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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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    $\begingroup$ @amWhy but are equivalent. $\endgroup$
    – Carol
    Feb 9, 2015 at 15:21
  • $\begingroup$ Yes, many more solutions: $(a,a,a)$, $(0,b,b)$, or $(a,-a,c)$ with $a\mid c$, etc. $\endgroup$ Feb 9, 2015 at 15:33
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    $\begingroup$ It might be interesting to look for cases with $a,b,c>0$ and $\gcd(a,b,c)=1$ $\endgroup$
    – Joffan
    Feb 9, 2015 at 15:38
  • $\begingroup$ As for the equivalent statement, use $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$. $\endgroup$ Feb 9, 2015 at 15:43

4 Answers 4

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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 

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

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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  • $\begingroup$ Could you please provide a reference for why $B-1$ and $B+2$ must be of the form $x^2+3y^2$? $\endgroup$ Oct 19, 2015 at 0:52
  • $\begingroup$ @Batominovski, it is my result, but is equivalent to the famous result of Legendre on indefinite ternary forms, math.stackexchange.com/questions/27471/… or pages 80-82 in Rational Quadratic Forms by Cassels, 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. $\endgroup$
    – Will Jagy
    Oct 19, 2015 at 1:04
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    $\begingroup$ @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. $\endgroup$
    – Will Jagy
    Oct 20, 2015 at 18:36
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    $\begingroup$ Thanks a bunch. I received the file. $\endgroup$ Oct 21, 2015 at 13:00
  • $\begingroup$ @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 $\endgroup$
    – user45220
    Oct 25, 2015 at 4:47
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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}
 ]
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  • $\begingroup$ Hello, thank you for this. Can I ask which code did you use? Thanks! $\endgroup$
    – user45220
    Feb 9, 2015 at 16:50
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    $\begingroup$ I have included the code in the answer. $\endgroup$ Feb 9, 2015 at 16:55
  • $\begingroup$ 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. $\endgroup$
    – user45220
    Feb 9, 2015 at 18:08
  • $\begingroup$ 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. $\endgroup$ Feb 9, 2015 at 18:12
  • $\begingroup$ 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). $\endgroup$
    – user45220
    Feb 9, 2015 at 18:14
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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.

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    $\begingroup$ +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. $\endgroup$ Feb 9, 2015 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.

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