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I think this problem is from an old math olympiad, but not sure. The problem is:

Find all integer solutions to

\begin{equation}x^2+y^2+z^2=2^{11}.\end{equation}

I know that this can be describe as "find all coordinates with integer values inside (or on) a sphere with radius $\sqrt{2^{11}}=\sqrt{2048}\approx45.25$. However, I don't know how to arrive at the solution. A solution or any hints would be highly appreciated. I also know that people have studied geometry/integer problems like these, for instance Gauss circle problem but that's 2D (circles) not 3D (spheres).

I have found solutions using some online diophantine equation solver to be the following (I added the plus-minuses):

\begin{align} x1 &= 0, & y1 &= \pm32, & z1 &= \pm32 \\ x2 &= \pm32, & y2& = 0, & z2 &= \pm32 \\ x3 &= \pm32, & y3& = \pm32, & z3 &= 0 \end{align}

Hope you understand me!

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  • $\begingroup$ I just made a spreadsheet with $x$ and $y$ running from $0$ to $45$ and checked by eye. I did not find any more. $\endgroup$ – Ross Millikan Jul 4 '19 at 23:25
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    $\begingroup$ There will only be $12$ solutions if $r^2 = 2^{2n+1}$, with $n$ being an integer. The $12$ solutions will be $x = 0, y = \pm 2^{n}, z = \pm 2^{n}$, with each of $x, y, z$ being equal to $0$. There will only be $6$ solutions if $r^2 = 2^{2n}$. See oeis.org/A005875 for the explanation. $\endgroup$ – Varun Vejalla Jul 4 '19 at 23:34
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    $\begingroup$ Let's @automatically generate an answer. $\endgroup$ – Oscar Lanzi Jul 4 '19 at 23:40
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    $\begingroup$ Look mod 4, a square of a number is either $1$ or $0$ mod $4$ which implies that all $x, y, z$ are even, continue until you get $x^2+y^2+z^2=2$ $\endgroup$ – kingW3 Jul 4 '19 at 23:42
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    $\begingroup$ :) I wish StackExchange worked like that, @OscarLanzi. $\endgroup$ – Varun Vejalla Jul 4 '19 at 23:50
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if $$ x^2 + y^2 + z^2 \equiv 0 \pmod 4 \; , \; $$ then all three of $x,y,z$ are even. In your case, this observation is repeated a few times. Put another way, if $$ x^2 + y^2 + z^2 = 4n \; , \; $$ then $$ \left( \frac{x}{2 } \right)^2 +\left( \frac{y}{2 } \right)^2 +\left( \frac{z}{2 } \right)^2 = n$$ and so on

Since $$ 2^{11} = 4^5 \cdot 2,$$ you wind up solving $u^2 + v^2 + w^2 = 2$ (twelve answers) and multiplying all three variables by $32$

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  • $\begingroup$ Tea? C'mon man, wait for tea to get ready then answer while savoring! :-) $\endgroup$ – Oscar Lanzi Jul 4 '19 at 23:56
  • $\begingroup$ @OscarLanzi the bad news is that the place that had the Pu-Er I like best is closing down; it was very convenient. Expensive, of course, I have that type a few times a month, most days it is inexpensive teabags. $\endgroup$ – Will Jagy Jul 4 '19 at 23:59

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