Show that equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has infinitely many solutions in integers $x,y,z$

Question

While solving some old INMO problems I found that one, and I am completely stuck at it. The problem is:

Show that the equation $x^2+y^2+z^2=(x-y)(y-z)(z-x)$ has infinitely many solutions in integers $x,y,z$.

I shall be thankful if you can provide me any hints or suggestions. Thanks.

Answer 1

In this type of problems when you have to prove that there are infinitely many solutions it's convenient to look for possible values in P.A. So, suppose that $z-y=y-x=k$, then $z-x=2k$ and our equation becomes $$x^2+(x+k)^2+(x+2k)^2=2k^3.$$

So, after a little calculus we get the equation $3x^2+(6k)x+(5k^2-2k^3)=0$. Using the formula for the quadratic equation gives us $$x=\frac{-6k\pm \sqrt{(6k)^2-12(5k^2-2k^3)}}{6}=-k\pm \frac{k\sqrt{6(k-1)}}{3}.$$

Now, since we want $x\in \mathbb{Z}$ we need that $6(k-1)=u^2$ for some $u\in \mathbb{Z}$. Then $6\mid u^2$, so $6\mid u$ (why?). Set $u=6t$, thus we get $k=6t^2+1$ and hence we get $x=-(6t^2+1)\pm (6t^2+1)(2t)$. If we take the plus sign we get $x=12t^3-6t^2+2t-1$. Now, since $y=x+k$ and $z=x+2k$, replacing $x$ gives us $y=12t^3+2t$ and $z=12t^3+6t^2+2t+1$.

Finally, you can check that $$(12t^3-6t^2+2t-1)^2+(12t^3+2t)^2+(12t^3+6t^2+2t+1)^2=2(6t^2+1)^3.$$

Hence, the equation has infintely many integer solutions.

Answer 2

Instead of solving

$$x^2+y^2+z^2 = (x-y)(y-z)(z-x) \tag{*1}$$

one look at a simpler problem first. Let's say you have found a non-trivial integer solution for $$(u-v)(v-w)(w-u)|u^2+v^2+w^2\tag{*2}$$

one can set $\lambda$ to the integer $\displaystyle\;\frac{u^2+v^2+w^2}{(u-v)(v-w)(w-u)}\;$ and $(x,y,z) = (\lambda u,\lambda v, \lambda w)$ will give us a solution for $(*1)$.

It turns out it isn't that hard to find solutions for $(*2)$. One just take any two non-zero integers $p, q$, set $(u,v,w) = (v-p,v,v+q)$ and looks for expression of $v$ which makes $$pq(p+q) \;|\; (v-p)^2 + v^2 + (v+q)^2 = 3v^2 -2(p-q)v + (p^2+q^2)$$

For $p = q = 1$, we find

$$(u,v,w) = (2t-1,2t,2t+1) \quad\implies\quad \lambda = \frac{(2t-1)^2 + (2t)^2 + (2t+1)^2}{(-1)(-1)(2)} = 6t^2+1 $$ This will give us a parametrized family of solution of $(*1)$ $$(x,y,z) = ((2t-1)(6t^2+1), 2t(6t^2+1), (2t+1)(6t^2+1))$$

This demonstrate the original equation $(*1)$ does have infinitely many solutions.

Others solutions can be constructed in similar manner. For example, take $p = 1, q = 2$, we find $v = 6t-1$ give us another family of solutions: $$(x,y,z) = ((6t-2)(18t^2-4t+1),(6t-1)(18t^2-4t+1),(6t+1)(18t^2-4t+1))$$ The more interesting question is whether there are some ways to systematically exhaust all solutions of $(*1)$ and I've no idea on that.

Answer 3

These are all solutions with $$ -2 \leq x < y < z < 1700. $$ For some reason there are just a few solutions with some positive entries and some negative, that probably has a short proof. Anyway, once all three are positive and distinct, it is easy to see that we can demand $x<y<z.$

   x^2 + y^2 + z^2   x      y      z   x^2 + y^2 + z^2 FACTORED
          6         -2     -1      1              6 =  2 3     gcd(x,y,z) 1
          2         -1      0      1              2 =  2     gcd(x,y,z) 1
          6         -1      1      2              6 =  2 3     gcd(x,y,z) 1
        686          7     14     21            686 =  2 7^3     gcd(x,y,z) 7
      20250         55     85    100          20250 =  2 3^4 5^3     gcd(x,y,z) 5
      20250         60     75    105          20250 =  2 3^4 5^3     gcd(x,y,z) 15
      31250         75    100    125          31250 =  2 5^6     gcd(x,y,z) 25
      73002        115    161    184          73002 =  2 3 23^3     gcd(x,y,z) 23
     140250        175    205    260         140250 =  2 3 5^3 11 17     gcd(x,y,z) 5
     156750        155    250    265         156750 =  2 3 5^3 11 19     gcd(x,y,z) 5
     332750        275    330    385         332750 =  2 5^3 11^3     gcd(x,y,z) 55
     384846        259    266    497         384846 =  2 3 7^3 11 17     gcd(x,y,z) 7
    1647750        650    715    845        1647750 =  2 3 5^3 13^3     gcd(x,y,z) 65
    1825346        679    776    873        1825346 =  2 97^3     gcd(x,y,z) 97
    2409066        736    943    989        2409066 =  2 3^2 11 23^3     gcd(x,y,z) 23
    3188646        891   1053   1134        3188646 =  2 3^13     gcd(x,y,z) 81
    3188646        918    999   1161        3188646 =  2 3^13     gcd(x,y,z) 27
    2676750        775    790   1205        2676750 =  2 3 5^3 43 83     gcd(x,y,z) 5
    3668250        860   1195   1225        3668250 =  2 3 5^3 67 73     gcd(x,y,z) 5
    4276866       1063   1184   1321        4276866 =  2 3 11^2 43 137     gcd(x,y,z) 1
    5343750       1150   1175   1625        5343750 =  2 3^2 5^6 19     gcd(x,y,z) 25
    6885902       1359   1510   1661        6885902 =  2 151^3     gcd(x,y,z) 151
  x^2 + y^2 + z^2   x      y      z    x^2 + y^2 + z^2 FACTORED