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I am trying to find integer solutions to $x + y + z = 3$, $xy + yz + xz + 2xyz = 2017$. Any help is appreciated.

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    this has been asked before but for some reason I can't find it. – 2017-01-20
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    Current contest question. Don't answer. – 2017-01-20
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    I thought you said it ended today or something – 2017-01-20
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    It says to submit solutions by Friday Jan 20th. So I'm not completely sure. – 2017-01-20
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    try $$x=47,y=-1,z=-43$$ – 2017-01-20
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    I doubt they are asking such a question for personal benefit, when the account has asked 5 contest questions in the past 20 hours. Why are people giving solutions? – 2017-01-20
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    @Kaynex How can one remove an answer and have it put back at a later date? – 2017-01-20
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    @CarlSchildkraut: You can delete an answer. When I do it stays in the system and I (and other users with lots of rep) still see it. You can then undelete it when you want, but you have to remember to do so. – 2017-01-20

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let $xyz = a$, then the polynom with the roots (x, y, z) is

$t^3 - 3t^2 + (2017-2*a)t - a$

$t^3 - 3t^2 + (2017-2*a)t - a = 0 $ => $t^3 - 3t^2 + 2017t = a(2t+1)$

Try to find the integer solutions. t = 0 => a=0;

else $t$ is coptime to $2t+1$ => $2t+1 | t^2 - 3t + 2017$ => $2t+1 | 2t^2 - 6t + 4034$ => $2t+1|7t - 4034$ => $2t+1|14t-8068$ => $2t+1|8075$

So, we can find small set of possible values of t, from it get small set of possible values of a, and solve the some systems like x+y+z = 3, xy+xz +zy = 2017-2a, xyz = a