Solve the Diophantine equation $2xy + 3y^{2} = 24$

Question

Solve the equation $$2xy + 3y^{2} = 24, \qquad x, y \in \mathbb{Z}$$

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I tried by stating that $x$ and $y$ cannot both be odd and that y cannot be odd. So either $x$ and $y$ are both even or $x$ is even and $y$ is odd. Then I was trying take the residues modulo $3$ and modulo $8$ but could not arrive at a conclusion. Please help me solve this further.

Answer 1

As you have rightly pointed out, y can never be even.

Now note that the equation can be written as $y∗(2x+3y)=24$. Thus $y$ must be an integer factor of $24$.

You will thus get the following results : - $y=2,4,6,8,12$.

So this will give you $(x,y)=\{(3,2), (-3, -2),(−3,4),(3,-4),(−7,6),(7,-6),(−17,12),(17,-12)\}$

Only 8 does not give any solution. So here is the solution you asked for.

Answer 2

Equation can be written as $y*(2x+3y)=24$. Thus $y$ must be an integer factor of $24$. This should get you to a manageable range of possibilities.