**Problem #1**: (360 Problems for Mathematical Contests - 2.1.19) Solve in the nonnegative integers the equation

$ x^2+8y^2+6xy-3x-6y = 3 $.

**Solution**: Write the equation as a quadratic in $ x $, that is

$ x^2+(6y-3)x+(8y^2-6y-3) = 0 $.

The discriminant of this quadratic is $ (6y-3)^2-4(8y^2-6y-3) = (2y-3)^2+12 $. Since we are only looking for integral solutions, $ (2y-3)^2+12 $ has to be the square of a rational, but since it is integral, it must be the square of an integer. So $ (2y-3)^2+12 = a^2 $, but the difference between two squares must be the sum of consecutive odd integers and the only way $ 12 $ can be written like that is $ 12 = 5+7 $. So $ 2y-3 = 2 $ (since the difference between $ 2^2 $ and $ 3^2 $ is $ 5 $) is the only possible value, but that has no solutions. Hence our original equation has no solutions. QED.

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**Problem #2**: (360 Problems for Mathematical Contests - 2.1.24) Solve in the nonnegative integers the equation

$ x+y+z+xyz = xy+yz+zx+2 $.

**Solution**: Factor the given equation to get $ (x-1)(y-1)(z-1) = 1 $. So we have either $ x-1 = y-1 = z-1 = 1 $ or two of them negative and the other positive. This gives us the solutions

$ (2,2,2); (2,0,0); (0,2,0); (0,0,2) $. QED.

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Comment: Two relatively simple diophantine equations (equations in which we are only looking for integral solutions), but they introduce important techniques. Rewriting as a quadratic and looking at the determinant is often a good way to go, especially if you can show the determinant is less than zero or something. Factoring is of course an extremely common method but a lot of the time the right way to factor may be difficult to find.

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Practice Problem: Try finding the factoring solution to the first problem (it exists).

We factor the first equation into (x+4y-3)(x+2y)=3 ...but x+2y is nonnegative, so it must be 3 or 1. Casework from here and we get no solutions.

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