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Old 2008-05-01, 13:51   #2
Zeta-Flux
 
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May 2003

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Here is a partial solution.

It is easy to reduce the equation to xy+xz+yz=a.

It is also easy to see that 2 cannot divide a.

Let x=-(4k^2+2k+a), y=8k^2+2k+2a, z=8k^2+6k+2a+1.

We compute that xy+xz+yz=a for any k.

On the other hand, taking k==-1 mod a, we see that x==-2 mod a, y==6 mod a, and z==3 mod a.

Quote:
Note that as k increases (for large enough k), x,y,z also increase in magnitude.
Thus, this solution works for any a not divisible by 2 or 3.
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