Skip to main content
Bounty Ended with Geoff Robinson's answer chosen by mikhail skopenkov
some recent progress added; edited body
Source Link

What is the solution of the equation $x^2+y^2+z^2+t^2=w^2$ in polynomials over C ("Pythagorean 5-tuples")?

There are simple formulas describing Pythagorean n-tuples for n=3,4,6:

  • n=3. The formula for solutions of $x^2+y^2=z^2$ [4]:

$x=d(p^2−q^2)$,

$y=2dpq$,

$z=d(p^2+q^2),$

where p,q,d are arbitrary polynomials.

  • n=4. Similarly, all Pythagorean 4-tuples of polynomials are given by the identity [4, Theorem 2.2], [3, Theorem 7.1]:

$(p^2+q^2-r^2-s^2)^2+(2pr+2qs)^2+(2ps-2qr)^2=(p^2+q^2+r^2+s^2)^2$

  • n=6. The following identity produces Pythagorean 6-tuples [3, Theorem 7.2]:

$(m^2-n^2)^2+(2mn)^2+(2(n_0m_1-m_1n_0+m_3n_2-m_2n_3))^2+$

$(2(n_0m_2-m_2n_0+m_1n_3-m_3n_1))^2+$

$(2(n_0m_3-m_3n_0+m_2n_1-m_1n_2))^2=(m^2+n^2)^2$

where $m=(m_1,m_2,m_3,m_4)$, $n=(n_1,n_2,n_3,n_4)$, and $mn$ is the usual scalar product.

These identities are somehow related to sl(2,R), sl(2,C), sl(2,H), but the case n=5 is missing in this description [3, Theorems 7.1 and 7.2].

The above formuli describe also Pythagorean n-tuples of integers. There are another descriptions of those; see [2] and [5, Chapter 5] for n<10, [6] for n<15, [1, Theorem 1 in Chapter 3] for arbitrary n, and also the answers below.

There are reasons to believe that Pythagorean 5-tuples cannot be described by a single polynomial identity. Thus identities producing a "large" set of solutions are also of interest, like $(-p^2+q^2+r^2+s^2)^2+(2pq)^2+(2pr)^2+(2ps)^2=(p^2+q^2+r^2+s^2)^2$

This identity does not give all solutions because it produces only reducible polynomials y,z,t (once p,q,r,s are nonconstant). Examples of solutions which cannot be obtained by the approaches in the answers below are also interesting.

Given a solution (x,y,z,t,w), methods to construct a new solution (x',y',z',t',w') are also of interest. For instance,

$x'=w+y+z$,

$y'=w+z+x$,

$z'=w+x+y$,

$t'=t$,

$w'=2w+x+y+z$

[see the answer of Ken Fan below for generalizations to other n] or

$x'+iy'+jz'+kt'=q(x+iy+jz+kt)q$,

where $q$ is arbitrary polynomial with quaternionic coefficients [see the answer of Geoff Robinson below].

--

[1] L.J. Mordell, Diophantine Equations, Academic Press, London, 1969

[2] D. Cass, P.J. Arpaia, MATRIX GENERATION OF PYTHAGOREAN n-TUPLES, Proc. AMS 109:1 (1990), 1-7

[3] J. Kocik, Clifford Algebras and Euclid’s Parametrization of Pythagorean Triples, Adv. Appl. Clifford Alg. 17:1 (2007), 71-93

[4] R. Dietz, J. Hoschek and B. Juttler, An algebraic approach to curves and surfaces on the sphere and on other quadrics, Computer Aided Geom. Design 10 (1993) 211-229

[5] V. Kac, Infinite-Dimensional Lie Algebras (3rd edn. ed.), CUP, 1990

[6] E. Vinberg, The groups of units of certain quadratic forms (Russian), Mat. Sbornik (N.S.) 87(129) (1972), 18–36

What is the solution of the equation $x^2+y^2+z^2+t^2=w^2$ in polynomials over C ("Pythagorean 5-tuples")?

There are simple formulas describing Pythagorean n-tuples for n=3,4,6:

  • n=3. The formula for solutions of $x^2+y^2=z^2$ [4]:

$x=d(p^2−q^2)$,

$y=2dpq$,

$z=d(p^2+q^2),$

where p,q,d are arbitrary polynomials.

  • n=4. Similarly, all Pythagorean 4-tuples of polynomials are given by the identity [4, Theorem 2.2], [3, Theorem 7.1]:

$(p^2+q^2-r^2-s^2)^2+(2pr+2qs)^2+(2ps-2qr)^2=(p^2+q^2+r^2+s^2)^2$

  • n=6. The following identity produces Pythagorean 6-tuples [3, Theorem 7.2]:

$(m^2-n^2)^2+(2mn)^2+(2(n_0m_1-m_1n_0+m_3n_2-m_2n_3))^2+$

$(2(n_0m_2-m_2n_0+m_1n_3-m_3n_1))^2+$

$(2(n_0m_3-m_3n_0+m_2n_1-m_1n_2))^2=(m^2+n^2)^2$

where $m=(m_1,m_2,m_3,m_4)$, $n=(n_1,n_2,n_3,n_4)$, and $mn$ is the usual scalar product.

These identities are somehow related to sl(2,R), sl(2,C), sl(2,H), but the case n=5 is missing in this description [3, Theorems 7.1 and 7.2].

The above formuli describe also Pythagorean n-tuples of integers. There are another descriptions of those; see [2] and [5, Chapter 5] for n<10, [1, Theorem 1 in Chapter 3] for arbitrary n, and also the answers below.

There are reasons to believe that Pythagorean 5-tuples cannot be described by a single polynomial identity. Thus identities producing a "large" set of solutions are also of interest, like $(-p^2+q^2+r^2+s^2)^2+(2pq)^2+(2pr)^2+(2ps)^2=(p^2+q^2+r^2+s^2)^2$

[1] L.J. Mordell, Diophantine Equations, Academic Press, London, 1969

[2] D. Cass, P.J. Arpaia, MATRIX GENERATION OF PYTHAGOREAN n-TUPLES, Proc. AMS 109:1 (1990), 1-7

[3] J. Kocik, Clifford Algebras and Euclid’s Parametrization of Pythagorean Triples, Adv. Appl. Clifford Alg. 17:1 (2007), 71-93

[4] R. Dietz, J. Hoschek and B. Juttler, An algebraic approach to curves and surfaces on the sphere and on other quadrics, Computer Aided Geom. Design 10 (1993) 211-229

[5] V. Kac, Infinite-Dimensional Lie Algebras (3rd edn. ed.), CUP, 1990

What is the solution of the equation $x^2+y^2+z^2+t^2=w^2$ in polynomials over C ("Pythagorean 5-tuples")?

There are simple formulas describing Pythagorean n-tuples for n=3,4,6:

  • n=3. The formula for solutions of $x^2+y^2=z^2$ [4]:

$x=d(p^2−q^2)$,

$y=2dpq$,

$z=d(p^2+q^2),$

where p,q,d are arbitrary polynomials.

  • n=4. Similarly, all Pythagorean 4-tuples of polynomials are given by the identity [4, Theorem 2.2], [3, Theorem 7.1]:

$(p^2+q^2-r^2-s^2)^2+(2pr+2qs)^2+(2ps-2qr)^2=(p^2+q^2+r^2+s^2)^2$

  • n=6. The following identity produces Pythagorean 6-tuples [3, Theorem 7.2]:

$(m^2-n^2)^2+(2mn)^2+(2(n_0m_1-m_1n_0+m_3n_2-m_2n_3))^2+$

$(2(n_0m_2-m_2n_0+m_1n_3-m_3n_1))^2+$

$(2(n_0m_3-m_3n_0+m_2n_1-m_1n_2))^2=(m^2+n^2)^2$

where $m=(m_1,m_2,m_3,m_4)$, $n=(n_1,n_2,n_3,n_4)$, and $mn$ is the usual scalar product.

These identities are somehow related to sl(2,R), sl(2,C), sl(2,H), but the case n=5 is missing in this description [3, Theorems 7.1 and 7.2].

The above formuli describe also Pythagorean n-tuples of integers. There are another descriptions of those; see [2] and [5, Chapter 5] for n<10, [6] for n<15, [1, Theorem 1 in Chapter 3] for arbitrary n, and also the answers below.

There are reasons to believe that Pythagorean 5-tuples cannot be described by a single polynomial identity. Thus identities producing a "large" set of solutions are also of interest, like $(-p^2+q^2+r^2+s^2)^2+(2pq)^2+(2pr)^2+(2ps)^2=(p^2+q^2+r^2+s^2)^2$

This identity does not give all solutions because it produces only reducible polynomials y,z,t (once p,q,r,s are nonconstant). Examples of solutions which cannot be obtained by the approaches in the answers below are also interesting.

Given a solution (x,y,z,t,w), methods to construct a new solution (x',y',z',t',w') are also of interest. For instance,

$x'=w+y+z$,

$y'=w+z+x$,

$z'=w+x+y$,

$t'=t$,

$w'=2w+x+y+z$

[see the answer of Ken Fan below for generalizations to other n] or

$x'+iy'+jz'+kt'=q(x+iy+jz+kt)q$,

where $q$ is arbitrary polynomial with quaternionic coefficients [see the answer of Geoff Robinson below].

--

[1] L.J. Mordell, Diophantine Equations, Academic Press, London, 1969

[2] D. Cass, P.J. Arpaia, MATRIX GENERATION OF PYTHAGOREAN n-TUPLES, Proc. AMS 109:1 (1990), 1-7

[3] J. Kocik, Clifford Algebras and Euclid’s Parametrization of Pythagorean Triples, Adv. Appl. Clifford Alg. 17:1 (2007), 71-93

[4] R. Dietz, J. Hoschek and B. Juttler, An algebraic approach to curves and surfaces on the sphere and on other quadrics, Computer Aided Geom. Design 10 (1993) 211-229

[5] V. Kac, Infinite-Dimensional Lie Algebras (3rd edn. ed.), CUP, 1990

[6] E. Vinberg, The groups of units of certain quadratic forms (Russian), Mat. Sbornik (N.S.) 87(129) (1972), 18–36

reference [5] added
Source Link

What is the solution of the equation $x^2+y^2+z^2+t^2=w^2$ in polynomials over C ("Pythagorean 5-tuples")?

There are simple formulas describing Pythagorean n-tuples for n=3,4,6:

  • n=3. The formula for solutions of $x^2+y^2=z^2$ [4]:

$x=d(p^2−q^2)$,

$y=2dpq$,

$z=d(p^2+q^2),$

where p,q,d are arbitrary polynomials.

  • n=4. Similarly, all Pythagorean 4-tuples of polynomials are given by the identity [4, Theorem 2.2], [3, Theorem 7.1]:

$(p^2+q^2-r^2-s^2)^2+(2pr+2qs)^2+(2ps-2qr)^2=(p^2+q^2+r^2+s^2)^2$

  • n=6. The following identity produces Pythagorean 6-tuples [3, Theorem 7.2]:

$(m^2-n^2)^2+(2mn)^2+(2(n_0m_1-m_1n_0+m_3n_2-m_2n_3))^2+$

$(2(n_0m_2-m_2n_0+m_1n_3-m_3n_1))^2+$

$(2(n_0m_3-m_3n_0+m_2n_1-m_1n_2))^2=(m^2+n^2)^2$

where $m=(m_1,m_2,m_3,m_4)$, $n=(n_1,n_2,n_3,n_4)$, and $mn$ is the usual scalar product.

These identities are somehow related to sl(2,R), sl(2,C), sl(2,H), but the case n=5 is missing in this description [3, Theorems 7.1 and 7.2].

The above formuli describe also Pythagorean n-tuples of integers. There are another descriptions of those; see [2] and [5, Chapter 5] for n<10, [1, Theorem 1 in Chapter 3] for arbitrary n, and also the answers below.

There are reasons to believe that Pythagorean 5-tuples cannot be described by a single polynomial identity. Thus identities producing a "large" set of solutions are also of interest, like $(-p^2+q^2+r^2+s^2)^2+(2pq)^2+(2pr)^2+(2ps)^2=(p^2+q^2+r^2+s^2)^2$

[1] L.J. Mordell, Diophantine Equations, Academic Press, London, 1969

[2] D. Cass, P.J. Arpaia, MATRIX GENERATION OF PYTHAGOREAN n-TUPLES, Proc. AMS 109:1 (1990), 1-7

[3] J. Kocik, Clifford Algebras and Euclid’s Parametrization of Pythagorean Triples, Adv. Appl. Clifford Alg. 17:1 (2007), 71-93

[4] R. Dietz, J. Hoschek and B. Juttler, An algebraic approach to curves and surfaces on the sphere and on other quadrics, Computer Aided Geom. Design 10 (1993) 211-229

[5] V. Kac, Infinite-Dimensional Lie Algebras (3rd edn. ed.), CUP, 1990

What is the solution of the equation $x^2+y^2+z^2+t^2=w^2$ in polynomials over C ("Pythagorean 5-tuples")?

There are simple formulas describing Pythagorean n-tuples for n=3,4,6:

  • n=3. The formula for solutions of $x^2+y^2=z^2$ [4]:

$x=d(p^2−q^2)$,

$y=2dpq$,

$z=d(p^2+q^2),$

where p,q,d are arbitrary polynomials.

  • n=4. Similarly, all Pythagorean 4-tuples of polynomials are given by the identity [4, Theorem 2.2], [3, Theorem 7.1]:

$(p^2+q^2-r^2-s^2)^2+(2pr+2qs)^2+(2ps-2qr)^2=(p^2+q^2+r^2+s^2)^2$

  • n=6. The following identity produces Pythagorean 6-tuples [3, Theorem 7.2]:

$(m^2-n^2)^2+(2mn)^2+(2(n_0m_1-m_1n_0+m_3n_2-m_2n_3))^2+$

$(2(n_0m_2-m_2n_0+m_1n_3-m_3n_1))^2+$

$(2(n_0m_3-m_3n_0+m_2n_1-m_1n_2))^2=(m^2+n^2)^2$

where $m=(m_1,m_2,m_3,m_4)$, $n=(n_1,n_2,n_3,n_4)$, and $mn$ is the usual scalar product.

These identities are somehow related to sl(2,R), sl(2,C), sl(2,H), but the case n=5 is missing in this description [3, Theorems 7.1 and 7.2].

The above formuli describe also Pythagorean n-tuples of integers. There are another descriptions of those; see [2] for n<10, [1, Theorem 1 in Chapter 3] for arbitrary n, and also the answers below.

There are reasons to believe that Pythagorean 5-tuples cannot be described by a single polynomial identity. Thus identities producing a "large" set of solutions are also of interest, like $(-p^2+q^2+r^2+s^2)^2+(2pq)^2+(2pr)^2+(2ps)^2=(p^2+q^2+r^2+s^2)^2$

[1] L.J. Mordell, Diophantine Equations, Academic Press, London, 1969

[2] D. Cass, P.J. Arpaia, MATRIX GENERATION OF PYTHAGOREAN n-TUPLES, Proc. AMS 109:1 (1990), 1-7

[3] J. Kocik, Clifford Algebras and Euclid’s Parametrization of Pythagorean Triples, Adv. Appl. Clifford Alg. 17:1 (2007), 71-93

[4] R. Dietz, J. Hoschek and B. Juttler, An algebraic approach to curves and surfaces on the sphere and on other quadrics, Computer Aided Geom. Design 10 (1993) 211-229

What is the solution of the equation $x^2+y^2+z^2+t^2=w^2$ in polynomials over C ("Pythagorean 5-tuples")?

There are simple formulas describing Pythagorean n-tuples for n=3,4,6:

  • n=3. The formula for solutions of $x^2+y^2=z^2$ [4]:

$x=d(p^2−q^2)$,

$y=2dpq$,

$z=d(p^2+q^2),$

where p,q,d are arbitrary polynomials.

  • n=4. Similarly, all Pythagorean 4-tuples of polynomials are given by the identity [4, Theorem 2.2], [3, Theorem 7.1]:

$(p^2+q^2-r^2-s^2)^2+(2pr+2qs)^2+(2ps-2qr)^2=(p^2+q^2+r^2+s^2)^2$

  • n=6. The following identity produces Pythagorean 6-tuples [3, Theorem 7.2]:

$(m^2-n^2)^2+(2mn)^2+(2(n_0m_1-m_1n_0+m_3n_2-m_2n_3))^2+$

$(2(n_0m_2-m_2n_0+m_1n_3-m_3n_1))^2+$

$(2(n_0m_3-m_3n_0+m_2n_1-m_1n_2))^2=(m^2+n^2)^2$

where $m=(m_1,m_2,m_3,m_4)$, $n=(n_1,n_2,n_3,n_4)$, and $mn$ is the usual scalar product.

These identities are somehow related to sl(2,R), sl(2,C), sl(2,H), but the case n=5 is missing in this description [3, Theorems 7.1 and 7.2].

The above formuli describe also Pythagorean n-tuples of integers. There are another descriptions of those; see [2] and [5, Chapter 5] for n<10, [1, Theorem 1 in Chapter 3] for arbitrary n, and also the answers below.

There are reasons to believe that Pythagorean 5-tuples cannot be described by a single polynomial identity. Thus identities producing a "large" set of solutions are also of interest, like $(-p^2+q^2+r^2+s^2)^2+(2pq)^2+(2pr)^2+(2ps)^2=(p^2+q^2+r^2+s^2)^2$

[1] L.J. Mordell, Diophantine Equations, Academic Press, London, 1969

[2] D. Cass, P.J. Arpaia, MATRIX GENERATION OF PYTHAGOREAN n-TUPLES, Proc. AMS 109:1 (1990), 1-7

[3] J. Kocik, Clifford Algebras and Euclid’s Parametrization of Pythagorean Triples, Adv. Appl. Clifford Alg. 17:1 (2007), 71-93

[4] R. Dietz, J. Hoschek and B. Juttler, An algebraic approach to curves and surfaces on the sphere and on other quadrics, Computer Aided Geom. Design 10 (1993) 211-229

[5] V. Kac, Infinite-Dimensional Lie Algebras (3rd edn. ed.), CUP, 1990

Bounty Started worth 100 reputation by mikhail skopenkov
description of Pythagorean n-tuples for n=3,4,6 added
Source Link

What is the solution of the equation x^2+y^2+z^2+t^2=w^2$x^2+y^2+z^2+t^2=w^2$ in integers and polynomials over C ("Pythagorean 5-tuples")?

There are simple formulas describing Pythagorean n-tuples for n=3,4,6:

  • n=3. The formula for solutions of $x^2+y^2=z^2$ [4]:

$x=d(p^2−q^2)$,

$y=2dpq$,

$z=d(p^2+q^2),$

where p,q,d are arbitrary polynomials.

  • n=4. Similarly, all Pythagorean 4-tuples of polynomials are given by the identity [4, Theorem 2.2], [3, Theorem 7.1]:

$(p^2+q^2-r^2-s^2)^2+(2pr+2qs)^2+(2ps-2qr)^2=(p^2+q^2+r^2+s^2)^2$

  • n=6. The following identity produces Pythagorean 6-tuples [3, Theorem 7.2]:

$(m^2-n^2)^2+(2mn)^2+(2(n_0m_1-m_1n_0+m_3n_2-m_2n_3))^2+$

$(2(n_0m_2-m_2n_0+m_1n_3-m_3n_1))^2+$

$(2(n_0m_3-m_3n_0+m_2n_1-m_1n_2))^2=(m^2+n^2)^2$

where $m=(m_1,m_2,m_3,m_4)$, $n=(n_1,n_2,n_3,n_4)$, and $mn$ is the usual scalar product.

These identities are somehow related to sl(2,R), sl(2,C), sl(2,H), but the case n=5 is missing in this description [3, Theorems 7.1 and 7.2], [4, Theorem 2.2]. There are different descriptions of

The above formuli describe also Pythagorean n-tuples of integers,. There are another descriptions of those; see [2] for n<10 and, [1, Theorem 1 in Chapter 3] for arbitrary n, and also the answers below. Formulas

There are reasons to believe that Pythagorean 5-tuples cannot be described by a single polynomial identity. Thus identities producing a "large" set of solutions are also of interest, like $(-p^2+q^2+r^2+s^2)^2+(2pq)^2+(2pr)^2+(2ps)^2=(p^2+q^2+r^2+s^2)^2$.

[1] L.J. Mordell, Diophantine Equations, Academic Press, London, 1969

[2] D. Cass, P.J. Arpaia, MATRIX GENERATION OF PYTHAGOREAN n-TUPLES, Proc. AMS 109:1 (1990), 1-7

[3] J. Kocik, Clifford Algebras and Euclid’s Parametrization of Pythagorean Triples, Adv. Appl. Clifford Alg. 17:1 (2007), 71-93

[4] R. Dietz, J. Hoschek and B. Juttler, An algebraic approach to curves and surfaces on the sphere and on other quadrics, Computer Aided Geom. Design 10 (1993) 211-229

What is the solution of the equation x^2+y^2+z^2+t^2=w^2 in integers and polynomials over C ("Pythagorean 5-tuples")?

There are simple formulas describing Pythagorean n-tuples for n=3,4,6, somehow related to sl(2,R), sl(2,C), sl(2,H), but the case n=5 is missing in this description [3, Theorems 7.1 and 7.2], [4, Theorem 2.2]. There are different descriptions of Pythagorean n-tuples of integers, see [2] for n<10 and [1, Theorem 1 in Chapter 3] for arbitrary n. Formulas producing a "large" set of solutions are also of interest, like $(-p^2+q^2+r^2+s^2)^2+(2pq)^2+(2pr)^2+(2ps)^2=(p^2+q^2+r^2+s^2)^2$.

[1] L.J. Mordell, Diophantine Equations, Academic Press, London, 1969

[2] D. Cass, P.J. Arpaia, MATRIX GENERATION OF PYTHAGOREAN n-TUPLES, Proc. AMS 109:1 (1990), 1-7

[3] J. Kocik, Clifford Algebras and Euclid’s Parametrization of Pythagorean Triples, Adv. Appl. Clifford Alg. 17:1 (2007), 71-93

[4] R. Dietz, J. Hoschek and B. Juttler, An algebraic approach to curves and surfaces on the sphere and on other quadrics, Computer Aided Geom. Design 10 (1993) 211-229

What is the solution of the equation $x^2+y^2+z^2+t^2=w^2$ in polynomials over C ("Pythagorean 5-tuples")?

There are simple formulas describing Pythagorean n-tuples for n=3,4,6:

  • n=3. The formula for solutions of $x^2+y^2=z^2$ [4]:

$x=d(p^2−q^2)$,

$y=2dpq$,

$z=d(p^2+q^2),$

where p,q,d are arbitrary polynomials.

  • n=4. Similarly, all Pythagorean 4-tuples of polynomials are given by the identity [4, Theorem 2.2], [3, Theorem 7.1]:

$(p^2+q^2-r^2-s^2)^2+(2pr+2qs)^2+(2ps-2qr)^2=(p^2+q^2+r^2+s^2)^2$

  • n=6. The following identity produces Pythagorean 6-tuples [3, Theorem 7.2]:

$(m^2-n^2)^2+(2mn)^2+(2(n_0m_1-m_1n_0+m_3n_2-m_2n_3))^2+$

$(2(n_0m_2-m_2n_0+m_1n_3-m_3n_1))^2+$

$(2(n_0m_3-m_3n_0+m_2n_1-m_1n_2))^2=(m^2+n^2)^2$

where $m=(m_1,m_2,m_3,m_4)$, $n=(n_1,n_2,n_3,n_4)$, and $mn$ is the usual scalar product.

These identities are somehow related to sl(2,R), sl(2,C), sl(2,H), but the case n=5 is missing in this description [3, Theorems 7.1 and 7.2].

The above formuli describe also Pythagorean n-tuples of integers. There are another descriptions of those; see [2] for n<10, [1, Theorem 1 in Chapter 3] for arbitrary n, and also the answers below.

There are reasons to believe that Pythagorean 5-tuples cannot be described by a single polynomial identity. Thus identities producing a "large" set of solutions are also of interest, like $(-p^2+q^2+r^2+s^2)^2+(2pq)^2+(2pr)^2+(2ps)^2=(p^2+q^2+r^2+s^2)^2$

[1] L.J. Mordell, Diophantine Equations, Academic Press, London, 1969

[2] D. Cass, P.J. Arpaia, MATRIX GENERATION OF PYTHAGOREAN n-TUPLES, Proc. AMS 109:1 (1990), 1-7

[3] J. Kocik, Clifford Algebras and Euclid’s Parametrization of Pythagorean Triples, Adv. Appl. Clifford Alg. 17:1 (2007), 71-93

[4] R. Dietz, J. Hoschek and B. Juttler, An algebraic approach to curves and surfaces on the sphere and on other quadrics, Computer Aided Geom. Design 10 (1993) 211-229

added 672 characters in body
Source Link
Loading
Source Link
Loading