ChallengeFind the sum of squares

anemone

Paris la ville de l'amour
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If $$\displaystyle x^3=12x+7y$$ and $$\displaystyle y^3=7x+12y$$, evaluate $$\displaystyle x^2+y^2$$.

MarkFL

MarkFL

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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 $$x^2+y^2$$.

anemone

anemone

Paris la ville de l'amour
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Aren't you my smartest Mark? Hehehe...

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

MarkFL