find all integer x and y that fulfils
231x^2 +528y^2=16896
Notice that if (x,y) is a solution then $\displaystyle (\pm x,\pm y)$ is a solution. This means we can only focus on positive values x and y. If $\displaystyle y\geq 6$ then $\displaystyle 528y^2 \geq 19008 > 16896$, and so no solution exists. Thus, you need to see for which $\displaystyle 0\leq y\leq 5$ we have $\displaystyle \sqrt{ \frac{16896 - 528y^2}{231} }$ is an integer.
Thanks for this, but here is another theory, that some made, but i dont get some of the steps he makes...
Let x^2=a, y^2=b
231a+528b=16896
16896=32*528 that will give (a,b)=(0,32) {why does he take 32*528?}
a=0+528n {how does he get this?}
b=32-231n {how does he get this?}
231a+528b = 231(0+528n)+528(32-231n) = 231*528n+528*32-528*231n
x^2=528n {how does he get this from the equation above?}
y^2=32-231n {how does he get this from the equation above?}
n<0 gives 528n<0, n>0 gives 32-231n<0
x^2 =0
y^2=32
Is then the answer x=0,y=sqrt(32)=5,6 {but that aint a Z (integer?)}??