I couldn't make any progress on this problem, can anyone help?
I found it's the same as: Find all integers $a,b,c$ such that $ab+bc+ca$ divides $a^2+b^2+c^2$.
I found a solution $a=-b=1$, and $c$ any integer.
Any more solutions?
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1@amWhy but are equivalent. – Carol Feb 09 '15 at 15:21
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Yes, many more solutions: $(a,a,a)$, $(0,b,b)$, or $(a,-a,c)$ with $a\mid c$, etc. – Dietrich Burde Feb 09 '15 at 15:33
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2It might be interesting to look for cases with $a,b,c>0$ and $\gcd(a,b,c)=1$ – Joffan Feb 09 '15 at 15:38
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As for the equivalent statement, use $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$. – Dietrich Burde Feb 09 '15 at 15:43
4 Answers
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Did not notice this question. The diophantine equation $$ x^2 + y^2 + z^2 = B (yz + zx + xy) $$ has integer solutions $(x,y,z)$ not all equal to zero (and allowed negative)), if any only if we may express both $$ B-1 = u_1^2 + 3 v_1^2, $$ $$ B+2 = u_2^2 + 3 v_2^2 $$ all in integers. The values that work are $$ B = 1,2,5,10, 14,... $$
See
$x^2+y^2+z^2=5(xy+yz+zx)$ -- Is this all solutions?
and my several answers at
Find a solution: $3(x^2+y^2+z^2)=10(xy+yz+zx)$
examples for $B = 5,10,14.$ For $B = 5$ we take the three values $$ ( 5 u^2 + 9 uv + 3 v^2, 3 u^2 -3 uv - v^2, - u^2 + uv + 5 v^2 ). $$ Next, take the triple in decreasing absolute value. Finally, if the first one is negative, negate all three. The result is a list that is not too repetitive, with $x \geq y \geq |z|,$ because $y$ turns out to be positive in this recipe. Sometimes $z$ is also positive, not often. Oh, nice rule, we get to take $u,v \geq 0.$
./isotropy_binaries_combined 1 5 300 | sort -n
x y z u v
5 3 -1 < 5, 9, 3 > 1 0
17 5 -1 < 5, 9, 3 > 1 1
41 5 3 < 5, 9, 3 > 2 1
59 47 -15 < 5, 9, 3 > 1 3
75 17 -1 < 5, 9, 3 > 3 1
89 83 -25 < 5, 9, 3 > 1 4
101 47 -15 < 5, 9, 3 > 2 3
111 17 5 < 5, 9, 3 > 3 2
129 125 -37 < 5, 9, 3 > 1 5
173 59 -15 < 5, 9, 3 > 5 1
185 131 -43 < 5, 9, 3 > 2 5
185 167 -51 < 5, 9, 3 > 1 6
201 83 -25 < 5, 9, 3 > 3 4
215 41 3 < 5, 9, 3 > 4 3
227 41 5 < 5, 9, 3 > 5 2
237 89 -25 < 5, 9, 3 > 6 1
251 215 -67 < 5, 9, 3 > 1 7
255 131 -43 < 5, 9, 3 > 3 5
293 255 -79 < 5, 9, 3 > 2 7
x y z u v
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For $B=10,$ we take $$ ( 5 u^2 + 8 uv + 2 v^2, 2 u^2 -4 uv - v^2, - u^2 + 2 uv + 5 v^2 ). $$
./isotropy_binaries_combined 1 10 300 | sort -n
x y z u v
5 2 -1 < 5, 8, 2 > 1 0
29 23 -10 < 5, 8, 2 > 1 2
38 5 -1 < 5, 8, 2 > 2 1
50 47 -19 < 5, 8, 2 > 1 3
71 5 2 < 5, 8, 2 > 3 1
86 53 -25 < 5, 8, 2 > 2 3
101 23 -10 < 5, 8, 2 > 3 2
134 95 -43 < 5, 8, 2 > 1 5
167 29 -10 < 5, 8, 2 > 5 1
173 95 -46 < 5, 8, 2 > 3 4
191 125 -58 < 5, 8, 2 > 1 6
194 53 -25 < 5, 8, 2 > 4 3
215 146 -67 < 5, 8, 2 > 3 5
230 47 -19 < 5, 8, 2 > 6 1
263 50 -19 < 5, 8, 2 > 5 3
269 230 -97 < 5, 8, 2 > 2 7
290 149 -73 < 5, 8, 2 > 4 5
x y z u v
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For $B=14,$ we take $$ ( 3 u^2 + 6 uv + 2 v^2, 2 u^2 -2 uv - v^2, - u^2 + 3 v^2 ). $$ No $uv$ term in the third form. Go Figure.
./isotropy_binaries_combined 1 14 300 | sort -n
x y z u v
3 2 -1 < 3, 6, 2 > 1 0
11 2 -1 < 3, 6, 2 > 1 1
23 11 -6 < 3, 6, 2 > 1 2
26 3 -1 < 3, 6, 2 > 2 1
47 11 -6 < 3, 6, 2 > 3 1
59 47 -22 < 3, 6, 2 > 1 4
66 23 -13 < 3, 6, 2 > 2 3
71 3 2 < 3, 6, 2 > 3 2
74 23 -13 < 3, 6, 2 > 4 1
83 74 -33 < 3, 6, 2 > 1 5
107 39 -22 < 3, 6, 2 > 5 1
111 107 -46 < 3, 6, 2 > 1 6
122 71 -37 < 3, 6, 2 > 2 5
131 39 -22 < 3, 6, 2 > 3 4
138 11 -1 < 3, 6, 2 > 4 3
146 143 -61 < 3, 6, 2 > 1 7
146 59 -33 < 3, 6, 2 > 6 1
167 66 -37 < 3, 6, 2 > 3 5
183 11 2 < 3, 6, 2 > 5 3
191 179 -78 < 3, 6, 2 > 1 8
191 83 -46 < 3, 6, 2 > 7 1
194 143 -69 < 3, 6, 2 > 2 7
218 59 -33 < 3, 6, 2 > 4 5
227 23 -6 < 3, 6, 2 > 5 4
239 66 -37 < 3, 6, 2 > 7 2
242 111 -61 < 3, 6, 2 > 8 1
242 219 -97 < 3, 6, 2 > 1 9
251 138 -73 < 3, 6, 2 > 3 7
282 239 -109 < 3, 6, 2 > 2 9
291 47 -22 < 3, 6, 2 > 7 3
299 183 -94 < 3, 6, 2 > 3 8
299 263 -118 < 3, 6, 2 > 1 10
x y z u v
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For $B = 29$ we need two parametrizations;
$$ ( 23 u^2 + 49 u v + 17 v^2, 17 u^2 -15 u v -9 v^2, -9 u^2 -3 u v + 23 v^2)$$
$$ ( 27 u^2 + 45 u v + 11 v^2, 11 u^2 -23 u v -7 v^2, -7 u^2 + 9 u v + 27 v^2) $$
./isotropy_binaries_combined 1 29 1111 | sort -n
x y z u v
23 17 -9 < 23, 49, 17 > 1 0
27 11 -7 < 27, 45, 11 > 1 0
83 29 -19 < 27, 45, 11 > 1 1
89 11 -7 < 23, 49, 17 > 1 1
207 29 -19 < 23, 49, 17 > 2 1
209 17 -9 < 27, 45, 11 > 2 1
263 261 -121 < 27, 45, 11 > 1 3
323 189 -109 < 23, 49, 17 > 1 3
371 99 -67 < 23, 49, 17 > 3 1
389 23 -9 < 27, 45, 11 > 3 1
461 383 -193 < 27, 45, 11 > 1 4
477 269 -157 < 27, 45, 11 > 2 3
491 347 -187 < 23, 49, 17 > 1 4
539 153 -103 < 23, 49, 17 > 2 3
557 99 -67 < 27, 45, 11 > 3 2
569 27 -7 < 23, 49, 17 > 3 2
693 551 -283 < 23, 49, 17 > 1 5
833 737 -361 < 27, 45, 11 > 2 5
911 153 -103 < 27, 45, 11 > 5 1
929 801 -397 < 23, 49, 17 > 1 6
959 477 -289 < 27, 45, 11 > 3 4
1007 509 -307 < 23, 49, 17 > 2 5
1019 693 -379 < 27, 45, 11 > 1 6
1067 251 -171 < 23, 49, 17 > 3 4
1071 239 -163 < 27, 45, 11 > 4 3
1109 27 11 < 23, 49, 17 > 4 3
x y z u v
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Could you please provide a reference for why $B-1$ and $B+2$ must be of the form $x^2+3y^2$? – Batominovski Oct 19 '15 at 00:52
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@Batominovski, it is my result, but is equivalent to the famous result of Legendre on indefinite ternary forms, http://math.stackexchange.com/questions/27471/proof-of-legendres-theorem-on-the-ternary-quadratic-form or pages 80-82 in Rational Quadratic Forms by Cassels, http://store.doverpublications.com/0486466701.html I also wrote up a twenty page article (well, not to publish) about this problem, i could send you a pdf. – Will Jagy Oct 19 '15 at 01:04
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1@Batominovski, sent. If you don't see anything, email me so I can reply with the pdf (my address in my profile). In any case, it is usual to delete comments with email addresses after successful use. I have not had any real problems with my address public, just some occasional individuals, no bots. – Will Jagy Oct 20 '15 at 18:36
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@WillJagy: Hi, thanks for this informative answer. If it isn't too much trouble, I too am interested in the pdf regarding the result at the beginning. My email is: [email protected] Thank you – user45220 Oct 25 '15 at 04:47
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Yes, there are lots of solutions. These are the ones with $f<g<h\le102$ and $\gcd(f,g,h)=1$. $$ {1,4,9}\\ {1,9,16}\\ {1,25,36}\\ {1,36,49}\\ {1,49,64}\\ {1,64,81}\\ {1,81,100}\\ {2,3,71}\\ {2,5,71}\\ {3,5,41}\\ {4,9,25}\\ {4,25,49}\\ {4,49,81}\\ {9,16,49}\\ {9,25,64}\\ {9,49,100}\\ {16,25,81} $$ Mathematica code used:
fmax = 100;
Do[
If[GCD[f, g, h] != 1, Continue[]];
If[Mod[f^2 + g^2 + h^2, f g + g h + h f] == 0, Print[{f, g, h}]],
{f, fmax}, {g, f + 1, fmax + 1}, {h, g + 1, fmax + 2}
]
Julián Aguirre
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Hello, thank you for this. Can I ask which code did you use? Thanks! – user45220 Feb 09 '15 at 16:50
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Hello again, I have figured how to make the code print $\frac{fg+gh+hf}{f^2+g^2+h^2}$. But how can I make it to print only distinct fractions? There are too many $\frac{1}{2}$. Thank you very much Sir. – user45220 Feb 09 '15 at 18:08
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You can save the quotient in a list. Every time a new solution is found check if the quotient is in the list. If it is,
Continue[]. If not, print it and add it to the list. – Julián Aguirre Feb 09 '15 at 18:12 -
Sorry, I do not know how to use code, I am a complete beginner. If you show me the new one I will accept your answer, because it is exactly what I need (when I thought about it). – user45220 Feb 09 '15 at 18:14
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I am not bargaining. Your question was about existence of solutions. I answered it, and I provided the code and gave instructions on how to modify it to your needs. If you do no like it's OK, do not accept it. Post your new question in a programming site. – Julián Aguirre Feb 09 '15 at 18:35
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I am very sorry, you are right. I will make a new question for this on mathematica exchange. Thank you for your help. – user45220 Feb 09 '15 at 18:39
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Since $(a+b+c)^2 = 2(ab+ac+bc)+(a^2+b^2+c^2)$, if $$ (a+b+c)^2 = k(ab+ac+bc), \tag{1}$$ then $k\geq 2$, and for $k=2$ we have only the trivial solution $(a,b,c)=(0,0,0)$.
Assuming $k=3$, we have: $$ a^2+b^2+c^2 = ab+ac+bc \tag{2}$$ and by the Cauchy-Schwarz inequality $(2)$ holds only for $a=b=c$.
Assuming $k=4$ we have the parametric solution: $$ (a,b,c) = (m^2,n^2,(n+m)^2) \tag{3}$$ so there are plenty of solutions, and even more can be computed by Vieta jumping.
Markov triples are deeply related.
Jack D'Aurizio
- 353,855
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1+1 Also for a fixed $k$ we should get (when solutions exist) a rational parametrization in the style of parametrization of Pythagorean triples. After all, we have a curve of genus zero. – Jyrki Lahtonen Feb 09 '15 at 18:05
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For such equations:
$$(a+b+c)^2=-t^2(ab+ac+bc)$$
$t$ - you can specify any, then decisions can be recorded.
$$a=p^2-2(t^2+t+2)ps+(2t^3+t^2+4t+4)s^2$$
$$b=-p^2+2(t^2-t+2)ps+(2t^3-t^2+4t-4)s^2$$
$$c=t(p^2-(t^2+4)s^2)$$
$p,s$ - integers asked us.
individ
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