Mixing
Is there a perfect square that is the sum of $3$ perfect squares?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
This is part of a bigger question, but it boils down to:
Is there a square number that is equal to the sum of three different square numbers?
I could only find a special case where two of the three are equal? https://pir2.forumeiros.com/t86615-soma-de-tres-quadrados (in portuguese).
Any clue?
number-theory square-numbers
edited Jul 27 at 3:38
user21820
35.8k440136
asked Jul 26 at 18:11
Leonardo Barichello
192
add a comment |Â
up vote
3
down vote
favorite
This is part of a bigger question, but it boils down to:
Is there a square number that is equal to the sum of three different square numbers?
I could only find a special case where two of the three are equal? https://pir2.forumeiros.com/t86615-soma-de-tres-quadrados (in portuguese).
Any clue?
number-theory square-numbers
edited Jul 27 at 3:38
user21820
35.8k440136
asked Jul 26 at 18:11
Leonardo Barichello
192
1
Yes, google pythagorean quartuples
– Peter
Jul 26 at 18:12
6
... or Pythagorean quadruple. For other interesting questions, see Euler brick.
– Joffan
Jul 26 at 18:18
6
$$(2ps)^2+(2ks)^2+(p^2+k^2-s^2)^2=(p^2+k^2+s^2)^2$$
– individ
Jul 26 at 18:20
1
In Python,[[(a,b,c,d) for (a,b,c,d) in itertools.product(range(1,21),repeat = 4) if a**2 + b**2 + c**2 - d**2 == 0]will instantly give you the 99 solutions with all terms <= 20.
– John Coleman
Jul 26 at 21:07
Please read how-to-ask. In particular, what is the bigger question?
– user21820
Jul 27 at 3:39
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
This is part of a bigger question, but it boils down to:
Is there a square number that is equal to the sum of three different square numbers?
I could only find a special case where two of the three are equal? https://pir2.forumeiros.com/t86615-soma-de-tres-quadrados (in portuguese).
Any clue?
number-theory square-numbers
edited Jul 27 at 3:38
user21820
35.8k440136
asked Jul 26 at 18:11
Leonardo Barichello
192
This is part of a bigger question, but it boils down to:
Is there a square number that is equal to the sum of three different square numbers?
I could only find a special case where two of the three are equal? https://pir2.forumeiros.com/t86615-soma-de-tres-quadrados (in portuguese).
Any clue?
number-theory square-numbers
edited Jul 27 at 3:38
user21820
35.8k440136
asked Jul 26 at 18:11
Leonardo Barichello
192
edited Jul 27 at 3:38
user21820
35.8k440136
edited Jul 27 at 3:38
user21820
35.8k440136
edited Jul 27 at 3:38
user21820
35.8k440136
35.8k440136
asked Jul 26 at 18:11
Leonardo Barichello
192
asked Jul 26 at 18:11
Leonardo Barichello
192
asked Jul 26 at 18:11
Leonardo Barichello
192
192
1
Yes, google pythagorean quartuples
– Peter
Jul 26 at 18:12
6
... or Pythagorean quadruple. For other interesting questions, see Euler brick.
– Joffan
Jul 26 at 18:18
6
$$(2ps)^2+(2ks)^2+(p^2+k^2-s^2)^2=(p^2+k^2+s^2)^2$$
– individ
Jul 26 at 18:20
1
In Python,[[(a,b,c,d) for (a,b,c,d) in itertools.product(range(1,21),repeat = 4) if a**2 + b**2 + c**2 - d**2 == 0]will instantly give you the 99 solutions with all terms <= 20.
– John Coleman
Jul 26 at 21:07
Please read how-to-ask. In particular, what is the bigger question?
– user21820
Jul 27 at 3:39
add a comment |Â
1
Yes, google pythagorean quartuples
– Peter
Jul 26 at 18:12
6
... or Pythagorean quadruple. For other interesting questions, see Euler brick.
– Joffan
Jul 26 at 18:18
6
$$(2ps)^2+(2ks)^2+(p^2+k^2-s^2)^2=(p^2+k^2+s^2)^2$$
– individ
Jul 26 at 18:20
1
In Python,[[(a,b,c,d) for (a,b,c,d) in itertools.product(range(1,21),repeat = 4) if a**2 + b**2 + c**2 - d**2 == 0]will instantly give you the 99 solutions with all terms <= 20.
– John Coleman
Jul 26 at 21:07
Please read how-to-ask. In particular, what is the bigger question?
– user21820
Jul 27 at 3:39
1
1
Yes, google pythagorean quartuples
– Peter
Jul 26 at 18:12
Yes, google pythagorean quartuples
– Peter
Jul 26 at 18:12
6
6
... or Pythagorean quadruple. For other interesting questions, see Euler brick.
– Joffan
Jul 26 at 18:18
... or Pythagorean quadruple. For other interesting questions, see Euler brick.
– Joffan
Jul 26 at 18:18
6
6
$$(2ps)^2+(2ks)^2+(p^2+k^2-s^2)^2=(p^2+k^2+s^2)^2$$
– individ
Jul 26 at 18:20
$$(2ps)^2+(2ks)^2+(p^2+k^2-s^2)^2=(p^2+k^2+s^2)^2$$
– individ
Jul 26 at 18:20
1
1
In Python,
[[(a,b,c,d) for (a,b,c,d) in itertools.product(range(1,21),repeat = 4) if a**2 + b**2 + c**2 - d**2 == 0] will instantly give you the 99 solutions with all terms <= 20.– John Coleman
Jul 26 at 21:07
In Python,
[[(a,b,c,d) for (a,b,c,d) in itertools.product(range(1,21),repeat = 4) if a**2 + b**2 + c**2 - d**2 == 0] will instantly give you the 99 solutions with all terms <= 20.– John Coleman
Jul 26 at 21:07
Please read how-to-ask. In particular, what is the bigger question?
– user21820
Jul 27 at 3:39
Please read how-to-ask. In particular, what is the bigger question?
– user21820
Jul 27 at 3:39
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
16
down vote
$$3^2+4^2+12^2=13^2$$
answered Jul 26 at 18:12
Lord Shark the Unknown
84.6k950111
add a comment |Â
up vote
12
down vote
We know that $(n+1)^2-n^2=2n+1$, so pick your favorite Pythagorean triple $a^2+b^2=c^2$ with $c$ odd. Let $c^2=2n+1, n=frac c^2-12$ and
$$a^2+b^2+n^2=(n+1)^2$$
If you pick a triple with $c$ even, we can use $(n+2)^2-n^2=4n+4$, so we can let $n=frac c^2-44$ and have $$a^2+b^2+n^2=(n+2)^2$$
If $c$ is even, $c^2$ is divisible by $4$, so the division will come out even.
answered Jul 26 at 18:34
Ross Millikan
275k21186351
add a comment |Â
up vote
5
down vote
It really is a four variable parametrization, coming from quaternion multiplication. The attribution in wikipedia is to the number theorist V. A. Lebesgue. He did publish on this in the 1850's. However, this was surely known to Euler.
Discussed in detail in Pall 1940
I used those techniques, quaternions, for related problems Find three numbers such that the sum of all three is a square and the sum of any two is a square and
http://math.stackexchange.com/questions/1964607/when-will-a-parametric-solution-generate-all-possible-solutions/1965805#1965805 I was quite proud of this one, it gives a complete integer parametrization for $a^2 + b^2 + c^2 = 3 d^2$ using material from Jones and Pall 1939
Meanwhile, the first acceptable proof that ALL quadruples arise this way (primitive ones, that is) was due to L. E. Dickson in 1920. There is a very nice article by Spira, let me find a link. I also have a pdf of Spira (1962).
answered Jul 26 at 19:04
Will Jagy
96.9k594195
add a comment |Â
up vote
2
down vote
There are many of them.
$$113^2 = 112^2+12^2 +9^2 $$
is an example.
answered Jul 26 at 18:20
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
2
down vote
Dickson's History of the Theory of Numbers,
volume 2,
starting on page 261
has a number of solutions.
Here is one due to Euler:
$(p^2 + 1)^2(q^2 + 1)^2
= (q^2 - 1)^2(p^2 + 1)^2 + 4q^2(p^2 - 1)^2 + (4pq)^2
$.
answered Jul 26 at 18:37
marty cohen
69.1k446122
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
16
down vote
$$3^2+4^2+12^2=13^2$$
answered Jul 26 at 18:12
Lord Shark the Unknown
84.6k950111
add a comment |Â
up vote
16
down vote
$$3^2+4^2+12^2=13^2$$
answered Jul 26 at 18:12
Lord Shark the Unknown
84.6k950111
add a comment |Â
up vote
16
down vote
up vote
16
down vote
$$3^2+4^2+12^2=13^2$$
answered Jul 26 at 18:12
Lord Shark the Unknown
84.6k950111
$$3^2+4^2+12^2=13^2$$
answered Jul 26 at 18:12
Lord Shark the Unknown
84.6k950111
answered Jul 26 at 18:12
Lord Shark the Unknown
84.6k950111
answered Jul 26 at 18:12
Lord Shark the Unknown
84.6k950111
answered Jul 26 at 18:12
Lord Shark the Unknown
84.6k950111
84.6k950111
add a comment |Â
add a comment |Â
up vote
12
down vote
We know that $(n+1)^2-n^2=2n+1$, so pick your favorite Pythagorean triple $a^2+b^2=c^2$ with $c$ odd. Let $c^2=2n+1, n=frac c^2-12$ and
$$a^2+b^2+n^2=(n+1)^2$$
If you pick a triple with $c$ even, we can use $(n+2)^2-n^2=4n+4$, so we can let $n=frac c^2-44$ and have $$a^2+b^2+n^2=(n+2)^2$$
If $c$ is even, $c^2$ is divisible by $4$, so the division will come out even.
answered Jul 26 at 18:34
Ross Millikan
275k21186351
add a comment |Â
up vote
12
down vote
We know that $(n+1)^2-n^2=2n+1$, so pick your favorite Pythagorean triple $a^2+b^2=c^2$ with $c$ odd. Let $c^2=2n+1, n=frac c^2-12$ and
$$a^2+b^2+n^2=(n+1)^2$$
If you pick a triple with $c$ even, we can use $(n+2)^2-n^2=4n+4$, so we can let $n=frac c^2-44$ and have $$a^2+b^2+n^2=(n+2)^2$$
If $c$ is even, $c^2$ is divisible by $4$, so the division will come out even.
answered Jul 26 at 18:34
Ross Millikan
275k21186351
add a comment |Â
up vote
12
down vote
up vote
12
down vote
We know that $(n+1)^2-n^2=2n+1$, so pick your favorite Pythagorean triple $a^2+b^2=c^2$ with $c$ odd. Let $c^2=2n+1, n=frac c^2-12$ and
$$a^2+b^2+n^2=(n+1)^2$$
If you pick a triple with $c$ even, we can use $(n+2)^2-n^2=4n+4$, so we can let $n=frac c^2-44$ and have $$a^2+b^2+n^2=(n+2)^2$$
If $c$ is even, $c^2$ is divisible by $4$, so the division will come out even.
answered Jul 26 at 18:34
Ross Millikan
275k21186351
We know that $(n+1)^2-n^2=2n+1$, so pick your favorite Pythagorean triple $a^2+b^2=c^2$ with $c$ odd. Let $c^2=2n+1, n=frac c^2-12$ and
$$a^2+b^2+n^2=(n+1)^2$$
If you pick a triple with $c$ even, we can use $(n+2)^2-n^2=4n+4$, so we can let $n=frac c^2-44$ and have $$a^2+b^2+n^2=(n+2)^2$$
If $c$ is even, $c^2$ is divisible by $4$, so the division will come out even.
answered Jul 26 at 18:34
Ross Millikan
275k21186351
answered Jul 26 at 18:34
Ross Millikan
275k21186351
answered Jul 26 at 18:34
Ross Millikan
275k21186351
answered Jul 26 at 18:34
Ross Millikan
275k21186351
275k21186351
add a comment |Â
add a comment |Â
up vote
5
down vote
It really is a four variable parametrization, coming from quaternion multiplication. The attribution in wikipedia is to the number theorist V. A. Lebesgue. He did publish on this in the 1850's. However, this was surely known to Euler.
Discussed in detail in Pall 1940
I used those techniques, quaternions, for related problems Find three numbers such that the sum of all three is a square and the sum of any two is a square and
http://math.stackexchange.com/questions/1964607/when-will-a-parametric-solution-generate-all-possible-solutions/1965805#1965805 I was quite proud of this one, it gives a complete integer parametrization for $a^2 + b^2 + c^2 = 3 d^2$ using material from Jones and Pall 1939
Meanwhile, the first acceptable proof that ALL quadruples arise this way (primitive ones, that is) was due to L. E. Dickson in 1920. There is a very nice article by Spira, let me find a link. I also have a pdf of Spira (1962).
answered Jul 26 at 19:04
Will Jagy
96.9k594195
add a comment |Â
up vote
5
down vote
It really is a four variable parametrization, coming from quaternion multiplication. The attribution in wikipedia is to the number theorist V. A. Lebesgue. He did publish on this in the 1850's. However, this was surely known to Euler.
Discussed in detail in Pall 1940
I used those techniques, quaternions, for related problems Find three numbers such that the sum of all three is a square and the sum of any two is a square and
http://math.stackexchange.com/questions/1964607/when-will-a-parametric-solution-generate-all-possible-solutions/1965805#1965805 I was quite proud of this one, it gives a complete integer parametrization for $a^2 + b^2 + c^2 = 3 d^2$ using material from Jones and Pall 1939
Meanwhile, the first acceptable proof that ALL quadruples arise this way (primitive ones, that is) was due to L. E. Dickson in 1920. There is a very nice article by Spira, let me find a link. I also have a pdf of Spira (1962).
answered Jul 26 at 19:04
Will Jagy
96.9k594195
add a comment |Â
up vote
5
down vote
up vote
5
down vote
It really is a four variable parametrization, coming from quaternion multiplication. The attribution in wikipedia is to the number theorist V. A. Lebesgue. He did publish on this in the 1850's. However, this was surely known to Euler.
Discussed in detail in Pall 1940
I used those techniques, quaternions, for related problems Find three numbers such that the sum of all three is a square and the sum of any two is a square and
http://math.stackexchange.com/questions/1964607/when-will-a-parametric-solution-generate-all-possible-solutions/1965805#1965805 I was quite proud of this one, it gives a complete integer parametrization for $a^2 + b^2 + c^2 = 3 d^2$ using material from Jones and Pall 1939
Meanwhile, the first acceptable proof that ALL quadruples arise this way (primitive ones, that is) was due to L. E. Dickson in 1920. There is a very nice article by Spira, let me find a link. I also have a pdf of Spira (1962).
answered Jul 26 at 19:04
Will Jagy
96.9k594195
It really is a four variable parametrization, coming from quaternion multiplication. The attribution in wikipedia is to the number theorist V. A. Lebesgue. He did publish on this in the 1850's. However, this was surely known to Euler.
Discussed in detail in Pall 1940
I used those techniques, quaternions, for related problems Find three numbers such that the sum of all three is a square and the sum of any two is a square and
http://math.stackexchange.com/questions/1964607/when-will-a-parametric-solution-generate-all-possible-solutions/1965805#1965805 I was quite proud of this one, it gives a complete integer parametrization for $a^2 + b^2 + c^2 = 3 d^2$ using material from Jones and Pall 1939
Meanwhile, the first acceptable proof that ALL quadruples arise this way (primitive ones, that is) was due to L. E. Dickson in 1920. There is a very nice article by Spira, let me find a link. I also have a pdf of Spira (1962).
answered Jul 26 at 19:04
Will Jagy
96.9k594195
edited Jul 27 at 18:07
answered Jul 26 at 19:04
Will Jagy
96.9k594195
answered Jul 26 at 19:04
Will Jagy
96.9k594195
answered Jul 26 at 19:04
Will Jagy
96.9k594195
96.9k594195
add a comment |Â
add a comment |Â
up vote
2
down vote
There are many of them.
$$113^2 = 112^2+12^2 +9^2 $$
is an example.
answered Jul 26 at 18:20
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
2
down vote
There are many of them.
$$113^2 = 112^2+12^2 +9^2 $$
is an example.
answered Jul 26 at 18:20
Mohammad Riazi-Kermani
27.3k41851
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There are many of them.
$$113^2 = 112^2+12^2 +9^2 $$
is an example.
answered Jul 26 at 18:20
Mohammad Riazi-Kermani
27.3k41851
There are many of them.
$$113^2 = 112^2+12^2 +9^2 $$
is an example.
answered Jul 26 at 18:20
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:20
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:20
Mohammad Riazi-Kermani
27.3k41851
answered Jul 26 at 18:20
Mohammad Riazi-Kermani
27.3k41851
27.3k41851
add a comment |Â
add a comment |Â
up vote
2
down vote
Dickson's History of the Theory of Numbers,
volume 2,
starting on page 261
has a number of solutions.
Here is one due to Euler:
$(p^2 + 1)^2(q^2 + 1)^2
= (q^2 - 1)^2(p^2 + 1)^2 + 4q^2(p^2 - 1)^2 + (4pq)^2
$.
answered Jul 26 at 18:37
marty cohen
69.1k446122
add a comment |Â
up vote
2
down vote
Dickson's History of the Theory of Numbers,
volume 2,
starting on page 261
has a number of solutions.
Here is one due to Euler:
$(p^2 + 1)^2(q^2 + 1)^2
= (q^2 - 1)^2(p^2 + 1)^2 + 4q^2(p^2 - 1)^2 + (4pq)^2
$.
answered Jul 26 at 18:37
marty cohen
69.1k446122
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Dickson's History of the Theory of Numbers,
volume 2,
starting on page 261
has a number of solutions.
Here is one due to Euler:
$(p^2 + 1)^2(q^2 + 1)^2
= (q^2 - 1)^2(p^2 + 1)^2 + 4q^2(p^2 - 1)^2 + (4pq)^2
$.
answered Jul 26 at 18:37
marty cohen
69.1k446122
Dickson's History of the Theory of Numbers,
volume 2,
starting on page 261
has a number of solutions.
Here is one due to Euler:
$(p^2 + 1)^2(q^2 + 1)^2
= (q^2 - 1)^2(p^2 + 1)^2 + 4q^2(p^2 - 1)^2 + (4pq)^2
$.
answered Jul 26 at 18:37
marty cohen
69.1k446122
answered Jul 26 at 18:37
marty cohen
69.1k446122
answered Jul 26 at 18:37
marty cohen
69.1k446122
answered Jul 26 at 18:37
marty cohen
69.1k446122
69.1k446122
add a comment |Â
add a comment |Â
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1
Yes, google pythagorean quartuples
– Peter
Jul 26 at 18:12
6
... or Pythagorean quadruple. For other interesting questions, see Euler brick.
– Joffan
Jul 26 at 18:18
6
$$(2ps)^2+(2ks)^2+(p^2+k^2-s^2)^2=(p^2+k^2+s^2)^2$$
– individ
Jul 26 at 18:20
1
In Python,
[[(a,b,c,d) for (a,b,c,d) in itertools.product(range(1,21),repeat = 4) if a**2 + b**2 + c**2 - d**2 == 0]will instantly give you the 99 solutions with all terms <= 20.– John Coleman
Jul 26 at 21:07
Please read how-to-ask. In particular, what is the bigger question?
– user21820
Jul 27 at 3:39