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- Feb 14, 2012

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Solve the system $x^4+y^4=8^2$ and $x-y=2$.

Attempt:

$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$

$(x+y)^4=(x^4+y^4)+4xy(x^2+y^2)+6x^2y^2$

$(x+y)^4=8^2+4xy((x-y)^2+2xy)+6x^2y^2$

$(x+y)^4=64+4xy((2)^2+2xy)+6x^2y^2$

$(x+y)^4=64+16xy+14x^2y^2$

Up to this point, I know that I need to form another equation (with $(x+y)^4$ the subject) that has only terms of xy...

$x-y=2$

$(x-y)^2=2^2$

$x^2+y^2-2xy=4$

$(x+y)^2-4xy=4$

$(x+y)^4=(4+4xy)^2=16+32xy+16x^2y^2$

$\therefore 64+16xy+14x^2y^2=16+32xy+16x^2y^2$

$x^2y^2+8xy-24=0$

And by using the quadratic formula to solve for xy, I get:

$\displaystyle xy=\frac{-8 \pm \sqrt {8^2-4(-24)(1)}}{2}=\frac{-8 \pm 4 \sqrt{10}}{2}=-4 \pm 2\sqrt{10}$

I then solve the values for y by multiplying each and every term of the equation $x-y=2$ by y, and obtained another quadratic equation in terms of y and from there, I managed to find all 4 pairs of answers accordingly.

Now, I was wondering if there are any other methods to tackle this problem effectively.

Thanks.