This problem is from Discrete Mathematics and its Applications:
Prove that there are no solutions in integers $x$ and $y$ to the equation $2x^2 + 5y^2 = 14$.
I am trying to use proof by contradiction, which is described by the book as
Suppose we want to prove that a statement $p$ is true. Furthermore, suppose that we can find a contradiction $q$ such that $\lnot p \implies q$ is true. Because $q$ is false, but $\lnot p \implies q$ is true, we can conclude that $\lnot p$ is false, which means that $p$ is true. How can we find a contradiction $q$ that might help us prove that $p$ is true in this way?
Because the statement $r \land \lnot r$ is a contradiction whenever $r$ is a proposition, we can prove that $p$ is true if we can show that $\lnot p \implies (r \land \lnot r)$ is true for some proposition $r$. Proofs of this type are called proofs by contradiction. Because a proof by contradiction does not prove a result directly, it is another type of indirect proof.
Here is my work/thought process:
My initial proposition, $p$, is that there are no solutions in integers $x$ and $y$ to the equation $2x^2 + 5y^2 = 14$. I know that by proof by contradiction, I have to assume that the proposition isn't true, $\lnot p$, meaning there is a solution to $x$ and $y$ in the equation and show that assuming this leads to a contradiction (something that always evaluates to false, no matter the input values).
First, I recognized that for the sum be even, $14$, the two components, $2x^2$ and $5y^2$ have to be even as well.
I am able to show that $2x^2$ is even from the definition of even, that is, there is some integer $k$ such that $2x^2 = 2k$. $k$ would be $x^2$. However I have a hard time showing that $5y^2$ cannot be even. I first tried the same definition, meaning $k = (5/2) y^2$ but this wouldn't be an integer. However it is possible for $5y^2$ to be even, say $y = 10$. Am I going about this the right way? Is the even + even justification appropriate for this situation?