hello,we get this here $(x=7\land y=21\land z=39)\lor (x=7\land y=39\land z=21)\lor (x=9\land y=9\land z=43)\lor (x=9\land y=29\land z=33)\lor (x=9\land y=33\land z=29)\lor (x=9\land y=43\land z=9)\lor (x=21\land y=7\land z=39)\lor (x=21\land y=27\land z=29)\lor (x=21\land y=29\land z=27)\lor (x=21\land y=39\land z=7)\lor (x=27\land y=21\land z=29)\lor (x=27\land y=29\land z=21)\lor (x=29\land y=9\land z=33)\lor (x=29\land y=21\land z=27)\lor (x=29\land y=27\land z=21)\lor (x=29\land y=33\land z=9)\lor (x=33\land y=9\land z=29)\lor (x=33\land y=29\land z=9)\lor (x=39\land y=7\land z=21)\lor (x=39\land y=21\land z=7)\lor (x=43\land y=9\land z=9)$ Sonnhard.

But how do we get there?

At least one is divisible by 3 and all are odd so he checked all the cases with $x$ odd I think

andrei.pantea wrote: At least one is divisible by 3 and all are odd so he checked all the cases with $x$ odd I think That's ridiculous bash though.

for x in range(45): for y in range(45): for z in range(45): if x**2+y**2+z**2==2011: print(x,y,z)for x in range(45): for y in range(45): for z in range(45): if x**2+y**2+z**2==2011: print(x,y,z)RunResetPop Out / Is this Cono Sur Math Olympiad?

Ankoganit wrote: for x in range(45): for y in range(45): for z in range(45): if x**2+y**2+z**2==2011: print(x,y,z)for x in range(45): for y in range(45): for z in range(45): if x**2+y**2+z**2==2011: print(x,y,z)RunResetPop Out / Is this Cono Sur Math Olympiad? Dude , what was the output of this program ?

7 21 39 7 39 21 9 9 43 9 29 33 9 33 29 9 43 9 21 7 39 21 27 29 21 29 27 21 39 7 27 21 29 27 29 21 29 9 33 29 21 27 29 27 21 29 33 9 33 9 29 33 29 9 39 7 21 39 21 7 43 9 9 This is the output

Swag00 wrote: Dude , what was the output of this program ? Just so you'd know, you can click the 'Run' button to see the output.

Ankoganit wrote: Swag00 wrote: Dude , what was the output of this program ? Just so you'd know, you can click the 'Run' button to see the output. but would you really do that if you were to give the cono sur olympiad ??

Swag00 wrote: Ankoganit wrote: for x in range(45): for y in range(45): for z in range(45): if x**2+y**2+z**2==2011: print(x,y,z)for x in range(45): for y in range(45): for z in range(45): if x**2+y**2+z**2==2011: print(x,y,z)RunResetPop Out / Is this Cono Sur Math Olympiad? Dude , what was the output of this program ? what's that? I wish I could do like that

That's Python Programming. It lets you fly.

Ankoganit wrote: Swag00 wrote: Dude , what was the output of this program ? Just so you'd know, you can click the 'Run' button to see the output. Oops, that was stupid of me

Ankoganit wrote: That's Python Programming. It lets you fly. Only in cyber olympiads I guess ...!

Looks like Ankoganit also preparing for iOi this year

See also here: https://artofproblemsolving.com/community/c6h1528721p9181210 https://artofproblemsolving.com/community/c6h586859p3472767 http://artofproblemsolving.com/community/c6h304361p2390397

Notice that the sum of two squares cannot be $3\pmod 4$, so all of $x,y,z$ must be odd. Since $2011\equiv 1\pmod 3$, two of $x,y,z$ are $0\pmod 3$, and the other one is not a multiple of $3$. WLOG that $3\mid x,y$. This implies $z^2 \equiv 4\pmod 9$, so $z\equiv \{2,7\}\pmod 9$. Since $z$ is odd and $z<43$, we have $z\in \{7,11,25,29,43\}$. Notice that $\nu_7(2011 - 11^2) $ and $\nu_7(2011 - 25^2)$ is $1$, so by the Sum of Two Squares Theorem, $z$ cannot be $11$ or $25$. Let $x = 3a$ and $y = 3b$. We have $a^2 + b^2 = \frac{2011 - z^2}{9}$. If $z=7$, then $a^2 + b^2 = 218$, which can be checked to give $(a,b)$ is a permutation of $(7,13)$. If $z=29$, then $a^2 + b^2 = 130$, which can be checked to give that $(a,b)$ is a permutation of $(3,11)$ or $(7,9)$. If $z=43$, then $a^2 + b^2 = 18$, which can be checked to give that $(a,b) = (3,3)$. Thus, we conclude that the solutions are of the form $(21,39,7), (9,33,29), (21,27,29), (9,9,43)$, and permutations, which obviously work.