Solutions for $a^2+b^2+c^2=d^2$
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I have to find an infinite set for $a,b,c,d$ such that $a^2+b^2+c^2=d^2$ and $a,b,c,dinmathbb N$
I have found one possible set which is $x,2x,2x,3x$ and $xinmathbb N$
But I want another infinite sets and how to reach them because I have found my solution by hit and trial .
elementary-number-theory
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up vote
3
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I have to find an infinite set for $a,b,c,d$ such that $a^2+b^2+c^2=d^2$ and $a,b,c,dinmathbb N$
I have found one possible set which is $x,2x,2x,3x$ and $xinmathbb N$
But I want another infinite sets and how to reach them because I have found my solution by hit and trial .
elementary-number-theory
@NickLiu $0notinmathbb{N}$
– Ian Miller
Jan 8 '17 at 15:52
Yes @Nick Liu c should be natural
– Atul Mishra
Jan 8 '17 at 15:52
Given a positive integer $N$ there is an integral solution of $X^2 + Y^2 + Z^2 = N$ as long as $N$ is not of the form $4^alpha(8beta - 1)$. Since no perfect square is of that form, then you can choose $din mathbb{N}$ and $a,b,c$ the solution of $X^2 + Y^2 + Z^2 = d^2$. Some of the values might be $0$ tho.
– Darth Geek
Jan 8 '17 at 15:57
2
$$a=2pk$$ $$b=2sk$$ $$c=p^2+s^2-k^2$$ $$d=p^2+s^2+k^2$$
– individ
Jan 8 '17 at 15:57
Possible duplicate of this.
– user 170039
Jan 8 '17 at 17:14
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have to find an infinite set for $a,b,c,d$ such that $a^2+b^2+c^2=d^2$ and $a,b,c,dinmathbb N$
I have found one possible set which is $x,2x,2x,3x$ and $xinmathbb N$
But I want another infinite sets and how to reach them because I have found my solution by hit and trial .
elementary-number-theory
I have to find an infinite set for $a,b,c,d$ such that $a^2+b^2+c^2=d^2$ and $a,b,c,dinmathbb N$
I have found one possible set which is $x,2x,2x,3x$ and $xinmathbb N$
But I want another infinite sets and how to reach them because I have found my solution by hit and trial .
elementary-number-theory
elementary-number-theory
edited Jan 8 '17 at 16:07
Vidyanshu Mishra
8,21932877
8,21932877
asked Jan 8 '17 at 15:46
Atul Mishra
1,884729
1,884729
@NickLiu $0notinmathbb{N}$
– Ian Miller
Jan 8 '17 at 15:52
Yes @Nick Liu c should be natural
– Atul Mishra
Jan 8 '17 at 15:52
Given a positive integer $N$ there is an integral solution of $X^2 + Y^2 + Z^2 = N$ as long as $N$ is not of the form $4^alpha(8beta - 1)$. Since no perfect square is of that form, then you can choose $din mathbb{N}$ and $a,b,c$ the solution of $X^2 + Y^2 + Z^2 = d^2$. Some of the values might be $0$ tho.
– Darth Geek
Jan 8 '17 at 15:57
2
$$a=2pk$$ $$b=2sk$$ $$c=p^2+s^2-k^2$$ $$d=p^2+s^2+k^2$$
– individ
Jan 8 '17 at 15:57
Possible duplicate of this.
– user 170039
Jan 8 '17 at 17:14
add a comment |
@NickLiu $0notinmathbb{N}$
– Ian Miller
Jan 8 '17 at 15:52
Yes @Nick Liu c should be natural
– Atul Mishra
Jan 8 '17 at 15:52
Given a positive integer $N$ there is an integral solution of $X^2 + Y^2 + Z^2 = N$ as long as $N$ is not of the form $4^alpha(8beta - 1)$. Since no perfect square is of that form, then you can choose $din mathbb{N}$ and $a,b,c$ the solution of $X^2 + Y^2 + Z^2 = d^2$. Some of the values might be $0$ tho.
– Darth Geek
Jan 8 '17 at 15:57
2
$$a=2pk$$ $$b=2sk$$ $$c=p^2+s^2-k^2$$ $$d=p^2+s^2+k^2$$
– individ
Jan 8 '17 at 15:57
Possible duplicate of this.
– user 170039
Jan 8 '17 at 17:14
@NickLiu $0notinmathbb{N}$
– Ian Miller
Jan 8 '17 at 15:52
@NickLiu $0notinmathbb{N}$
– Ian Miller
Jan 8 '17 at 15:52
Yes @Nick Liu c should be natural
– Atul Mishra
Jan 8 '17 at 15:52
Yes @Nick Liu c should be natural
– Atul Mishra
Jan 8 '17 at 15:52
Given a positive integer $N$ there is an integral solution of $X^2 + Y^2 + Z^2 = N$ as long as $N$ is not of the form $4^alpha(8beta - 1)$. Since no perfect square is of that form, then you can choose $din mathbb{N}$ and $a,b,c$ the solution of $X^2 + Y^2 + Z^2 = d^2$. Some of the values might be $0$ tho.
– Darth Geek
Jan 8 '17 at 15:57
Given a positive integer $N$ there is an integral solution of $X^2 + Y^2 + Z^2 = N$ as long as $N$ is not of the form $4^alpha(8beta - 1)$. Since no perfect square is of that form, then you can choose $din mathbb{N}$ and $a,b,c$ the solution of $X^2 + Y^2 + Z^2 = d^2$. Some of the values might be $0$ tho.
– Darth Geek
Jan 8 '17 at 15:57
2
2
$$a=2pk$$ $$b=2sk$$ $$c=p^2+s^2-k^2$$ $$d=p^2+s^2+k^2$$
– individ
Jan 8 '17 at 15:57
$$a=2pk$$ $$b=2sk$$ $$c=p^2+s^2-k^2$$ $$d=p^2+s^2+k^2$$
– individ
Jan 8 '17 at 15:57
Possible duplicate of this.
– user 170039
Jan 8 '17 at 17:14
Possible duplicate of this.
– user 170039
Jan 8 '17 at 17:14
add a comment |
5 Answers
5
active
oldest
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up vote
8
down vote
accepted
There is a parametrization of all primitive soutions, $gcd(a,b,c,d)=1.$ Take any four integers $m,n,p,q$ with $gcd(m,n,p,q)=1$ and $m+n+p+q$ odd. Those are the only rules. Then
$$ a = m^2 + n^2 - p^2 - q^2, $$
$$ b = 2 (mq+np), $$
$$ c = 2(nq -mp), $$
$$ d = m^2 + n^2 + p^2 + q^2 $$
satisfy
$$ a^2 + b^2 + c^2 = d^2. $$
This is simply quaternion multiplication. It seems likely that the formula was known to Euler, who wrote out the multiplication formula for sums of four squares, which amounts to quaternion multiplication. However, people mostly mention this item as due to the number theorist V. A. Lebesgue in the mid 1800's. Apparently the first correct proof that this gives all primitive solutions was Dickson in 1920, so I am not sure what Lebesgue did.
gp-pari
? a = m^2 + n^2 - p^2 - q^2
%6 = m^2 + (n^2 + (-p^2 - q^2))
? b = 2 *( m * q + n * p)
%7 = 2*q*m + 2*p*n
? c = 2 * ( n * q - m * p)
%8 = -2*p*m + 2*q*n
? d = m^2 + n^2 + p^2 + q^2
%9 = m^2 + (n^2 + (p^2 + q^2))
? a^2 + b^2 + c^2 - d^2
%10 = 0
?
@IanMiller "But I want another infinite sets and how to reach them because I have found my solution by hit and trial ." This question gives a recipe for finding infinitely many disjoint infinite families of solutions which form a partition of the full set of solutions.
– John Gowers
Jan 8 '17 at 16:45
@IanMiller for every $(m, n, p, q)$ you get a solution $(a, b, c, d)$ using the parametrization in the answer. Note that when $p = n = 0$, we have the usual parametrization for primitive Pythagorean triples
– Alex Macedo
Jan 8 '17 at 16:54
Sorry. I typoed when I was trying it. Very sorry. Removing my stupid comments.
– Ian Miller
Jan 8 '17 at 17:09
add a comment |
up vote
8
down vote
I'll give one obvious example that I found, and that is
If
$$p^2+q^2=2rstag1$$
Then$$(p+q)^2+(p-q)^2+(r-s)^2=(r+s)^2tag2$$
For example, if $(p,q,r,s)=(4,2,5,2)$, then$$6^2+2^2+3^2=7^2tag3$$
Interestingly enough, Euler did the same work for these sums of squares and he gave an explicit formula.
$$(m^2+n^2-p^2-q^2)^2+4(mq+np)^2+4(nq-mp)^2=(m^2+n^2+p^2+q^2)^2tag4$$
For arbitrary $m,n,p,qinmathbb{Z}$.
add a comment |
up vote
5
down vote
As you noticed, given a solution you can multiply it by any natural greater than $1$ to get another solution, so we can concentrate on solutions that do not have a common divisor. You can just keep finding solutions by hand. A more systematic approach is to write $a^2+b^2=d^2-c^2=(d+c)(d-c)$ The two terms on the right are either both odd or both even. You can just choose $a,b$ with one odd and one even and $a gt b$, then factor $a^2+b^2$. So $a=4,b=1$ gives $d+c=17, d-c=1$ and you can solve the equations to get $d=9,c=8$ and a new solution is $4,1,8,9$. You can always use the factor with $1$ giving $a,b,frac 12(a^2+b^2-1),frac12(a^2+b^2+1)$ as a fundamental solution whenever $a,b$ are of opposite parity. If $a^2+b^2$ is not prime or the square of a prime there will be other solutions. So for $a=8, b=1$ we have $a^2+b^2=65=5 cdot 13$, and we find $8,1,4,9$ as a solution.So $a=4,b=1$ gives $d+c=17, d-c=1$ and you can solve the equations to get $d=9,c=8$. Unfortunately here it give a permutation of the solution we already have. If we choose $a=9,b=2$ we get $a^2+b^2=85=5 cdot 17$ and find the solution $9,2,6,11$ as well as $9,2,42,43$
Impressed with the last two lines
– Atul Mishra
Jan 8 '17 at 16:00
3
But d will be fractional if a & b are of opposite parity@rossmilikan
– Atul Mishra
Jan 8 '17 at 16:02
@AtulMishra: I missed the $pm1$ that makes them integral. Still working a bit.
– Ross Millikan
Jan 8 '17 at 16:13
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up vote
3
down vote
Neater solution. (not obvious where I got formula from)
Note that: $(b^2-a^2)^2+(2ab)^2+(2cd)^2-(c^2+d^2)^2=(a^2+b^2+c^2-d^2)(a^2+b^2-c^2+d^2)$.
So if you have a solution $(a,b,c,d)$ to $a^2+b^2+c^2-d^2=0$ then another solution is:
$$bigg(b^2-a^2,2ab,2cd,c^2+d^2bigg)$$
Applying this to your solution $(1,2,2,3)$ gives $(3,4,12,13)$.
Note the order of terms could be juggled around for further applications of this. The only requirement is that $b>a$.
$(3,4,12,13)$ becomes $(24,12,313)$.
$(3,12,4,13)$ becomes $(35,72,104,185)$.
$(4,12,3,13)$ becomes $(128,96,78,178)$.
Ugly Solution. (more obvious where formula comes from)
Pythagoras's theorem is a related problem only involving three terms: $a^2+b^2=c^2$. It has a fundamental solution of $x^2-y^2, 2xy, x^2+y^2$. We can built upon this.
If we tackle your problem $a^2+b^2+c^2=d^2$ then we can start with $a^2+b^2=(x^2-y^2)^2, c=2xy, d=x^2+y^2$. Note that solving $a^2+b^2=(x^2-y^2)^2$ is another case of Pythagoras: $a=k^2-l^2, b=2kl, x^2-y^2=k^2+l^2$. Note that $x^2-y^2=k^2+l^2$ is the same as your original equation to which you have a solution so we can build another from there: $x=3,k=2,l=1,y=2$ (Note we need $k>l$).
So with $k=2,l=1,y=2,x=3$ then we get $a=3,b=4,c=12,d=13$ giving $3^2+4^2+12^2=13^2$.
So in general if you have a solution $(a,b,c,d)$ (with $a<b$) you can form a new solution: $(b^2-a^2,2ab,2cd,c^2+d^2)$.
add a comment |
up vote
3
down vote
It is well known that the identity $(2n)^2+(n^2-1)^2=(n^2+1)^2$ produces a class of Pythagorean Triples. We can apply the same algorithm twice by replacing $nto n^2+1$, as $$color{Green}{(4n)^2+(2(n^2-1))^2+((n^2+1)^2-1)^2=((n^2+1)^2+1)^2},,,,forall ninBbb{N}.$$ In fact this solution has an interesting property which does not holds for others :)
Also there is another similar identity which looks rather simple, $$color{red}{n^2+(n+1)^2+(n^2+n)^2=(n^2+n+1)^2},,,,forall ninBbb{N}.$$
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
There is a parametrization of all primitive soutions, $gcd(a,b,c,d)=1.$ Take any four integers $m,n,p,q$ with $gcd(m,n,p,q)=1$ and $m+n+p+q$ odd. Those are the only rules. Then
$$ a = m^2 + n^2 - p^2 - q^2, $$
$$ b = 2 (mq+np), $$
$$ c = 2(nq -mp), $$
$$ d = m^2 + n^2 + p^2 + q^2 $$
satisfy
$$ a^2 + b^2 + c^2 = d^2. $$
This is simply quaternion multiplication. It seems likely that the formula was known to Euler, who wrote out the multiplication formula for sums of four squares, which amounts to quaternion multiplication. However, people mostly mention this item as due to the number theorist V. A. Lebesgue in the mid 1800's. Apparently the first correct proof that this gives all primitive solutions was Dickson in 1920, so I am not sure what Lebesgue did.
gp-pari
? a = m^2 + n^2 - p^2 - q^2
%6 = m^2 + (n^2 + (-p^2 - q^2))
? b = 2 *( m * q + n * p)
%7 = 2*q*m + 2*p*n
? c = 2 * ( n * q - m * p)
%8 = -2*p*m + 2*q*n
? d = m^2 + n^2 + p^2 + q^2
%9 = m^2 + (n^2 + (p^2 + q^2))
? a^2 + b^2 + c^2 - d^2
%10 = 0
?
@IanMiller "But I want another infinite sets and how to reach them because I have found my solution by hit and trial ." This question gives a recipe for finding infinitely many disjoint infinite families of solutions which form a partition of the full set of solutions.
– John Gowers
Jan 8 '17 at 16:45
@IanMiller for every $(m, n, p, q)$ you get a solution $(a, b, c, d)$ using the parametrization in the answer. Note that when $p = n = 0$, we have the usual parametrization for primitive Pythagorean triples
– Alex Macedo
Jan 8 '17 at 16:54
Sorry. I typoed when I was trying it. Very sorry. Removing my stupid comments.
– Ian Miller
Jan 8 '17 at 17:09
add a comment |
up vote
8
down vote
accepted
There is a parametrization of all primitive soutions, $gcd(a,b,c,d)=1.$ Take any four integers $m,n,p,q$ with $gcd(m,n,p,q)=1$ and $m+n+p+q$ odd. Those are the only rules. Then
$$ a = m^2 + n^2 - p^2 - q^2, $$
$$ b = 2 (mq+np), $$
$$ c = 2(nq -mp), $$
$$ d = m^2 + n^2 + p^2 + q^2 $$
satisfy
$$ a^2 + b^2 + c^2 = d^2. $$
This is simply quaternion multiplication. It seems likely that the formula was known to Euler, who wrote out the multiplication formula for sums of four squares, which amounts to quaternion multiplication. However, people mostly mention this item as due to the number theorist V. A. Lebesgue in the mid 1800's. Apparently the first correct proof that this gives all primitive solutions was Dickson in 1920, so I am not sure what Lebesgue did.
gp-pari
? a = m^2 + n^2 - p^2 - q^2
%6 = m^2 + (n^2 + (-p^2 - q^2))
? b = 2 *( m * q + n * p)
%7 = 2*q*m + 2*p*n
? c = 2 * ( n * q - m * p)
%8 = -2*p*m + 2*q*n
? d = m^2 + n^2 + p^2 + q^2
%9 = m^2 + (n^2 + (p^2 + q^2))
? a^2 + b^2 + c^2 - d^2
%10 = 0
?
@IanMiller "But I want another infinite sets and how to reach them because I have found my solution by hit and trial ." This question gives a recipe for finding infinitely many disjoint infinite families of solutions which form a partition of the full set of solutions.
– John Gowers
Jan 8 '17 at 16:45
@IanMiller for every $(m, n, p, q)$ you get a solution $(a, b, c, d)$ using the parametrization in the answer. Note that when $p = n = 0$, we have the usual parametrization for primitive Pythagorean triples
– Alex Macedo
Jan 8 '17 at 16:54
Sorry. I typoed when I was trying it. Very sorry. Removing my stupid comments.
– Ian Miller
Jan 8 '17 at 17:09
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
There is a parametrization of all primitive soutions, $gcd(a,b,c,d)=1.$ Take any four integers $m,n,p,q$ with $gcd(m,n,p,q)=1$ and $m+n+p+q$ odd. Those are the only rules. Then
$$ a = m^2 + n^2 - p^2 - q^2, $$
$$ b = 2 (mq+np), $$
$$ c = 2(nq -mp), $$
$$ d = m^2 + n^2 + p^2 + q^2 $$
satisfy
$$ a^2 + b^2 + c^2 = d^2. $$
This is simply quaternion multiplication. It seems likely that the formula was known to Euler, who wrote out the multiplication formula for sums of four squares, which amounts to quaternion multiplication. However, people mostly mention this item as due to the number theorist V. A. Lebesgue in the mid 1800's. Apparently the first correct proof that this gives all primitive solutions was Dickson in 1920, so I am not sure what Lebesgue did.
gp-pari
? a = m^2 + n^2 - p^2 - q^2
%6 = m^2 + (n^2 + (-p^2 - q^2))
? b = 2 *( m * q + n * p)
%7 = 2*q*m + 2*p*n
? c = 2 * ( n * q - m * p)
%8 = -2*p*m + 2*q*n
? d = m^2 + n^2 + p^2 + q^2
%9 = m^2 + (n^2 + (p^2 + q^2))
? a^2 + b^2 + c^2 - d^2
%10 = 0
?
There is a parametrization of all primitive soutions, $gcd(a,b,c,d)=1.$ Take any four integers $m,n,p,q$ with $gcd(m,n,p,q)=1$ and $m+n+p+q$ odd. Those are the only rules. Then
$$ a = m^2 + n^2 - p^2 - q^2, $$
$$ b = 2 (mq+np), $$
$$ c = 2(nq -mp), $$
$$ d = m^2 + n^2 + p^2 + q^2 $$
satisfy
$$ a^2 + b^2 + c^2 = d^2. $$
This is simply quaternion multiplication. It seems likely that the formula was known to Euler, who wrote out the multiplication formula for sums of four squares, which amounts to quaternion multiplication. However, people mostly mention this item as due to the number theorist V. A. Lebesgue in the mid 1800's. Apparently the first correct proof that this gives all primitive solutions was Dickson in 1920, so I am not sure what Lebesgue did.
gp-pari
? a = m^2 + n^2 - p^2 - q^2
%6 = m^2 + (n^2 + (-p^2 - q^2))
? b = 2 *( m * q + n * p)
%7 = 2*q*m + 2*p*n
? c = 2 * ( n * q - m * p)
%8 = -2*p*m + 2*q*n
? d = m^2 + n^2 + p^2 + q^2
%9 = m^2 + (n^2 + (p^2 + q^2))
? a^2 + b^2 + c^2 - d^2
%10 = 0
?
edited Jan 8 '17 at 17:17
answered Jan 8 '17 at 16:17
Will Jagy
101k597198
101k597198
@IanMiller "But I want another infinite sets and how to reach them because I have found my solution by hit and trial ." This question gives a recipe for finding infinitely many disjoint infinite families of solutions which form a partition of the full set of solutions.
– John Gowers
Jan 8 '17 at 16:45
@IanMiller for every $(m, n, p, q)$ you get a solution $(a, b, c, d)$ using the parametrization in the answer. Note that when $p = n = 0$, we have the usual parametrization for primitive Pythagorean triples
– Alex Macedo
Jan 8 '17 at 16:54
Sorry. I typoed when I was trying it. Very sorry. Removing my stupid comments.
– Ian Miller
Jan 8 '17 at 17:09
add a comment |
@IanMiller "But I want another infinite sets and how to reach them because I have found my solution by hit and trial ." This question gives a recipe for finding infinitely many disjoint infinite families of solutions which form a partition of the full set of solutions.
– John Gowers
Jan 8 '17 at 16:45
@IanMiller for every $(m, n, p, q)$ you get a solution $(a, b, c, d)$ using the parametrization in the answer. Note that when $p = n = 0$, we have the usual parametrization for primitive Pythagorean triples
– Alex Macedo
Jan 8 '17 at 16:54
Sorry. I typoed when I was trying it. Very sorry. Removing my stupid comments.
– Ian Miller
Jan 8 '17 at 17:09
@IanMiller "But I want another infinite sets and how to reach them because I have found my solution by hit and trial ." This question gives a recipe for finding infinitely many disjoint infinite families of solutions which form a partition of the full set of solutions.
– John Gowers
Jan 8 '17 at 16:45
@IanMiller "But I want another infinite sets and how to reach them because I have found my solution by hit and trial ." This question gives a recipe for finding infinitely many disjoint infinite families of solutions which form a partition of the full set of solutions.
– John Gowers
Jan 8 '17 at 16:45
@IanMiller for every $(m, n, p, q)$ you get a solution $(a, b, c, d)$ using the parametrization in the answer. Note that when $p = n = 0$, we have the usual parametrization for primitive Pythagorean triples
– Alex Macedo
Jan 8 '17 at 16:54
@IanMiller for every $(m, n, p, q)$ you get a solution $(a, b, c, d)$ using the parametrization in the answer. Note that when $p = n = 0$, we have the usual parametrization for primitive Pythagorean triples
– Alex Macedo
Jan 8 '17 at 16:54
Sorry. I typoed when I was trying it. Very sorry. Removing my stupid comments.
– Ian Miller
Jan 8 '17 at 17:09
Sorry. I typoed when I was trying it. Very sorry. Removing my stupid comments.
– Ian Miller
Jan 8 '17 at 17:09
add a comment |
up vote
8
down vote
I'll give one obvious example that I found, and that is
If
$$p^2+q^2=2rstag1$$
Then$$(p+q)^2+(p-q)^2+(r-s)^2=(r+s)^2tag2$$
For example, if $(p,q,r,s)=(4,2,5,2)$, then$$6^2+2^2+3^2=7^2tag3$$
Interestingly enough, Euler did the same work for these sums of squares and he gave an explicit formula.
$$(m^2+n^2-p^2-q^2)^2+4(mq+np)^2+4(nq-mp)^2=(m^2+n^2+p^2+q^2)^2tag4$$
For arbitrary $m,n,p,qinmathbb{Z}$.
add a comment |
up vote
8
down vote
I'll give one obvious example that I found, and that is
If
$$p^2+q^2=2rstag1$$
Then$$(p+q)^2+(p-q)^2+(r-s)^2=(r+s)^2tag2$$
For example, if $(p,q,r,s)=(4,2,5,2)$, then$$6^2+2^2+3^2=7^2tag3$$
Interestingly enough, Euler did the same work for these sums of squares and he gave an explicit formula.
$$(m^2+n^2-p^2-q^2)^2+4(mq+np)^2+4(nq-mp)^2=(m^2+n^2+p^2+q^2)^2tag4$$
For arbitrary $m,n,p,qinmathbb{Z}$.
add a comment |
up vote
8
down vote
up vote
8
down vote
I'll give one obvious example that I found, and that is
If
$$p^2+q^2=2rstag1$$
Then$$(p+q)^2+(p-q)^2+(r-s)^2=(r+s)^2tag2$$
For example, if $(p,q,r,s)=(4,2,5,2)$, then$$6^2+2^2+3^2=7^2tag3$$
Interestingly enough, Euler did the same work for these sums of squares and he gave an explicit formula.
$$(m^2+n^2-p^2-q^2)^2+4(mq+np)^2+4(nq-mp)^2=(m^2+n^2+p^2+q^2)^2tag4$$
For arbitrary $m,n,p,qinmathbb{Z}$.
I'll give one obvious example that I found, and that is
If
$$p^2+q^2=2rstag1$$
Then$$(p+q)^2+(p-q)^2+(r-s)^2=(r+s)^2tag2$$
For example, if $(p,q,r,s)=(4,2,5,2)$, then$$6^2+2^2+3^2=7^2tag3$$
Interestingly enough, Euler did the same work for these sums of squares and he gave an explicit formula.
$$(m^2+n^2-p^2-q^2)^2+4(mq+np)^2+4(nq-mp)^2=(m^2+n^2+p^2+q^2)^2tag4$$
For arbitrary $m,n,p,qinmathbb{Z}$.
edited Jan 8 '17 at 17:09
answered Jan 8 '17 at 16:27
Frank
3,7551631
3,7551631
add a comment |
add a comment |
up vote
5
down vote
As you noticed, given a solution you can multiply it by any natural greater than $1$ to get another solution, so we can concentrate on solutions that do not have a common divisor. You can just keep finding solutions by hand. A more systematic approach is to write $a^2+b^2=d^2-c^2=(d+c)(d-c)$ The two terms on the right are either both odd or both even. You can just choose $a,b$ with one odd and one even and $a gt b$, then factor $a^2+b^2$. So $a=4,b=1$ gives $d+c=17, d-c=1$ and you can solve the equations to get $d=9,c=8$ and a new solution is $4,1,8,9$. You can always use the factor with $1$ giving $a,b,frac 12(a^2+b^2-1),frac12(a^2+b^2+1)$ as a fundamental solution whenever $a,b$ are of opposite parity. If $a^2+b^2$ is not prime or the square of a prime there will be other solutions. So for $a=8, b=1$ we have $a^2+b^2=65=5 cdot 13$, and we find $8,1,4,9$ as a solution.So $a=4,b=1$ gives $d+c=17, d-c=1$ and you can solve the equations to get $d=9,c=8$. Unfortunately here it give a permutation of the solution we already have. If we choose $a=9,b=2$ we get $a^2+b^2=85=5 cdot 17$ and find the solution $9,2,6,11$ as well as $9,2,42,43$
Impressed with the last two lines
– Atul Mishra
Jan 8 '17 at 16:00
3
But d will be fractional if a & b are of opposite parity@rossmilikan
– Atul Mishra
Jan 8 '17 at 16:02
@AtulMishra: I missed the $pm1$ that makes them integral. Still working a bit.
– Ross Millikan
Jan 8 '17 at 16:13
add a comment |
up vote
5
down vote
As you noticed, given a solution you can multiply it by any natural greater than $1$ to get another solution, so we can concentrate on solutions that do not have a common divisor. You can just keep finding solutions by hand. A more systematic approach is to write $a^2+b^2=d^2-c^2=(d+c)(d-c)$ The two terms on the right are either both odd or both even. You can just choose $a,b$ with one odd and one even and $a gt b$, then factor $a^2+b^2$. So $a=4,b=1$ gives $d+c=17, d-c=1$ and you can solve the equations to get $d=9,c=8$ and a new solution is $4,1,8,9$. You can always use the factor with $1$ giving $a,b,frac 12(a^2+b^2-1),frac12(a^2+b^2+1)$ as a fundamental solution whenever $a,b$ are of opposite parity. If $a^2+b^2$ is not prime or the square of a prime there will be other solutions. So for $a=8, b=1$ we have $a^2+b^2=65=5 cdot 13$, and we find $8,1,4,9$ as a solution.So $a=4,b=1$ gives $d+c=17, d-c=1$ and you can solve the equations to get $d=9,c=8$. Unfortunately here it give a permutation of the solution we already have. If we choose $a=9,b=2$ we get $a^2+b^2=85=5 cdot 17$ and find the solution $9,2,6,11$ as well as $9,2,42,43$
Impressed with the last two lines
– Atul Mishra
Jan 8 '17 at 16:00
3
But d will be fractional if a & b are of opposite parity@rossmilikan
– Atul Mishra
Jan 8 '17 at 16:02
@AtulMishra: I missed the $pm1$ that makes them integral. Still working a bit.
– Ross Millikan
Jan 8 '17 at 16:13
add a comment |
up vote
5
down vote
up vote
5
down vote
As you noticed, given a solution you can multiply it by any natural greater than $1$ to get another solution, so we can concentrate on solutions that do not have a common divisor. You can just keep finding solutions by hand. A more systematic approach is to write $a^2+b^2=d^2-c^2=(d+c)(d-c)$ The two terms on the right are either both odd or both even. You can just choose $a,b$ with one odd and one even and $a gt b$, then factor $a^2+b^2$. So $a=4,b=1$ gives $d+c=17, d-c=1$ and you can solve the equations to get $d=9,c=8$ and a new solution is $4,1,8,9$. You can always use the factor with $1$ giving $a,b,frac 12(a^2+b^2-1),frac12(a^2+b^2+1)$ as a fundamental solution whenever $a,b$ are of opposite parity. If $a^2+b^2$ is not prime or the square of a prime there will be other solutions. So for $a=8, b=1$ we have $a^2+b^2=65=5 cdot 13$, and we find $8,1,4,9$ as a solution.So $a=4,b=1$ gives $d+c=17, d-c=1$ and you can solve the equations to get $d=9,c=8$. Unfortunately here it give a permutation of the solution we already have. If we choose $a=9,b=2$ we get $a^2+b^2=85=5 cdot 17$ and find the solution $9,2,6,11$ as well as $9,2,42,43$
As you noticed, given a solution you can multiply it by any natural greater than $1$ to get another solution, so we can concentrate on solutions that do not have a common divisor. You can just keep finding solutions by hand. A more systematic approach is to write $a^2+b^2=d^2-c^2=(d+c)(d-c)$ The two terms on the right are either both odd or both even. You can just choose $a,b$ with one odd and one even and $a gt b$, then factor $a^2+b^2$. So $a=4,b=1$ gives $d+c=17, d-c=1$ and you can solve the equations to get $d=9,c=8$ and a new solution is $4,1,8,9$. You can always use the factor with $1$ giving $a,b,frac 12(a^2+b^2-1),frac12(a^2+b^2+1)$ as a fundamental solution whenever $a,b$ are of opposite parity. If $a^2+b^2$ is not prime or the square of a prime there will be other solutions. So for $a=8, b=1$ we have $a^2+b^2=65=5 cdot 13$, and we find $8,1,4,9$ as a solution.So $a=4,b=1$ gives $d+c=17, d-c=1$ and you can solve the equations to get $d=9,c=8$. Unfortunately here it give a permutation of the solution we already have. If we choose $a=9,b=2$ we get $a^2+b^2=85=5 cdot 17$ and find the solution $9,2,6,11$ as well as $9,2,42,43$
edited Jan 8 '17 at 16:13
answered Jan 8 '17 at 15:56
Ross Millikan
288k23195365
288k23195365
Impressed with the last two lines
– Atul Mishra
Jan 8 '17 at 16:00
3
But d will be fractional if a & b are of opposite parity@rossmilikan
– Atul Mishra
Jan 8 '17 at 16:02
@AtulMishra: I missed the $pm1$ that makes them integral. Still working a bit.
– Ross Millikan
Jan 8 '17 at 16:13
add a comment |
Impressed with the last two lines
– Atul Mishra
Jan 8 '17 at 16:00
3
But d will be fractional if a & b are of opposite parity@rossmilikan
– Atul Mishra
Jan 8 '17 at 16:02
@AtulMishra: I missed the $pm1$ that makes them integral. Still working a bit.
– Ross Millikan
Jan 8 '17 at 16:13
Impressed with the last two lines
– Atul Mishra
Jan 8 '17 at 16:00
Impressed with the last two lines
– Atul Mishra
Jan 8 '17 at 16:00
3
3
But d will be fractional if a & b are of opposite parity@rossmilikan
– Atul Mishra
Jan 8 '17 at 16:02
But d will be fractional if a & b are of opposite parity@rossmilikan
– Atul Mishra
Jan 8 '17 at 16:02
@AtulMishra: I missed the $pm1$ that makes them integral. Still working a bit.
– Ross Millikan
Jan 8 '17 at 16:13
@AtulMishra: I missed the $pm1$ that makes them integral. Still working a bit.
– Ross Millikan
Jan 8 '17 at 16:13
add a comment |
up vote
3
down vote
Neater solution. (not obvious where I got formula from)
Note that: $(b^2-a^2)^2+(2ab)^2+(2cd)^2-(c^2+d^2)^2=(a^2+b^2+c^2-d^2)(a^2+b^2-c^2+d^2)$.
So if you have a solution $(a,b,c,d)$ to $a^2+b^2+c^2-d^2=0$ then another solution is:
$$bigg(b^2-a^2,2ab,2cd,c^2+d^2bigg)$$
Applying this to your solution $(1,2,2,3)$ gives $(3,4,12,13)$.
Note the order of terms could be juggled around for further applications of this. The only requirement is that $b>a$.
$(3,4,12,13)$ becomes $(24,12,313)$.
$(3,12,4,13)$ becomes $(35,72,104,185)$.
$(4,12,3,13)$ becomes $(128,96,78,178)$.
Ugly Solution. (more obvious where formula comes from)
Pythagoras's theorem is a related problem only involving three terms: $a^2+b^2=c^2$. It has a fundamental solution of $x^2-y^2, 2xy, x^2+y^2$. We can built upon this.
If we tackle your problem $a^2+b^2+c^2=d^2$ then we can start with $a^2+b^2=(x^2-y^2)^2, c=2xy, d=x^2+y^2$. Note that solving $a^2+b^2=(x^2-y^2)^2$ is another case of Pythagoras: $a=k^2-l^2, b=2kl, x^2-y^2=k^2+l^2$. Note that $x^2-y^2=k^2+l^2$ is the same as your original equation to which you have a solution so we can build another from there: $x=3,k=2,l=1,y=2$ (Note we need $k>l$).
So with $k=2,l=1,y=2,x=3$ then we get $a=3,b=4,c=12,d=13$ giving $3^2+4^2+12^2=13^2$.
So in general if you have a solution $(a,b,c,d)$ (with $a<b$) you can form a new solution: $(b^2-a^2,2ab,2cd,c^2+d^2)$.
add a comment |
up vote
3
down vote
Neater solution. (not obvious where I got formula from)
Note that: $(b^2-a^2)^2+(2ab)^2+(2cd)^2-(c^2+d^2)^2=(a^2+b^2+c^2-d^2)(a^2+b^2-c^2+d^2)$.
So if you have a solution $(a,b,c,d)$ to $a^2+b^2+c^2-d^2=0$ then another solution is:
$$bigg(b^2-a^2,2ab,2cd,c^2+d^2bigg)$$
Applying this to your solution $(1,2,2,3)$ gives $(3,4,12,13)$.
Note the order of terms could be juggled around for further applications of this. The only requirement is that $b>a$.
$(3,4,12,13)$ becomes $(24,12,313)$.
$(3,12,4,13)$ becomes $(35,72,104,185)$.
$(4,12,3,13)$ becomes $(128,96,78,178)$.
Ugly Solution. (more obvious where formula comes from)
Pythagoras's theorem is a related problem only involving three terms: $a^2+b^2=c^2$. It has a fundamental solution of $x^2-y^2, 2xy, x^2+y^2$. We can built upon this.
If we tackle your problem $a^2+b^2+c^2=d^2$ then we can start with $a^2+b^2=(x^2-y^2)^2, c=2xy, d=x^2+y^2$. Note that solving $a^2+b^2=(x^2-y^2)^2$ is another case of Pythagoras: $a=k^2-l^2, b=2kl, x^2-y^2=k^2+l^2$. Note that $x^2-y^2=k^2+l^2$ is the same as your original equation to which you have a solution so we can build another from there: $x=3,k=2,l=1,y=2$ (Note we need $k>l$).
So with $k=2,l=1,y=2,x=3$ then we get $a=3,b=4,c=12,d=13$ giving $3^2+4^2+12^2=13^2$.
So in general if you have a solution $(a,b,c,d)$ (with $a<b$) you can form a new solution: $(b^2-a^2,2ab,2cd,c^2+d^2)$.
add a comment |
up vote
3
down vote
up vote
3
down vote
Neater solution. (not obvious where I got formula from)
Note that: $(b^2-a^2)^2+(2ab)^2+(2cd)^2-(c^2+d^2)^2=(a^2+b^2+c^2-d^2)(a^2+b^2-c^2+d^2)$.
So if you have a solution $(a,b,c,d)$ to $a^2+b^2+c^2-d^2=0$ then another solution is:
$$bigg(b^2-a^2,2ab,2cd,c^2+d^2bigg)$$
Applying this to your solution $(1,2,2,3)$ gives $(3,4,12,13)$.
Note the order of terms could be juggled around for further applications of this. The only requirement is that $b>a$.
$(3,4,12,13)$ becomes $(24,12,313)$.
$(3,12,4,13)$ becomes $(35,72,104,185)$.
$(4,12,3,13)$ becomes $(128,96,78,178)$.
Ugly Solution. (more obvious where formula comes from)
Pythagoras's theorem is a related problem only involving three terms: $a^2+b^2=c^2$. It has a fundamental solution of $x^2-y^2, 2xy, x^2+y^2$. We can built upon this.
If we tackle your problem $a^2+b^2+c^2=d^2$ then we can start with $a^2+b^2=(x^2-y^2)^2, c=2xy, d=x^2+y^2$. Note that solving $a^2+b^2=(x^2-y^2)^2$ is another case of Pythagoras: $a=k^2-l^2, b=2kl, x^2-y^2=k^2+l^2$. Note that $x^2-y^2=k^2+l^2$ is the same as your original equation to which you have a solution so we can build another from there: $x=3,k=2,l=1,y=2$ (Note we need $k>l$).
So with $k=2,l=1,y=2,x=3$ then we get $a=3,b=4,c=12,d=13$ giving $3^2+4^2+12^2=13^2$.
So in general if you have a solution $(a,b,c,d)$ (with $a<b$) you can form a new solution: $(b^2-a^2,2ab,2cd,c^2+d^2)$.
Neater solution. (not obvious where I got formula from)
Note that: $(b^2-a^2)^2+(2ab)^2+(2cd)^2-(c^2+d^2)^2=(a^2+b^2+c^2-d^2)(a^2+b^2-c^2+d^2)$.
So if you have a solution $(a,b,c,d)$ to $a^2+b^2+c^2-d^2=0$ then another solution is:
$$bigg(b^2-a^2,2ab,2cd,c^2+d^2bigg)$$
Applying this to your solution $(1,2,2,3)$ gives $(3,4,12,13)$.
Note the order of terms could be juggled around for further applications of this. The only requirement is that $b>a$.
$(3,4,12,13)$ becomes $(24,12,313)$.
$(3,12,4,13)$ becomes $(35,72,104,185)$.
$(4,12,3,13)$ becomes $(128,96,78,178)$.
Ugly Solution. (more obvious where formula comes from)
Pythagoras's theorem is a related problem only involving three terms: $a^2+b^2=c^2$. It has a fundamental solution of $x^2-y^2, 2xy, x^2+y^2$. We can built upon this.
If we tackle your problem $a^2+b^2+c^2=d^2$ then we can start with $a^2+b^2=(x^2-y^2)^2, c=2xy, d=x^2+y^2$. Note that solving $a^2+b^2=(x^2-y^2)^2$ is another case of Pythagoras: $a=k^2-l^2, b=2kl, x^2-y^2=k^2+l^2$. Note that $x^2-y^2=k^2+l^2$ is the same as your original equation to which you have a solution so we can build another from there: $x=3,k=2,l=1,y=2$ (Note we need $k>l$).
So with $k=2,l=1,y=2,x=3$ then we get $a=3,b=4,c=12,d=13$ giving $3^2+4^2+12^2=13^2$.
So in general if you have a solution $(a,b,c,d)$ (with $a<b$) you can form a new solution: $(b^2-a^2,2ab,2cd,c^2+d^2)$.
edited Jan 8 '17 at 16:41
answered Jan 8 '17 at 16:34
Ian Miller
10.3k11437
10.3k11437
add a comment |
add a comment |
up vote
3
down vote
It is well known that the identity $(2n)^2+(n^2-1)^2=(n^2+1)^2$ produces a class of Pythagorean Triples. We can apply the same algorithm twice by replacing $nto n^2+1$, as $$color{Green}{(4n)^2+(2(n^2-1))^2+((n^2+1)^2-1)^2=((n^2+1)^2+1)^2},,,,forall ninBbb{N}.$$ In fact this solution has an interesting property which does not holds for others :)
Also there is another similar identity which looks rather simple, $$color{red}{n^2+(n+1)^2+(n^2+n)^2=(n^2+n+1)^2},,,,forall ninBbb{N}.$$
add a comment |
up vote
3
down vote
It is well known that the identity $(2n)^2+(n^2-1)^2=(n^2+1)^2$ produces a class of Pythagorean Triples. We can apply the same algorithm twice by replacing $nto n^2+1$, as $$color{Green}{(4n)^2+(2(n^2-1))^2+((n^2+1)^2-1)^2=((n^2+1)^2+1)^2},,,,forall ninBbb{N}.$$ In fact this solution has an interesting property which does not holds for others :)
Also there is another similar identity which looks rather simple, $$color{red}{n^2+(n+1)^2+(n^2+n)^2=(n^2+n+1)^2},,,,forall ninBbb{N}.$$
add a comment |
up vote
3
down vote
up vote
3
down vote
It is well known that the identity $(2n)^2+(n^2-1)^2=(n^2+1)^2$ produces a class of Pythagorean Triples. We can apply the same algorithm twice by replacing $nto n^2+1$, as $$color{Green}{(4n)^2+(2(n^2-1))^2+((n^2+1)^2-1)^2=((n^2+1)^2+1)^2},,,,forall ninBbb{N}.$$ In fact this solution has an interesting property which does not holds for others :)
Also there is another similar identity which looks rather simple, $$color{red}{n^2+(n+1)^2+(n^2+n)^2=(n^2+n+1)^2},,,,forall ninBbb{N}.$$
It is well known that the identity $(2n)^2+(n^2-1)^2=(n^2+1)^2$ produces a class of Pythagorean Triples. We can apply the same algorithm twice by replacing $nto n^2+1$, as $$color{Green}{(4n)^2+(2(n^2-1))^2+((n^2+1)^2-1)^2=((n^2+1)^2+1)^2},,,,forall ninBbb{N}.$$ In fact this solution has an interesting property which does not holds for others :)
Also there is another similar identity which looks rather simple, $$color{red}{n^2+(n+1)^2+(n^2+n)^2=(n^2+n+1)^2},,,,forall ninBbb{N}.$$
edited Jan 8 '17 at 18:37
answered Jan 8 '17 at 16:32
Bumblebee
9,57512551
9,57512551
add a comment |
add a comment |
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@NickLiu $0notinmathbb{N}$
– Ian Miller
Jan 8 '17 at 15:52
Yes @Nick Liu c should be natural
– Atul Mishra
Jan 8 '17 at 15:52
Given a positive integer $N$ there is an integral solution of $X^2 + Y^2 + Z^2 = N$ as long as $N$ is not of the form $4^alpha(8beta - 1)$. Since no perfect square is of that form, then you can choose $din mathbb{N}$ and $a,b,c$ the solution of $X^2 + Y^2 + Z^2 = d^2$. Some of the values might be $0$ tho.
– Darth Geek
Jan 8 '17 at 15:57
2
$$a=2pk$$ $$b=2sk$$ $$c=p^2+s^2-k^2$$ $$d=p^2+s^2+k^2$$
– individ
Jan 8 '17 at 15:57
Possible duplicate of this.
– user 170039
Jan 8 '17 at 17:14