i'm wrong? let y = 1. the pell equation 2001z² - 2000x² = 1 have the solution (1,1), so, have infinite solutions suficiently bigger!

You not see condition of $x,y,z$. Try again

e.lopes seems to be right though... all conditions seem to be satisfied

N.T.TUAN wrote: You not see condition of $x,y,z$. Try again hm.. i think that i'm right, my friend!

Maybe But problem said that find ALL

All non-trivial ($z\not =0$) solutions give by: \[\frac{x}{z}=\frac{r^{2}-2r-2000}{r^{2}+2000}, \ \frac{y}{z}=\frac{2000-4000r-r^{2}}{2000+r^{2}}, \ r\in Q.\]

Therefore answer of problem original is...?

All positive integer solutions give by $x=s|a^{2}-2ab-2000b^{2}|, \ y=s|2000b^{2}-4000ab-a^{2}|,\ z=s|a^{2}+2000b^{2}|,$ were $a,b\in Z, \ 2000*2001s\in N.$

It is easy to check that the below identity is true: $n(m^{2}+2m-n)^{2}+(m^{2}-2mn+n)^{2}= (n+1)(m^{2}+n)^{2}$. Setting $n=2000$ and $m=9000$ will give $x=m^{2}+2m-n>m^{2}+n=z$, $z=m^{2}+n=(45 \times 2000)^{2}+2000 > 2000^{2}\times 2025 > 1999 \times 2000 \times 2001$. And $y= m^{2}-2mn+n = (45 \times 2000)^{2}-2 \times 45 \times 2000+2000 < 2000^{3}-2000= 1999 \times 2000 \times 2001$. So we have triples that satisfy the given condition.