How can I complete this proof by contradiction? 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?
 A: Yes, a proof by contradiction can be given. Suppose you have found a pair $(x,y)$ satisfying the equation.
Since $14$ and $2x^2$ are even, also $5y^2$ must be even as well. Therefore $y$ is even and so $y=2z$ for some integer $z$.
This implies $2x^2+20z^2=14$ that simplifies to $x^2+10z^2=7$. But $10z^2>7$ if $z\ne0$, so we must have $z=0$ and so $x^2=7$, a contradiction.

About your argument, there's a glitch. Since $14$ is even, two integers that sum to $14$ are either both even or both odd. However, since $2x^2$ is clearly even, you can conclude (as I did above) that also $5y^2$ must be even.
A: HINT: You’ll have a very hard time proving it this way. I recommend a different approach altogether. Note that $x^2$ and $y^2$ are non-negative, so $2x^2$ and $5y^2$ are at most $14$. Thus, if $y$ is an integer, then $y$ must be one of three integers; what are they? And what do they force $2x^2$ to be?
Added: And you can use your observation that $y$ must be even to reduce the possibilities still further.
A: Roundabout proof (not the one you should use :-) ):
$x^2 \equiv \{0,1,4,5,6,9\}\mod 10$
$\Rightarrow 2x^2 \equiv \{0,2,8,10,12,18\}\mod 20$
$y^2 \equiv \{0,1\}\mod 4$
$\Rightarrow 5y^2 \equiv \{0,5\}\mod 20$
$\Rightarrow 2x^2+5y^2 \equiv \{0,2,3,5,7,8,10,12,13,15,17,18\}\mod 20$
$\Rightarrow 2x^2+5y^2 \not\equiv 14\mod 20$
$\Rightarrow 2x^2+5y^2 \neq 14$
A: If you want to make your proof work, first note that $y = 0$ is not an option because then you would have $x^2 = 7$. As you noted, $5y^2$ must be even, so $y$ must be even, hence $y^2$ is at least 4. This gives a sum which is too big. (at least 20 when your target is 14)
