Didn't bother to answer, since it seemed obvious that this was a sphere with imaginary radius: (x + 3/2)^2 + (y + 3/2)^2 + (z + 3/2)^2 = -7/4. R. On Fri, 9 Oct 2009, Gareth McCaughan wrote:

On Thursday 08 October 2009 21:10:02 Dan Asimov wrote:

...

Find all rational solutions of

x^2 + y^2 + z^2 + 3(x + y + z) + 5 = 0.

Since no one else has bitten yet, here's a solution after some spoiler space.

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There are no rational solutions.

Let x=a/n etc. with minimal n. Then we're solving x^2 + y^2 + z^2 + 3n(x+y+z) + 5n^2 = 0; not all of x,y,z,n are even.

If 4|n then 4|x^2+y^2+z^2, whence x,y,z are all even, contradiction.

If n is odd then x^2+3nx etc. are all even, so the LHS is odd, hence nonzero; contradiction.

So n=2m, m odd. Complete the squares, writing u=x+3m etc.; we get u^2+v^2+w^2 = 7m^2 but the RHS is 7 (mod 8) which the LHS cannot be.

(We could replace the last three paragraphs with "Now just try all possibilities for x,y,z,n mod 4 and observe that none of them works.")