Find positive integer solutions of following equation:
$x^4+2x^7-x^{14}-y^2=7$
This is particular form of equation $x^4+2ax^7-x^{14}-y^2=7$, where $a=1$. Here I argue about $ (a)$, such that this equation can have solutions in $\mathbb Z$.I found if $a=y$, then we can factorize the equation and find some equations which make system of equations which may result in integer solutions.In fact with $a=y$ we have:
$(x^7+x^2-y)(-x^7+x^2+y)=7$
Therefore solutions must satisfy one of these following system of equations:
$\big\{^{x^7+x^2-y=1, 7}_{-x^7+x^2+y=7, 7}$
$\big\{^{x^7+x^2-y=-1, -7}_{-x^7+x^2+y=-7, -1}$
Or we may solve equation for $(y)$ and get $(x, y)=(2, 131), (2, 125), (-2, -131), (-2, -125)$.
I doubt there are more integer solutions when $a≠y$.
We have $$y^2+x^4(x^{10}-2x^3+1)+7=0,$$ which is impossible because $$x^{10}-2x^3+1\geq x^6-2x^3+1=(x^3-1)^2\geq0.$$ So, $$y^2+x^4(x^{10}-2x^3+1)+7\geq y^2+7>0,$$ which says that our equation has no solutions.