Challenge Find the sum of squares

anemone

Paris la ville de l'amour
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#1
If \(\displaystyle x^3=12x+7y\) and \(\displaystyle y^3=7x+12y\), evaluate \(\displaystyle x^2+y^2\).
 
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MarkFL

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#2
Adding the two given equations, we have:

\(\displaystyle x^3+y^3=19x+19y\)

Factor:

\(\displaystyle (x+y)\left(x^2-xy+y^2\right)=19(x+y)\)

If \(x+y=0\) then the equations become:

1.) \(\displaystyle x^3=5x\)

\(\displaystyle x(x^2-5)=0\)

\(\displaystyle (x,y)=(0,0),\,(\pm\sqrt{5},\mp\sqrt{5})\)

And so \(x^2+y^2\) has the possible values of 0 and 10. We get the same solutions from the second equation. Now, in the case that \(x+y\ne0\) we may write:

\(\displaystyle x^2-xy+y^2=19\)

Now, if we subtract the second given equation from the first, we get:

\(\displaystyle x^3-y^3=5x-5y\)

Factor:

\(\displaystyle (x-y)\left(x^2+xy+y^2\right)=5(x-y)\)

If \(x-y=0\), then the equations become:

\(\displaystyle x^3=19x\)

\(\displaystyle x(x^2-19)=0\)

\(\displaystyle (x,y)=(0,0),\,(\pm\sqrt{19},\pm\sqrt{19})\)

And so \(x^2+y^2\) has the possible values of 0 and 38. We get the same solutions from the second equation. Now, in the case that \(x-y\ne0\) we may write:

\(\displaystyle x^2+xy+y^2=5\)

Adding this to the previous result where \(x+y\ne0\) we obtain:

\(\displaystyle 2(x^2+y^2)=24\)

\(\displaystyle x^2+y^2=12\)

And so we find 0, 10, 12, and 38 are all possible values of \(x^2+y^2\). :D
 
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anemone

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#3
Aren't you my smartest Mark? Hehehe...:)(Cool)

Your solution is of course correct and thanks again for participating my challenge math problem!
 
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