1. ## Algebra help

find all integer x and y that fulfils

231x^2 +528y^2=16896

Originally Posted by johnnyboiy
find all integer x and y that fulfils

231x^2 +528y^2=16896
This is an eqn of an ellipse. Please see attached graph. The integers that satisfy this eqn are shown RED points in graph.

3. Originally Posted by johnnyboiy
find all integer x and y that fulfils

231x^2 +528y^2=16896
Notice that if (x,y) is a solution then $(\pm x,\pm y)$ is a solution. This means we can only focus on positive values x and y. If $y\geq 6$ then $528y^2 \geq 19008 > 16896$, and so no solution exists. Thus, you need to see for which $0\leq y\leq 5$ we have $\sqrt{ \frac{16896 - 528y^2}{231} }$ is an integer.

4. Originally Posted by ThePerfectHacker
Notice that if (x,y) is a solution then $(\pm x,\pm y)$ is a solution. This means we can only focus on positive values x and y. If $y\geq 6$ then $528y^2 \geq 19008 > 16896$, and so no solution exists. Thus, you need to see for which $0\leq y\leq 5$ we have $\sqrt{ \frac{16896 - 528y^2}{231} }$ 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?)}??