Distinct integer solutions to the equation $(x^2+y^2+z^2)/(x+y+z)=2n$ [closed]
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I’ve been set this problem by my teacher, and had no idea where to start. $x, y, z$ and $n$ must be unique positive integers. I’ve found solutions where there is an equality for any of $x, y$ and $z$ but I haven’t been able to for unique positive integers, and would appreciate help.
elementary-number-theory diophantine-equations
asked Jan 24 at 16:24
T BurchT Burch
12
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closed as unclear what you're asking by Namaste, Dietrich Burde, Shailesh, max_zorn, Leucippus Jan 25 at 4:18
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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I’ve been set this problem by my teacher, and had no idea where to start. $x, y, z$ and $n$ must be unique positive integers. I’ve found solutions where there is an equality for any of $x, y$ and $z$ but I haven’t been able to for unique positive integers, and would appreciate help.
elementary-number-theory diophantine-equations
asked Jan 24 at 16:24
T BurchT Burch
12
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closed as unclear what you're asking by Namaste, Dietrich Burde, Shailesh, max_zorn, Leucippus Jan 25 at 4:18
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
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I think there is a problem in the formulation of your problem. One can prove that given some $n$ there are at most finitely many integer solutions $x$, $y$ and $z$. Maybe for some $n$ there is only one integer solution for $x$, $y$ and $z$?
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– quarague
Jan 24 at 16:35
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I think you must mean that $x,y,$ and $z$ must be distinct positive integers?
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– TonyK
Jan 24 at 18:37
add a comment |
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I’ve been set this problem by my teacher, and had no idea where to start. $x, y, z$ and $n$ must be unique positive integers. I’ve found solutions where there is an equality for any of $x, y$ and $z$ but I haven’t been able to for unique positive integers, and would appreciate help.
elementary-number-theory diophantine-equations
asked Jan 24 at 16:24
T BurchT Burch
12
$endgroup$
I’ve been set this problem by my teacher, and had no idea where to start. $x, y, z$ and $n$ must be unique positive integers. I’ve found solutions where there is an equality for any of $x, y$ and $z$ but I haven’t been able to for unique positive integers, and would appreciate help.
elementary-number-theory diophantine-equations
elementary-number-theory diophantine-equations
asked Jan 24 at 16:24
T BurchT Burch
12
asked Jan 24 at 16:24
T BurchT Burch
12
edited Jan 24 at 18:48
T Burch
asked Jan 24 at 16:24
T BurchT Burch
12
asked Jan 24 at 16:24
T BurchT Burch
12
asked Jan 24 at 16:24
T BurchT Burch
12
12
closed as unclear what you're asking by Namaste, Dietrich Burde, Shailesh, max_zorn, Leucippus Jan 25 at 4:18
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Namaste, Dietrich Burde, Shailesh, max_zorn, Leucippus Jan 25 at 4:18
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
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I think there is a problem in the formulation of your problem. One can prove that given some $n$ there are at most finitely many integer solutions $x$, $y$ and $z$. Maybe for some $n$ there is only one integer solution for $x$, $y$ and $z$?
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– quarague
Jan 24 at 16:35
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I think you must mean that $x,y,$ and $z$ must be distinct positive integers?
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– TonyK
Jan 24 at 18:37
add a comment |
2
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I think there is a problem in the formulation of your problem. One can prove that given some $n$ there are at most finitely many integer solutions $x$, $y$ and $z$. Maybe for some $n$ there is only one integer solution for $x$, $y$ and $z$?
$endgroup$
– quarague
Jan 24 at 16:35
$begingroup$
I think you must mean that $x,y,$ and $z$ must be distinct positive integers?
$endgroup$
– TonyK
Jan 24 at 18:37
2
2
$begingroup$
I think there is a problem in the formulation of your problem. One can prove that given some $n$ there are at most finitely many integer solutions $x$, $y$ and $z$. Maybe for some $n$ there is only one integer solution for $x$, $y$ and $z$?
$endgroup$
– quarague
Jan 24 at 16:35
$begingroup$
I think there is a problem in the formulation of your problem. One can prove that given some $n$ there are at most finitely many integer solutions $x$, $y$ and $z$. Maybe for some $n$ there is only one integer solution for $x$, $y$ and $z$?
$endgroup$
– quarague
Jan 24 at 16:35
$begingroup$
I think you must mean that $x,y,$ and $z$ must be distinct positive integers?
$endgroup$
– TonyK
Jan 24 at 18:37
$begingroup$
I think you must mean that $x,y,$ and $z$ must be distinct positive integers?
$endgroup$
– TonyK
Jan 24 at 18:37
add a comment |
3 Answers
3
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oldest
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There are infinitely many non-trivial solutions $x=y=z$ with $x$ even and $n=frac{x}{2}$:
$$
frac{x^2+y^2+z^2}{x+y+z}=frac{3x^2}{3x}=x=2n.
$$
So they cannot be unique.
Edit: The title now says distinct integer solutions (in the body it is still says unique integer solutions. In this case consider, for example,
$$
(x,y,z,n)=(18,12,8,7).
$$
This solves your equation in distinct integers.
answered Jan 24 at 16:30
Dietrich BurdeDietrich Burde
80.3k647104
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This one is quite difficult; at least, difficult to find all primitive integer solutions and prove that we have all. Taking $a=x-n,b=y-n,c=z-n,$ we are led to
$$ a^2 + b^2 + c^2 = 3n^2. $$
In order to get $gcd(a,b,c,n)=1,$ it is necessary that all be odd.
I think I will leave in the zeros in the formulas for $a,b,c.$ They serve as spacers.
$$ n = p^2 + q^2 + r^2 + s^2 $$
$$ a = p^2 + q^2 - r^2 - s^2 + 0 , p q - 2 p r + 2 q r + 2 p s + 2 q s + 0 , r s $$
$$ b = p^2 - q^2 + r^2 - s^2 + 2 p q - 0 , p r + 2 q r - 2 p s + 0 , q s + 2 r s$$
$$ c = p^2 - q^2 - r^2 + s^2 - 2 p q + 2 p r + 0 , q r + 0 , p s + 2 q s + 2 r s$$
In the table below, for each line just take the eight solutions $x=n pm a, y=n pm b,z=n pm c.$ Some of those will have $x,y,z$ positive.
=========================
n x y z p q r s
1 ; 2 2 2 x^2+y^2+z^2: 12 x+y+z: 6 * 2n: 12 ; 1 0 0 0
3 ; 8 4 4 x^2+y^2+z^2: 96 x+y+z: 16 * 2n: 96 ; 0 1 -1 -1
5 ; 12 10 6 x^2+y^2+z^2: 280 x+y+z: 28 * 2n: 280 ; 0 1 0 -2
7 ; 18 12 8 x^2+y^2+z^2: 532 x+y+z: 38 * 2n: 532 ; 1 2 1 1
9 ; 20 20 10 x^2+y^2+z^2: 900 x+y+z: 50 * 2n: 900 ; 0 2 2 1
9 ; 22 16 14 x^2+y^2+z^2: 936 x+y+z: 52 * 2n: 936 ; 0 2 -1 -2
11 ; 24 24 16 x^2+y^2+z^2: 1408 x+y+z: 64 * 2n: 1408 ; 0 1 1 -3
11 ; 28 18 16 x^2+y^2+z^2: 1364 x+y+z: 62 * 2n: 1364 ; 3 1 0 1
11 ; 30 12 12 x^2+y^2+z^2: 1188 x+y+z: 54 * 2n: 1188 ; 3 0 -1 1
13 ; 30 26 20 x^2+y^2+z^2: 1976 x+y+z: 76 * 2n: 1976 ; 0 2 0 -3
13 ; 32 24 18 x^2+y^2+z^2: 1924 x+y+z: 74 * 2n: 1924 ; 2 -2 -2 -1
15 ; 34 32 20 x^2+y^2+z^2: 2580 x+y+z: 86 * 2n: 2580 ; 3 2 1 1
15 ; 38 26 20 x^2+y^2+z^2: 2520 x+y+z: 84 * 2n: 2520 ; 1 -1 3 -2
15 ; 40 22 16 x^2+y^2+z^2: 2340 x+y+z: 78 * 2n: 2340 ; 1 -3 -2 -1
17 ; 40 30 30 x^2+y^2+z^2: 3400 x+y+z: 100 * 2n: 3400 ; 0 3 -2 -2
17 ; 40 34 24 x^2+y^2+z^2: 3332 x+y+z: 98 * 2n: 3332 ; 4 0 -1 0
17 ; 42 28 28 x^2+y^2+z^2: 3332 x+y+z: 98 * 2n: 3332 ; 0 3 2 2
17 ; 46 22 18 x^2+y^2+z^2: 2924 x+y+z: 86 * 2n: 2924 ; 3 -2 -2 0
19 ; 42 42 24 x^2+y^2+z^2: 4104 x+y+z: 108 * 2n: 4104 ; 1 -3 3 0
19 ; 44 36 32 x^2+y^2+z^2: 4256 x+y+z: 112 * 2n: 4256 ; 0 3 -1 -3
19 ; 48 30 30 x^2+y^2+z^2: 4104 x+y+z: 108 * 2n: 4104 ; 1 0 3 -3
19 ; 50 30 20 x^2+y^2+z^2: 3800 x+y+z: 100 * 2n: 3800 ; 1 1 1 -4
21 ; 46 44 34 x^2+y^2+z^2: 5208 x+y+z: 124 * 2n: 5208 ; 0 2 1 -4
21 ; 50 40 32 x^2+y^2+z^2: 5124 x+y+z: 122 * 2n: 5124 ; 2 3 2 2
21 ; 52 40 22 x^2+y^2+z^2: 4788 x+y+z: 114 * 2n: 4788 ; 4 2 0 1
23 ; 52 48 34 x^2+y^2+z^2: 6164 x+y+z: 134 * 2n: 6164 ; 1 -3 -3 -2
23 ; 54 48 24 x^2+y^2+z^2: 5796 x+y+z: 126 * 2n: 5796 ; 3 -2 -3 -1
23 ; 58 42 24 x^2+y^2+z^2: 5704 x+y+z: 124 * 2n: 5704 ; 1 -3 3 2
23 ; 60 36 30 x^2+y^2+z^2: 5796 x+y+z: 126 * 2n: 5796 ; 3 3 1 2
25 ; 56 50 42 x^2+y^2+z^2: 7400 x+y+z: 148 * 2n: 7400 ; 0 3 0 -4
25 ; 60 44 42 x^2+y^2+z^2: 7300 x+y+z: 146 * 2n: 7300 ; 4 -2 -2 -1
25 ; 60 48 36 x^2+y^2+z^2: 7200 x+y+z: 144 * 2n: 7200 ; 1 -2 4 -2
25 ; 66 38 30 x^2+y^2+z^2: 6700 x+y+z: 134 * 2n: 6700 ; 1 4 2 2
25 ; 68 30 26 x^2+y^2+z^2: 6200 x+y+z: 124 * 2n: 6200 ; 2 -1 4 -2
n x y z p q r s
==========================
answered Jan 24 at 17:26
Will JagyWill Jagy
104k5102201
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Above equation shown below:
begin{equation}
(x^2+y^2+z^2)/(x+y+z)=2n tag{A}
end{equation}
Equation $(A)$ has parametric solution:
$x=b(a+b)$
$y=a(7a+b)$
$z=2ab$
$n=(1/2)*(7a^2-2ab+b^2)$
For $(a,b)=(1,17)$ we get:
$(x,y,z,n)= (306,24,34,131)$
edited Jan 26 at 17:32
Community♦
1
answered Jan 25 at 2:26
SamSam
1
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Hi Sam, thank you for your answer and welcome to the forum. I'm sorry to tell you that your answer is not useful, because you have no development and any important explication in your answer.
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– El borito
Jan 25 at 2:48
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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There are infinitely many non-trivial solutions $x=y=z$ with $x$ even and $n=frac{x}{2}$:
$$
frac{x^2+y^2+z^2}{x+y+z}=frac{3x^2}{3x}=x=2n.
$$
So they cannot be unique.
Edit: The title now says distinct integer solutions (in the body it is still says unique integer solutions. In this case consider, for example,
$$
(x,y,z,n)=(18,12,8,7).
$$
This solves your equation in distinct integers.
answered Jan 24 at 16:30
Dietrich BurdeDietrich Burde
80.3k647104
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add a comment |
$begingroup$
There are infinitely many non-trivial solutions $x=y=z$ with $x$ even and $n=frac{x}{2}$:
$$
frac{x^2+y^2+z^2}{x+y+z}=frac{3x^2}{3x}=x=2n.
$$
So they cannot be unique.
Edit: The title now says distinct integer solutions (in the body it is still says unique integer solutions. In this case consider, for example,
$$
(x,y,z,n)=(18,12,8,7).
$$
This solves your equation in distinct integers.
answered Jan 24 at 16:30
Dietrich BurdeDietrich Burde
80.3k647104
$endgroup$
add a comment |
$begingroup$
There are infinitely many non-trivial solutions $x=y=z$ with $x$ even and $n=frac{x}{2}$:
$$
frac{x^2+y^2+z^2}{x+y+z}=frac{3x^2}{3x}=x=2n.
$$
So they cannot be unique.
Edit: The title now says distinct integer solutions (in the body it is still says unique integer solutions. In this case consider, for example,
$$
(x,y,z,n)=(18,12,8,7).
$$
This solves your equation in distinct integers.
answered Jan 24 at 16:30
Dietrich BurdeDietrich Burde
80.3k647104
$endgroup$
There are infinitely many non-trivial solutions $x=y=z$ with $x$ even and $n=frac{x}{2}$:
$$
frac{x^2+y^2+z^2}{x+y+z}=frac{3x^2}{3x}=x=2n.
$$
So they cannot be unique.
Edit: The title now says distinct integer solutions (in the body it is still says unique integer solutions. In this case consider, for example,
$$
(x,y,z,n)=(18,12,8,7).
$$
This solves your equation in distinct integers.
answered Jan 24 at 16:30
Dietrich BurdeDietrich Burde
80.3k647104
edited Jan 24 at 20:45
answered Jan 24 at 16:30
Dietrich BurdeDietrich Burde
80.3k647104
answered Jan 24 at 16:30
Dietrich BurdeDietrich Burde
80.3k647104
answered Jan 24 at 16:30
Dietrich BurdeDietrich Burde
80.3k647104
80.3k647104
add a comment |
add a comment |
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This one is quite difficult; at least, difficult to find all primitive integer solutions and prove that we have all. Taking $a=x-n,b=y-n,c=z-n,$ we are led to
$$ a^2 + b^2 + c^2 = 3n^2. $$
In order to get $gcd(a,b,c,n)=1,$ it is necessary that all be odd.
I think I will leave in the zeros in the formulas for $a,b,c.$ They serve as spacers.
$$ n = p^2 + q^2 + r^2 + s^2 $$
$$ a = p^2 + q^2 - r^2 - s^2 + 0 , p q - 2 p r + 2 q r + 2 p s + 2 q s + 0 , r s $$
$$ b = p^2 - q^2 + r^2 - s^2 + 2 p q - 0 , p r + 2 q r - 2 p s + 0 , q s + 2 r s$$
$$ c = p^2 - q^2 - r^2 + s^2 - 2 p q + 2 p r + 0 , q r + 0 , p s + 2 q s + 2 r s$$
In the table below, for each line just take the eight solutions $x=n pm a, y=n pm b,z=n pm c.$ Some of those will have $x,y,z$ positive.
=========================
n x y z p q r s
1 ; 2 2 2 x^2+y^2+z^2: 12 x+y+z: 6 * 2n: 12 ; 1 0 0 0
3 ; 8 4 4 x^2+y^2+z^2: 96 x+y+z: 16 * 2n: 96 ; 0 1 -1 -1
5 ; 12 10 6 x^2+y^2+z^2: 280 x+y+z: 28 * 2n: 280 ; 0 1 0 -2
7 ; 18 12 8 x^2+y^2+z^2: 532 x+y+z: 38 * 2n: 532 ; 1 2 1 1
9 ; 20 20 10 x^2+y^2+z^2: 900 x+y+z: 50 * 2n: 900 ; 0 2 2 1
9 ; 22 16 14 x^2+y^2+z^2: 936 x+y+z: 52 * 2n: 936 ; 0 2 -1 -2
11 ; 24 24 16 x^2+y^2+z^2: 1408 x+y+z: 64 * 2n: 1408 ; 0 1 1 -3
11 ; 28 18 16 x^2+y^2+z^2: 1364 x+y+z: 62 * 2n: 1364 ; 3 1 0 1
11 ; 30 12 12 x^2+y^2+z^2: 1188 x+y+z: 54 * 2n: 1188 ; 3 0 -1 1
13 ; 30 26 20 x^2+y^2+z^2: 1976 x+y+z: 76 * 2n: 1976 ; 0 2 0 -3
13 ; 32 24 18 x^2+y^2+z^2: 1924 x+y+z: 74 * 2n: 1924 ; 2 -2 -2 -1
15 ; 34 32 20 x^2+y^2+z^2: 2580 x+y+z: 86 * 2n: 2580 ; 3 2 1 1
15 ; 38 26 20 x^2+y^2+z^2: 2520 x+y+z: 84 * 2n: 2520 ; 1 -1 3 -2
15 ; 40 22 16 x^2+y^2+z^2: 2340 x+y+z: 78 * 2n: 2340 ; 1 -3 -2 -1
17 ; 40 30 30 x^2+y^2+z^2: 3400 x+y+z: 100 * 2n: 3400 ; 0 3 -2 -2
17 ; 40 34 24 x^2+y^2+z^2: 3332 x+y+z: 98 * 2n: 3332 ; 4 0 -1 0
17 ; 42 28 28 x^2+y^2+z^2: 3332 x+y+z: 98 * 2n: 3332 ; 0 3 2 2
17 ; 46 22 18 x^2+y^2+z^2: 2924 x+y+z: 86 * 2n: 2924 ; 3 -2 -2 0
19 ; 42 42 24 x^2+y^2+z^2: 4104 x+y+z: 108 * 2n: 4104 ; 1 -3 3 0
19 ; 44 36 32 x^2+y^2+z^2: 4256 x+y+z: 112 * 2n: 4256 ; 0 3 -1 -3
19 ; 48 30 30 x^2+y^2+z^2: 4104 x+y+z: 108 * 2n: 4104 ; 1 0 3 -3
19 ; 50 30 20 x^2+y^2+z^2: 3800 x+y+z: 100 * 2n: 3800 ; 1 1 1 -4
21 ; 46 44 34 x^2+y^2+z^2: 5208 x+y+z: 124 * 2n: 5208 ; 0 2 1 -4
21 ; 50 40 32 x^2+y^2+z^2: 5124 x+y+z: 122 * 2n: 5124 ; 2 3 2 2
21 ; 52 40 22 x^2+y^2+z^2: 4788 x+y+z: 114 * 2n: 4788 ; 4 2 0 1
23 ; 52 48 34 x^2+y^2+z^2: 6164 x+y+z: 134 * 2n: 6164 ; 1 -3 -3 -2
23 ; 54 48 24 x^2+y^2+z^2: 5796 x+y+z: 126 * 2n: 5796 ; 3 -2 -3 -1
23 ; 58 42 24 x^2+y^2+z^2: 5704 x+y+z: 124 * 2n: 5704 ; 1 -3 3 2
23 ; 60 36 30 x^2+y^2+z^2: 5796 x+y+z: 126 * 2n: 5796 ; 3 3 1 2
25 ; 56 50 42 x^2+y^2+z^2: 7400 x+y+z: 148 * 2n: 7400 ; 0 3 0 -4
25 ; 60 44 42 x^2+y^2+z^2: 7300 x+y+z: 146 * 2n: 7300 ; 4 -2 -2 -1
25 ; 60 48 36 x^2+y^2+z^2: 7200 x+y+z: 144 * 2n: 7200 ; 1 -2 4 -2
25 ; 66 38 30 x^2+y^2+z^2: 6700 x+y+z: 134 * 2n: 6700 ; 1 4 2 2
25 ; 68 30 26 x^2+y^2+z^2: 6200 x+y+z: 124 * 2n: 6200 ; 2 -1 4 -2
n x y z p q r s
==========================
answered Jan 24 at 17:26
Will JagyWill Jagy
104k5102201
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add a comment |
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This one is quite difficult; at least, difficult to find all primitive integer solutions and prove that we have all. Taking $a=x-n,b=y-n,c=z-n,$ we are led to
$$ a^2 + b^2 + c^2 = 3n^2. $$
In order to get $gcd(a,b,c,n)=1,$ it is necessary that all be odd.
I think I will leave in the zeros in the formulas for $a,b,c.$ They serve as spacers.
$$ n = p^2 + q^2 + r^2 + s^2 $$
$$ a = p^2 + q^2 - r^2 - s^2 + 0 , p q - 2 p r + 2 q r + 2 p s + 2 q s + 0 , r s $$
$$ b = p^2 - q^2 + r^2 - s^2 + 2 p q - 0 , p r + 2 q r - 2 p s + 0 , q s + 2 r s$$
$$ c = p^2 - q^2 - r^2 + s^2 - 2 p q + 2 p r + 0 , q r + 0 , p s + 2 q s + 2 r s$$
In the table below, for each line just take the eight solutions $x=n pm a, y=n pm b,z=n pm c.$ Some of those will have $x,y,z$ positive.
=========================
n x y z p q r s
1 ; 2 2 2 x^2+y^2+z^2: 12 x+y+z: 6 * 2n: 12 ; 1 0 0 0
3 ; 8 4 4 x^2+y^2+z^2: 96 x+y+z: 16 * 2n: 96 ; 0 1 -1 -1
5 ; 12 10 6 x^2+y^2+z^2: 280 x+y+z: 28 * 2n: 280 ; 0 1 0 -2
7 ; 18 12 8 x^2+y^2+z^2: 532 x+y+z: 38 * 2n: 532 ; 1 2 1 1
9 ; 20 20 10 x^2+y^2+z^2: 900 x+y+z: 50 * 2n: 900 ; 0 2 2 1
9 ; 22 16 14 x^2+y^2+z^2: 936 x+y+z: 52 * 2n: 936 ; 0 2 -1 -2
11 ; 24 24 16 x^2+y^2+z^2: 1408 x+y+z: 64 * 2n: 1408 ; 0 1 1 -3
11 ; 28 18 16 x^2+y^2+z^2: 1364 x+y+z: 62 * 2n: 1364 ; 3 1 0 1
11 ; 30 12 12 x^2+y^2+z^2: 1188 x+y+z: 54 * 2n: 1188 ; 3 0 -1 1
13 ; 30 26 20 x^2+y^2+z^2: 1976 x+y+z: 76 * 2n: 1976 ; 0 2 0 -3
13 ; 32 24 18 x^2+y^2+z^2: 1924 x+y+z: 74 * 2n: 1924 ; 2 -2 -2 -1
15 ; 34 32 20 x^2+y^2+z^2: 2580 x+y+z: 86 * 2n: 2580 ; 3 2 1 1
15 ; 38 26 20 x^2+y^2+z^2: 2520 x+y+z: 84 * 2n: 2520 ; 1 -1 3 -2
15 ; 40 22 16 x^2+y^2+z^2: 2340 x+y+z: 78 * 2n: 2340 ; 1 -3 -2 -1
17 ; 40 30 30 x^2+y^2+z^2: 3400 x+y+z: 100 * 2n: 3400 ; 0 3 -2 -2
17 ; 40 34 24 x^2+y^2+z^2: 3332 x+y+z: 98 * 2n: 3332 ; 4 0 -1 0
17 ; 42 28 28 x^2+y^2+z^2: 3332 x+y+z: 98 * 2n: 3332 ; 0 3 2 2
17 ; 46 22 18 x^2+y^2+z^2: 2924 x+y+z: 86 * 2n: 2924 ; 3 -2 -2 0
19 ; 42 42 24 x^2+y^2+z^2: 4104 x+y+z: 108 * 2n: 4104 ; 1 -3 3 0
19 ; 44 36 32 x^2+y^2+z^2: 4256 x+y+z: 112 * 2n: 4256 ; 0 3 -1 -3
19 ; 48 30 30 x^2+y^2+z^2: 4104 x+y+z: 108 * 2n: 4104 ; 1 0 3 -3
19 ; 50 30 20 x^2+y^2+z^2: 3800 x+y+z: 100 * 2n: 3800 ; 1 1 1 -4
21 ; 46 44 34 x^2+y^2+z^2: 5208 x+y+z: 124 * 2n: 5208 ; 0 2 1 -4
21 ; 50 40 32 x^2+y^2+z^2: 5124 x+y+z: 122 * 2n: 5124 ; 2 3 2 2
21 ; 52 40 22 x^2+y^2+z^2: 4788 x+y+z: 114 * 2n: 4788 ; 4 2 0 1
23 ; 52 48 34 x^2+y^2+z^2: 6164 x+y+z: 134 * 2n: 6164 ; 1 -3 -3 -2
23 ; 54 48 24 x^2+y^2+z^2: 5796 x+y+z: 126 * 2n: 5796 ; 3 -2 -3 -1
23 ; 58 42 24 x^2+y^2+z^2: 5704 x+y+z: 124 * 2n: 5704 ; 1 -3 3 2
23 ; 60 36 30 x^2+y^2+z^2: 5796 x+y+z: 126 * 2n: 5796 ; 3 3 1 2
25 ; 56 50 42 x^2+y^2+z^2: 7400 x+y+z: 148 * 2n: 7400 ; 0 3 0 -4
25 ; 60 44 42 x^2+y^2+z^2: 7300 x+y+z: 146 * 2n: 7300 ; 4 -2 -2 -1
25 ; 60 48 36 x^2+y^2+z^2: 7200 x+y+z: 144 * 2n: 7200 ; 1 -2 4 -2
25 ; 66 38 30 x^2+y^2+z^2: 6700 x+y+z: 134 * 2n: 6700 ; 1 4 2 2
25 ; 68 30 26 x^2+y^2+z^2: 6200 x+y+z: 124 * 2n: 6200 ; 2 -1 4 -2
n x y z p q r s
==========================
answered Jan 24 at 17:26
Will JagyWill Jagy
104k5102201
$endgroup$
add a comment |
$begingroup$
This one is quite difficult; at least, difficult to find all primitive integer solutions and prove that we have all. Taking $a=x-n,b=y-n,c=z-n,$ we are led to
$$ a^2 + b^2 + c^2 = 3n^2. $$
In order to get $gcd(a,b,c,n)=1,$ it is necessary that all be odd.
I think I will leave in the zeros in the formulas for $a,b,c.$ They serve as spacers.
$$ n = p^2 + q^2 + r^2 + s^2 $$
$$ a = p^2 + q^2 - r^2 - s^2 + 0 , p q - 2 p r + 2 q r + 2 p s + 2 q s + 0 , r s $$
$$ b = p^2 - q^2 + r^2 - s^2 + 2 p q - 0 , p r + 2 q r - 2 p s + 0 , q s + 2 r s$$
$$ c = p^2 - q^2 - r^2 + s^2 - 2 p q + 2 p r + 0 , q r + 0 , p s + 2 q s + 2 r s$$
In the table below, for each line just take the eight solutions $x=n pm a, y=n pm b,z=n pm c.$ Some of those will have $x,y,z$ positive.
=========================
n x y z p q r s
1 ; 2 2 2 x^2+y^2+z^2: 12 x+y+z: 6 * 2n: 12 ; 1 0 0 0
3 ; 8 4 4 x^2+y^2+z^2: 96 x+y+z: 16 * 2n: 96 ; 0 1 -1 -1
5 ; 12 10 6 x^2+y^2+z^2: 280 x+y+z: 28 * 2n: 280 ; 0 1 0 -2
7 ; 18 12 8 x^2+y^2+z^2: 532 x+y+z: 38 * 2n: 532 ; 1 2 1 1
9 ; 20 20 10 x^2+y^2+z^2: 900 x+y+z: 50 * 2n: 900 ; 0 2 2 1
9 ; 22 16 14 x^2+y^2+z^2: 936 x+y+z: 52 * 2n: 936 ; 0 2 -1 -2
11 ; 24 24 16 x^2+y^2+z^2: 1408 x+y+z: 64 * 2n: 1408 ; 0 1 1 -3
11 ; 28 18 16 x^2+y^2+z^2: 1364 x+y+z: 62 * 2n: 1364 ; 3 1 0 1
11 ; 30 12 12 x^2+y^2+z^2: 1188 x+y+z: 54 * 2n: 1188 ; 3 0 -1 1
13 ; 30 26 20 x^2+y^2+z^2: 1976 x+y+z: 76 * 2n: 1976 ; 0 2 0 -3
13 ; 32 24 18 x^2+y^2+z^2: 1924 x+y+z: 74 * 2n: 1924 ; 2 -2 -2 -1
15 ; 34 32 20 x^2+y^2+z^2: 2580 x+y+z: 86 * 2n: 2580 ; 3 2 1 1
15 ; 38 26 20 x^2+y^2+z^2: 2520 x+y+z: 84 * 2n: 2520 ; 1 -1 3 -2
15 ; 40 22 16 x^2+y^2+z^2: 2340 x+y+z: 78 * 2n: 2340 ; 1 -3 -2 -1
17 ; 40 30 30 x^2+y^2+z^2: 3400 x+y+z: 100 * 2n: 3400 ; 0 3 -2 -2
17 ; 40 34 24 x^2+y^2+z^2: 3332 x+y+z: 98 * 2n: 3332 ; 4 0 -1 0
17 ; 42 28 28 x^2+y^2+z^2: 3332 x+y+z: 98 * 2n: 3332 ; 0 3 2 2
17 ; 46 22 18 x^2+y^2+z^2: 2924 x+y+z: 86 * 2n: 2924 ; 3 -2 -2 0
19 ; 42 42 24 x^2+y^2+z^2: 4104 x+y+z: 108 * 2n: 4104 ; 1 -3 3 0
19 ; 44 36 32 x^2+y^2+z^2: 4256 x+y+z: 112 * 2n: 4256 ; 0 3 -1 -3
19 ; 48 30 30 x^2+y^2+z^2: 4104 x+y+z: 108 * 2n: 4104 ; 1 0 3 -3
19 ; 50 30 20 x^2+y^2+z^2: 3800 x+y+z: 100 * 2n: 3800 ; 1 1 1 -4
21 ; 46 44 34 x^2+y^2+z^2: 5208 x+y+z: 124 * 2n: 5208 ; 0 2 1 -4
21 ; 50 40 32 x^2+y^2+z^2: 5124 x+y+z: 122 * 2n: 5124 ; 2 3 2 2
21 ; 52 40 22 x^2+y^2+z^2: 4788 x+y+z: 114 * 2n: 4788 ; 4 2 0 1
23 ; 52 48 34 x^2+y^2+z^2: 6164 x+y+z: 134 * 2n: 6164 ; 1 -3 -3 -2
23 ; 54 48 24 x^2+y^2+z^2: 5796 x+y+z: 126 * 2n: 5796 ; 3 -2 -3 -1
23 ; 58 42 24 x^2+y^2+z^2: 5704 x+y+z: 124 * 2n: 5704 ; 1 -3 3 2
23 ; 60 36 30 x^2+y^2+z^2: 5796 x+y+z: 126 * 2n: 5796 ; 3 3 1 2
25 ; 56 50 42 x^2+y^2+z^2: 7400 x+y+z: 148 * 2n: 7400 ; 0 3 0 -4
25 ; 60 44 42 x^2+y^2+z^2: 7300 x+y+z: 146 * 2n: 7300 ; 4 -2 -2 -1
25 ; 60 48 36 x^2+y^2+z^2: 7200 x+y+z: 144 * 2n: 7200 ; 1 -2 4 -2
25 ; 66 38 30 x^2+y^2+z^2: 6700 x+y+z: 134 * 2n: 6700 ; 1 4 2 2
25 ; 68 30 26 x^2+y^2+z^2: 6200 x+y+z: 124 * 2n: 6200 ; 2 -1 4 -2
n x y z p q r s
==========================
answered Jan 24 at 17:26
Will JagyWill Jagy
104k5102201
$endgroup$
This one is quite difficult; at least, difficult to find all primitive integer solutions and prove that we have all. Taking $a=x-n,b=y-n,c=z-n,$ we are led to
$$ a^2 + b^2 + c^2 = 3n^2. $$
In order to get $gcd(a,b,c,n)=1,$ it is necessary that all be odd.
I think I will leave in the zeros in the formulas for $a,b,c.$ They serve as spacers.
$$ n = p^2 + q^2 + r^2 + s^2 $$
$$ a = p^2 + q^2 - r^2 - s^2 + 0 , p q - 2 p r + 2 q r + 2 p s + 2 q s + 0 , r s $$
$$ b = p^2 - q^2 + r^2 - s^2 + 2 p q - 0 , p r + 2 q r - 2 p s + 0 , q s + 2 r s$$
$$ c = p^2 - q^2 - r^2 + s^2 - 2 p q + 2 p r + 0 , q r + 0 , p s + 2 q s + 2 r s$$
In the table below, for each line just take the eight solutions $x=n pm a, y=n pm b,z=n pm c.$ Some of those will have $x,y,z$ positive.
=========================
n x y z p q r s
1 ; 2 2 2 x^2+y^2+z^2: 12 x+y+z: 6 * 2n: 12 ; 1 0 0 0
3 ; 8 4 4 x^2+y^2+z^2: 96 x+y+z: 16 * 2n: 96 ; 0 1 -1 -1
5 ; 12 10 6 x^2+y^2+z^2: 280 x+y+z: 28 * 2n: 280 ; 0 1 0 -2
7 ; 18 12 8 x^2+y^2+z^2: 532 x+y+z: 38 * 2n: 532 ; 1 2 1 1
9 ; 20 20 10 x^2+y^2+z^2: 900 x+y+z: 50 * 2n: 900 ; 0 2 2 1
9 ; 22 16 14 x^2+y^2+z^2: 936 x+y+z: 52 * 2n: 936 ; 0 2 -1 -2
11 ; 24 24 16 x^2+y^2+z^2: 1408 x+y+z: 64 * 2n: 1408 ; 0 1 1 -3
11 ; 28 18 16 x^2+y^2+z^2: 1364 x+y+z: 62 * 2n: 1364 ; 3 1 0 1
11 ; 30 12 12 x^2+y^2+z^2: 1188 x+y+z: 54 * 2n: 1188 ; 3 0 -1 1
13 ; 30 26 20 x^2+y^2+z^2: 1976 x+y+z: 76 * 2n: 1976 ; 0 2 0 -3
13 ; 32 24 18 x^2+y^2+z^2: 1924 x+y+z: 74 * 2n: 1924 ; 2 -2 -2 -1
15 ; 34 32 20 x^2+y^2+z^2: 2580 x+y+z: 86 * 2n: 2580 ; 3 2 1 1
15 ; 38 26 20 x^2+y^2+z^2: 2520 x+y+z: 84 * 2n: 2520 ; 1 -1 3 -2
15 ; 40 22 16 x^2+y^2+z^2: 2340 x+y+z: 78 * 2n: 2340 ; 1 -3 -2 -1
17 ; 40 30 30 x^2+y^2+z^2: 3400 x+y+z: 100 * 2n: 3400 ; 0 3 -2 -2
17 ; 40 34 24 x^2+y^2+z^2: 3332 x+y+z: 98 * 2n: 3332 ; 4 0 -1 0
17 ; 42 28 28 x^2+y^2+z^2: 3332 x+y+z: 98 * 2n: 3332 ; 0 3 2 2
17 ; 46 22 18 x^2+y^2+z^2: 2924 x+y+z: 86 * 2n: 2924 ; 3 -2 -2 0
19 ; 42 42 24 x^2+y^2+z^2: 4104 x+y+z: 108 * 2n: 4104 ; 1 -3 3 0
19 ; 44 36 32 x^2+y^2+z^2: 4256 x+y+z: 112 * 2n: 4256 ; 0 3 -1 -3
19 ; 48 30 30 x^2+y^2+z^2: 4104 x+y+z: 108 * 2n: 4104 ; 1 0 3 -3
19 ; 50 30 20 x^2+y^2+z^2: 3800 x+y+z: 100 * 2n: 3800 ; 1 1 1 -4
21 ; 46 44 34 x^2+y^2+z^2: 5208 x+y+z: 124 * 2n: 5208 ; 0 2 1 -4
21 ; 50 40 32 x^2+y^2+z^2: 5124 x+y+z: 122 * 2n: 5124 ; 2 3 2 2
21 ; 52 40 22 x^2+y^2+z^2: 4788 x+y+z: 114 * 2n: 4788 ; 4 2 0 1
23 ; 52 48 34 x^2+y^2+z^2: 6164 x+y+z: 134 * 2n: 6164 ; 1 -3 -3 -2
23 ; 54 48 24 x^2+y^2+z^2: 5796 x+y+z: 126 * 2n: 5796 ; 3 -2 -3 -1
23 ; 58 42 24 x^2+y^2+z^2: 5704 x+y+z: 124 * 2n: 5704 ; 1 -3 3 2
23 ; 60 36 30 x^2+y^2+z^2: 5796 x+y+z: 126 * 2n: 5796 ; 3 3 1 2
25 ; 56 50 42 x^2+y^2+z^2: 7400 x+y+z: 148 * 2n: 7400 ; 0 3 0 -4
25 ; 60 44 42 x^2+y^2+z^2: 7300 x+y+z: 146 * 2n: 7300 ; 4 -2 -2 -1
25 ; 60 48 36 x^2+y^2+z^2: 7200 x+y+z: 144 * 2n: 7200 ; 1 -2 4 -2
25 ; 66 38 30 x^2+y^2+z^2: 6700 x+y+z: 134 * 2n: 6700 ; 1 4 2 2
25 ; 68 30 26 x^2+y^2+z^2: 6200 x+y+z: 124 * 2n: 6200 ; 2 -1 4 -2
n x y z p q r s
==========================
answered Jan 24 at 17:26
Will JagyWill Jagy
104k5102201
edited Jan 25 at 17:44
answered Jan 24 at 17:26
Will JagyWill Jagy
104k5102201
answered Jan 24 at 17:26
Will JagyWill Jagy
104k5102201
answered Jan 24 at 17:26
Will JagyWill Jagy
104k5102201
104k5102201
add a comment |
add a comment |
$begingroup$
Above equation shown below:
begin{equation}
(x^2+y^2+z^2)/(x+y+z)=2n tag{A}
end{equation}
Equation $(A)$ has parametric solution:
$x=b(a+b)$
$y=a(7a+b)$
$z=2ab$
$n=(1/2)*(7a^2-2ab+b^2)$
For $(a,b)=(1,17)$ we get:
$(x,y,z,n)= (306,24,34,131)$
edited Jan 26 at 17:32
Community♦
1
answered Jan 25 at 2:26
SamSam
1
$endgroup$
$begingroup$
Hi Sam, thank you for your answer and welcome to the forum. I'm sorry to tell you that your answer is not useful, because you have no development and any important explication in your answer.
$endgroup$
– El borito
Jan 25 at 2:48
add a comment |
$begingroup$
Above equation shown below:
begin{equation}
(x^2+y^2+z^2)/(x+y+z)=2n tag{A}
end{equation}
Equation $(A)$ has parametric solution:
$x=b(a+b)$
$y=a(7a+b)$
$z=2ab$
$n=(1/2)*(7a^2-2ab+b^2)$
For $(a,b)=(1,17)$ we get:
$(x,y,z,n)= (306,24,34,131)$
edited Jan 26 at 17:32
Community♦
1
answered Jan 25 at 2:26
SamSam
1
$endgroup$
$begingroup$
Hi Sam, thank you for your answer and welcome to the forum. I'm sorry to tell you that your answer is not useful, because you have no development and any important explication in your answer.
$endgroup$
– El borito
Jan 25 at 2:48
add a comment |
$begingroup$
Above equation shown below:
begin{equation}
(x^2+y^2+z^2)/(x+y+z)=2n tag{A}
end{equation}
Equation $(A)$ has parametric solution:
$x=b(a+b)$
$y=a(7a+b)$
$z=2ab$
$n=(1/2)*(7a^2-2ab+b^2)$
For $(a,b)=(1,17)$ we get:
$(x,y,z,n)= (306,24,34,131)$
edited Jan 26 at 17:32
Community♦
1
answered Jan 25 at 2:26
SamSam
1
$endgroup$
Above equation shown below:
begin{equation}
(x^2+y^2+z^2)/(x+y+z)=2n tag{A}
end{equation}
Equation $(A)$ has parametric solution:
$x=b(a+b)$
$y=a(7a+b)$
$z=2ab$
$n=(1/2)*(7a^2-2ab+b^2)$
For $(a,b)=(1,17)$ we get:
$(x,y,z,n)= (306,24,34,131)$
edited Jan 26 at 17:32
Community♦
1
answered Jan 25 at 2:26
SamSam
1
edited Jan 26 at 17:32
Community♦
1
edited Jan 26 at 17:32
Community♦
1
edited Jan 26 at 17:32
Community♦
1
1
answered Jan 25 at 2:26
SamSam
1
answered Jan 25 at 2:26
SamSam
1
answered Jan 25 at 2:26
SamSam
1
1
$begingroup$
Hi Sam, thank you for your answer and welcome to the forum. I'm sorry to tell you that your answer is not useful, because you have no development and any important explication in your answer.
$endgroup$
– El borito
Jan 25 at 2:48
add a comment |
$begingroup$
Hi Sam, thank you for your answer and welcome to the forum. I'm sorry to tell you that your answer is not useful, because you have no development and any important explication in your answer.
$endgroup$
– El borito
Jan 25 at 2:48
$begingroup$
Hi Sam, thank you for your answer and welcome to the forum. I'm sorry to tell you that your answer is not useful, because you have no development and any important explication in your answer.
$endgroup$
– El borito
Jan 25 at 2:48
$begingroup$
Hi Sam, thank you for your answer and welcome to the forum. I'm sorry to tell you that your answer is not useful, because you have no development and any important explication in your answer.
$endgroup$
– El borito
Jan 25 at 2:48
add a comment |
2
$begingroup$
I think there is a problem in the formulation of your problem. One can prove that given some $n$ there are at most finitely many integer solutions $x$, $y$ and $z$. Maybe for some $n$ there is only one integer solution for $x$, $y$ and $z$?
$endgroup$
– quarague
Jan 24 at 16:35
$begingroup$
I think you must mean that $x,y,$ and $z$ must be distinct positive integers?
$endgroup$
– TonyK
Jan 24 at 18:37