Trivial solution when solving in integers Suppose we want to solve $4(x+y)^{2}-3xy-6(x+y)=0$ where $x$ and $y$ are both integers. Why we only get the trivial solution?
 A: The equation can be arranged as a quadratic equation of x:  $4x^2+x(5y-6)+(4y^2-6y)=0$.
The discriminant(D)= $(5y-6)^2-4.4.(4y^2-6y)=36+36y-29y^2$
Now D needs to be  ≥0 as x is real.
$36+36y-29y^2 ≥0  ⇔  29y^2-36y-36  ≤0 $
=>$\frac{18-6\sqrt{38}}{29}≤y≤\frac{18+6\sqrt{38}}{29}$ 
as $(t-a)(t-b)≤0$(where a≤b) => (a≤t  and t≤b) or (b≤t  and t≤a).
Clearly, a≤t≤b as the other option is impossible.
Now, $\frac{18-6\sqrt{38}}{29}>-1$  and  $\frac{18+6\sqrt{38}}{29}<2$ (by observation)
$=>-1 < y < 2$  as y being integer can be 0 or 1 to make x real. 
If $y=0,D=36$, if $y=1,D=43$ .
Also D needs to be perfect square as x is integer.
So, y must be 0 to make x real ad integral.
Observation: the function in the LHS of the equation is symmetric w.r.t. x,y. The calculation & conclusion won't change if we interchange x,y. 
A: We use one of my favourite identities, $4xy=(x+y)^2-(x-y)^2$. Let $x+y=s$ and $x-y=t$.  
To avoid fractions, multiply our equation through by $4$. We get 
$$16(x+y)^2-12xy-24(x+y)=0,$$
which can be rewritten as 
$$13s^2 +3t^2-24s=0.\tag{$1$}$$
For Equation $(1)$ to hold, we need $13s^2-24s \le 0$.  The real zeros of $13s^2-24s$  are at $s=0$ and $s=24/13$, so the only possible integer values of $s$ are $s=0$ and $s=1$.
If $s=0$, then for $(1)$ to hold we need $t=0$, giving $x=y=0$. If $s=1$, we need $3t^2=11$, which is impossible.
