^2 + (a_1d)^2 = (a_1a_2)^2.$ This is how $65^2$ comes to be the sum of two squares in four different ways: take $a=4,b=3,c=12,d=5.$
$endgroup$
– David K
Dec 17 '18 at 12:00
$begingroup$
Of course if you have $a^2 + b^2 = a_1^2$ and $c^2 + d^2 = a_2^2$ then you also have $(a_2a)^2 + (a_2b)^2 = (a_1a_2)^2$ and $(a_1c)^2 + (a_1d)^2 = (a_1a_2)^2.$ This is how $65^2$ comes to be the sum of two squares in four different ways: take $a=4,b=3,c=12,d=5.$
$endgroup$
– David K
Dec 17 '18 at 12:00
add a comment |
$begingroup$
HINT
$3,4,5$ is a PPT, meaning that any multiple (like $6,8,10$) is a PT
$5,12,13$ is a PPT as well ...
answered Dec 17 '18 at 1:51
Bram28Bram28
63.7k44793
$endgroup$
1
$begingroup$
Why the downvote? Not a good hint?
$endgroup$
– Bram28
Dec 17 '18 at 2:02
add a comment |
$begingroup$
HINT
$3,4,5$ is a PPT, meaning that any multiple (like $6,8,10$) is a PT
$5,12,13$ is a PPT as well ...
answered Dec 17 '18 at 1:51
Bram28Bram28
63.7k44793
$endgroup$
1
$begingroup$
Why the downvote? Not a good hint?
$endgroup$
– Bram28
Dec 17 '18 at 2:02
add a comment |
$begingroup$
HINT
$3,4,5$ is a PPT, meaning that any multiple (like $6,8,10$) is a PT
$5,12,13$ is a PPT as well ...
answered Dec 17 '18 at 1:51
Bram28Bram28
63.7k44793
$endgroup$
HINT
$3,4,5$ is a PPT, meaning that any multiple (like $6,8,10$) is a PT
$5,12,13$ is a PPT as well ...
answered Dec 17 '18 at 1:51
Bram28Bram28
63.7k44793
answered Dec 17 '18 at 1:51
Bram28Bram28
63.7k44793
answered Dec 17 '18 at 1:51
Bram28Bram28
63.7k44793
answered Dec 17 '18 at 1:51
Bram28Bram28
63.7k44793
63.7k44793
1
$begingroup$
Why the downvote? Not a good hint?
$endgroup$
– Bram28
Dec 17 '18 at 2:02
add a comment |
1
$begingroup$
Why the downvote? Not a good hint?
$endgroup$
– Bram28
Dec 17 '18 at 2:02
1
1
$begingroup$
Why the downvote? Not a good hint?
$endgroup$
– Bram28
Dec 17 '18 at 2:02
$begingroup$
Why the downvote? Not a good hint?
$endgroup$
– Bram28
Dec 17 '18 at 2:02
add a comment |
$begingroup$
For example
begin{eqnarray*}
65^2=63^2+16^2=33^2+56^2.
end{eqnarray*}
answered Dec 17 '18 at 1:51
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
$begingroup$
In fact this happens infinitely often, see this sequence of the Online Encyclopedia of Integer Sequences.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 2:26
$begingroup$
In addition, there are the obvious $65^2=26^2+52^2=25^2+60^2$ (multiples of $3,4,5$ and $5,12,13$). So that's not just two, but four ways for $65^2.$
$endgroup$
– David K
Dec 17 '18 at 2:43
$begingroup$
@BrianHopkins Most of the numbers in that list are not squares, but of course some are.
$endgroup$
– David K
Dec 17 '18 at 2:46
1
$begingroup$
@DavidK Oops, yes: of the many related sequences, I think A009177 is more relevant.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 3:15
$begingroup$
Thanks for the solution and illuminating discussion
$endgroup$
– Narayanan Raman
Dec 30 '18 at 1:03
add a comment |
$begingroup$
For example
begin{eqnarray*}
65^2=63^2+16^2=33^2+56^2.
end{eqnarray*}
answered Dec 17 '18 at 1:51
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
$begingroup$
In fact this happens infinitely often, see this sequence of the Online Encyclopedia of Integer Sequences.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 2:26
$begingroup$
In addition, there are the obvious $65^2=26^2+52^2=25^2+60^2$ (multiples of $3,4,5$ and $5,12,13$). So that's not just two, but four ways for $65^2.$
$endgroup$
– David K
Dec 17 '18 at 2:43
$begingroup$
@BrianHopkins Most of the numbers in that list are not squares, but of course some are.
$endgroup$
– David K
Dec 17 '18 at 2:46
1
$begingroup$
@DavidK Oops, yes: of the many related sequences, I think A009177 is more relevant.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 3:15
$begingroup$
Thanks for the solution and illuminating discussion
$endgroup$
– Narayanan Raman
Dec 30 '18 at 1:03
add a comment |
$begingroup$
For example
begin{eqnarray*}
65^2=63^2+16^2=33^2+56^2.
end{eqnarray*}
answered Dec 17 '18 at 1:51
Donald SplutterwitDonald Splutterwit
22.9k21446
$endgroup$
For example
begin{eqnarray*}
65^2=63^2+16^2=33^2+56^2.
end{eqnarray*}
answered Dec 17 '18 at 1:51
Donald SplutterwitDonald Splutterwit
22.9k21446
answered Dec 17 '18 at 1:51
Donald SplutterwitDonald Splutterwit
22.9k21446
answered Dec 17 '18 at 1:51
Donald SplutterwitDonald Splutterwit
22.9k21446
answered Dec 17 '18 at 1:51
Donald SplutterwitDonald Splutterwit
22.9k21446
22.9k21446
$begingroup$
In fact this happens infinitely often, see this sequence of the Online Encyclopedia of Integer Sequences.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 2:26
$begingroup$
In addition, there are the obvious $65^2=26^2+52^2=25^2+60^2$ (multiples of $3,4,5$ and $5,12,13$). So that's not just two, but four ways for $65^2.$
$endgroup$
– David K
Dec 17 '18 at 2:43
$begingroup$
@BrianHopkins Most of the numbers in that list are not squares, but of course some are.
$endgroup$
– David K
Dec 17 '18 at 2:46
1
$begingroup$
@DavidK Oops, yes: of the many related sequences, I think A009177 is more relevant.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 3:15
$begingroup$
Thanks for the solution and illuminating discussion
$endgroup$
– Narayanan Raman
Dec 30 '18 at 1:03
add a comment |
$begingroup$
In fact this happens infinitely often, see this sequence of the Online Encyclopedia of Integer Sequences.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 2:26
$begingroup$
In addition, there are the obvious $65^2=26^2+52^2=25^2+60^2$ (multiples of $3,4,5$ and $5,12,13$). So that's not just two, but four ways for $65^2.$
$endgroup$
– David K
Dec 17 '18 at 2:43
$begingroup$
@BrianHopkins Most of the numbers in that list are not squares, but of course some are.
$endgroup$
– David K
Dec 17 '18 at 2:46
1
$begingroup$
@DavidK Oops, yes: of the many related sequences, I think A009177 is more relevant.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 3:15
$begingroup$
Thanks for the solution and illuminating discussion
$endgroup$
– Narayanan Raman
Dec 30 '18 at 1:03
$begingroup$
In fact this happens infinitely often, see this sequence of the Online Encyclopedia of Integer Sequences.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 2:26
$begingroup$
In fact this happens infinitely often, see this sequence of the Online Encyclopedia of Integer Sequences.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 2:26
$begingroup$
In addition, there are the obvious $65^2=26^2+52^2=25^2+60^2$ (multiples of $3,4,5$ and $5,12,13$). So that's not just two, but four ways for $65^2.$
$endgroup$
– David K
Dec 17 '18 at 2:43
$begingroup$
In addition, there are the obvious $65^2=26^2+52^2=25^2+60^2$ (multiples of $3,4,5$ and $5,12,13$). So that's not just two, but four ways for $65^2.$
$endgroup$
– David K
Dec 17 '18 at 2:43
$begingroup$
@BrianHopkins Most of the numbers in that list are not squares, but of course some are.
$endgroup$
– David K
Dec 17 '18 at 2:46
$begingroup$
@BrianHopkins Most of the numbers in that list are not squares, but of course some are.
$endgroup$
– David K
Dec 17 '18 at 2:46
1
1
$begingroup$
@DavidK Oops, yes: of the many related sequences, I think A009177 is more relevant.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 3:15
$begingroup$
@DavidK Oops, yes: of the many related sequences, I think A009177 is more relevant.
$endgroup$
– Brian Hopkins
Dec 17 '18 at 3:15
$begingroup$
Thanks for the solution and illuminating discussion
$endgroup$
– Narayanan Raman
Dec 30 '18 at 1:03
$begingroup$
Thanks for the solution and illuminating discussion
$endgroup$
– Narayanan Raman
Dec 30 '18 at 1:03
add a comment |
3
$begingroup$
Number of integer solutions of $x^2 + y^2 = k$ gives all the information you need. And then some. Voting to close as a duplicate.
$endgroup$
– Jyrki Lahtonen
Dec 17 '18 at 5:24