1

Here is the question

If $x + y = 4$ and $xy = 2$, then find $x^6+ y^6$.

This is from a previous timed competition, so fastest answers are the best answers.

I've tried using sum of cubes, but I dont know what to do after $(x^2+y^2)(x^4-x^2y^2+y^4)$ . The only other way I can think of is solving for x and y, but that wouldn't be too quick. Any help?

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    Simple quadratic formula works I think. I didn't bother to solve it to the end, however $x = y-4$, meaning $\left(y-4\right)y = 2$, $y^2 - 4y + 2 = 0$, by quadratic formula $y = 2 \pm \sqrt{2}, x = 2 \mp \sqrt{2}$. From here just manually calculate $x^6,y^6$, and their sum – AyamGorengPedes Nov 06 '20 at 22:04
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    Unrelated, but I used to do math Olympiads, and more often than not, trying to be 'smart' about the problem just 'enough', then brute forcing the rest of the way is quicker compared to being smart thoroughly. Not elegant at all but it works. – AyamGorengPedes Nov 06 '20 at 22:08

5 Answers5

4

Since $(x,y)$ are roots of $\ x^2-sx+p=0$

Then consider this equation as the characteristic equation of the linear induction relation $$U_{n+1}=sU_u-pU_{n-1}$$

Which solution is $\ U_n=x^n+y^n\ $ with initial values $\begin{cases}U_0=2\\U_1=4\end{cases}$

Then go on calculating successive values:

  • $U_2=4U_1-2U_0=12$
  • $U_3=4U_2-2U_1=40$
  • $U_4=4U_3-2U_2=136$
  • $U_5=4U_4-2U_3=464$
  • $U_6=4U_5-2U_4=1584$

The advantage of this method, is that it is a no-brainer even if you are asked to get $x^{17}+y^{17}$ for instance, it is quite easy to calculate more terms.

zwim
  • 28,563
1

$x^2+y^2=(x+y)^2-2xy=16-4=12.$

$x^4+y^4=(x^2+y^2)^2-2(xy)^2=12^2-2\cdot 2^2=144-8=136$.

Now you can use your factorization:

$x^6+y^6=(x^2+y^2)[(x^4+y^4)-(xy)^2]$.

Tito Eliatron
  • 6,612
1

Another way is to write out

$$ (x+y)^3=(x^3+y^3)+3xy(x+y), $$

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

anon
  • 151,657
1

$(x+y) ^2 =16$$\Rightarrow$$ x^2 +y^2 +2xy=16$$\Rightarrow$$ x^2 +y^2 =12$

Because $xy=2$

$x^2 +y^2 =12$ $\Rightarrow $ $ (x^2 +y^2)^3 =12^3$$\Rightarrow $$x^6+y^6+3(xy)^2(x^2 +y^2) =1728$$\Rightarrow $$x^6+y^6=1728 - 3(4)(12)=1584$

Finally :

$x^6+y^6 =1584$

0

Your sum is$$\begin{align}\sum_\pm(2\pm\sqrt{2})^6&=2(2^6+\sqrt{2}^6+15(2^4\sqrt{2}^2+2^2\sqrt{2}^4))\\&=2(64+8+15(2\sqrt{2})^2(2^2+2))\\&=2(72+15\cdot8\cdot6)\\&=1584.\end{align}$$

J.G.
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