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How can we generate all integer solutions of the equation

$$(qr+rp+pq)(x^2+y^2+z^2) = (p^2+q^2+r^2)(yz+zx+xy),$$

given that $p,q,r$ are integers?

Clearly if any one of $(x,y,z), (x,z,y), (y,z,x), (y,x,z), (z,x,y), (z,y,x)$, is a solution, so are the others, so let's assume $x \leq y \leq z$. Also assume that $x,y,z$ are relatively prime, since then $(kx,ky,kz)$ is a solution for every integer $k$.

Example: if $(p,q,r) = (0,1,2)$, the solutions include $(2,4,15), (3,14,40), (4,30,77), (5,6,28), (5,52,126)$.

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    $\begingroup$ So there are "obvious" solutions when both $xy+yz+zx=pq+qr+rs=0$ or $x+y+z=p+q+r=0$ . Otherwise note that you can rewrite the above as being $\sum (\frac{x}{x+y+z})^2=\sum (\frac{p}{p+q+r})^2$. Now you can think about this as looking for rational points on the intersection of the sphere $a^2+b^2+c^2=1$ with $a+b+c=1$. $\endgroup$
    – Vlad Matei
    Commented Dec 16, 2019 at 18:34
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    $\begingroup$ To finish off you can write these solutions in parametric form. Namely note that this intersection is an ellipse so it should have a nice form. To do it substitute $c=1-a-b$. This give $a^2+b^2+ab-a-b=0$ and the equation of the underlying ellipse is $3X^2+Y^2=4$ where $X=2a+b-1$ and $Y=3b-1$. $\endgroup$
    – Vlad Matei
    Commented Dec 16, 2019 at 18:44
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    $\begingroup$ To parametrize the ellipse note that you have the obvious rational point $(0,2)$ and any other rational point on the ellipse will determine a line with this rational point of rational slope. Thus we can write all the solutions as $X=t(Y-2)$ for $t\in\mathbb{Q}$. This gives $Y=\cfrac{6t^2-2}{3t^2+1}$ and $X=\cfrac{4t}{3t^2+1}$. $\endgroup$
    – Vlad Matei
    Commented Dec 16, 2019 at 18:53

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$(qr+rp+pq)(x^2+y^2+z^2)=(p^2+q^2+r^2)(yz+zx+xy)$

$m,n,p,q,r$ are arbitrary.
Substitute $ x=t+p, y=mt+q, z=nt+r$ to above equation, then we get $$\scriptsize{t = \frac{(r^2q+2qrp-p^3-q^3-q^2p+r^2p-p^2q)n+(-rp^2+2qrp-r^3-p^3+q^2p+q^2r-r^2p)m+rp^2+p^2q-r^2q-q^2r-r^3-q^3+2qrp}{(-qr-pq-rp)n^2+((q^2+p^2+r^2)m+q^2+p^2+r^2)n+(-qr-pq-rp)m^2+(q^2+p^2+r^2)m-qr-pq-rp}.}$$ Thus, we get a parametric solution below.

$x = (p^2q+qrp+rp^2)m^2+((-r^2p-p^3-pq^2)n+r^3-q^2r-2qrp-2pq^2+rp^2)m+(p^2q+qrp+rp^2)n^2+(q^3-r^2q-2qrp-2r^2p+p^2q)n-qrp+r^2q+q^2r+q^3+r^3.$

$y = (r^2p+p^3+r^3-qrp+rp^2)m^2+((-r^2p-2qrp+pq^2+p^3-2r^2q)n-2qrp-rp^2+q^2r-2p^2q+r^3)m+(pq^2+q^2r+qrp)n^2+(-r^2q-q^3-p^2q)n+pq^2+q^2r+qrp.$

$z = (r^2p+r^2q+qrp)m^2+((-pq^2+r^2p+p^3-2qrp-2q^2r)n-rp^2-r^3-q^2r)m+(pq^2+p^3+q^3-qrp+p^2q)n^2+(q^3-p^2q-2rp^2-2qrp+r^2q)n+r^2p+r^2q+qrp.$

Example for $(p,q,r)=(0,1,2).$
$(x,y,z)>1, gcd(x,y,z)=1$
$$\frac{(q^2+r^2+p^2)}{(qr+pq+rp)}=\frac{(x^2+y^2+z^2)}{(yz+zx+xy)}=5/2. $$ $[ m, n , x , y, z ]=[ 0, 0, 15, 2, 4], [ 0, 3, 6, 5, 28],[ 0, 4, 3, 14, 40],[ 1, 2, 15, 2, 4],[ 1, 5, 6, 5, 28],[ 2, 4, 15, 2, 4],[ 3, 0, 33, 104, 10],[ 3, 1, 30, 77, 4],[ 4, 0, 39, 170, 28],[ 4, 2, 33, 104, 10],[ 4, 3, 30, 77, 4],[ 5, 1, 42, 209, 40],[ 5, 2, 39, 170, 28],[ 5, 4, 33, 104, 10],[ 5, 5, 30, 77, 4].$

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To illustrate finding all primitive integer solutions, let us take $(p,q,r) = (106,92,1)$ In this case, all $x,y,z$ will be positive, and we may simply put them in decreasing order. There are three distinct parametrizations, each component a quadratic form in (coprime) integer variables $u,v,$ just like Pythagorean Triples...

$$ 106 u^2 + 197 uv + 92 v^2 \; , \; \; 92 u^2 -13 uv + v^2 \; , \; \; u^2 + 15 uv + 106 v^2 \; . $$

$$ 124 u^2 + 179 uv + 65 v^2 \; , \; \; 65 u^2 -49 uv + 10 v^2 \; , \; \; 10 u^2 + 69 uv + 124 v^2 \; . $$

$$ 130 u^2 + 159 uv + 49 v^2 \; , \; \; 49 u^2 -61 uv + 20 v^2 \; , \; \; 20 u^2 + 101 uv + 130 v^2 \; . $$

Take a pair $(u,v)$ and produce the triples given by each recipe above. If any triple comes out nonprimitive (some gcd > 1) just discard that. The primitive version will be constructed by one of the other recipes. Oh, and put in decreasing order. Call that triple $(x,y,z).$ Below, the triple in brackets is the first quadratic form in that recipe, such as <106,197,92>.

    x          y         z       first quad. form     u  v
   106        92         1      < 106, 197, 92 >      1  0    
   124        65        10      < 124, 179, 65 >      1  0    
   130        49        20      < 130, 159, 49 >      1  0    
   338       251         8      < 130, 159, 49 >      1  1    
   368       203        26      < 124, 179, 65 >      1  1    
   395       122        80      < 106, 197, 92 >      1  1    
   887       412        94      < 130, 159, 49 >      2  1    
   919       302       172      < 124, 179, 65 >      2  1    
  1333      1246         8      < 124, 179, 65 >      1  3    
  1493      1048        46      < 130, 159, 49 >      1  3    
  1525      1000        62      < 106, 197, 92 >      1  3    
  1637       790       160      < 106, 197, 92 >      3  1    
  1696       613       278      < 130, 159, 49 >      3  1    
  1718       448       421      < 124, 179, 65 >      3  1    
  1915      1856        10      < 130, 159, 49 >      2  3    
  2155      1570        56      < 124, 179, 65 >      2  3    
  2270      1880        29      < 124, 179, 65 >      1  4    
  2320      1306       155      < 130, 159, 49 >      3  2    
  2434      1048       299      < 106, 197, 92 >      2  3    
  2450      1000       331      < 124, 179, 65 >      3  2    
  2504      1550       125      < 130, 159, 49 >      1  4    
  2504       754       523      < 106, 197, 92 >      3  2
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  • $\begingroup$ for additional information on this approach, including what "ratios" $\frac{yz+zx+xy}{x^2 + y^2 + z^2}$ are possible, see mathoverflow.net/questions/225781/… $\endgroup$
    – Will Jagy
    Commented Dec 17, 2019 at 22:42
  • $\begingroup$ If the "ratio" is below one half, there are integer triples with some positive and some negative entries. When the ratio is above one half, such as the $50/99$ I display in my answer, all $x,y,z$ will be positive. $\endgroup$
    – Will Jagy
    Commented Dec 18, 2019 at 1:35

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