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The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.

The sum of two numbers are $a$, the sum of the 3rd powers is $10a^3$. Calculate the sum of the 4th powers, expressed in $a$.

Is there a shorter/simpler solution than the one presented below? It feels there is some ‘trick’ to it. The solution presented below is more a ‘straight forward’ one.


Solution

We have \begin{gather*} \left\{ \begin{aligned} x+y&=a\\ x^3+y^3&=10a^3 \end{aligned} \right. \quad\Leftrightarrow\quad x^3+(a-x)^3=10a^3 \quad\Leftrightarrow\quad x^2-ax-3a^2=0 \end{gather*} which has the solutions $$ x_{1,2}=\tfrac{1}{2}(1\pm\sqrt{13}\,)a \qquad \Rightarrow \qquad y_{1,2}=\tfrac{1}{2}(1\mp\sqrt{13}\,)a. $$

Since $$ (1+z)^4+(1-z)^4=2(1+6z^2+z^4) $$ we have \begin{align*} x_1^4+y_1^4 & =\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4} =\tfrac{a^4}{16}\cdot2\bigl(1+6z^2+z^4\bigr)\big|_{z=\sqrt{13}} \\&=\tfrac{a^4}{8}(1+6\cdot13+13^2) =\tfrac{a^4}{8}\cdot248 =31a^4 \end{align*} and, as above, $$ x_2^4+y_2^4 =\bigl(\tfrac{1}{2}(1-\sqrt{13}\,)a\bigr)^{\!4}+\bigl(\tfrac{1}{2}(1+\sqrt{13}\,)a\bigr)^{\!4} =31a^4. $$

Hence, the answer is $31a^4$.


The original exam

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4 Answers 4

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Since $x+y=a$,

$$10a^3=x^3+y^3=(x+y)^3-3xy(x+y)=a^3-3axy\,,$$

and $xy=-3a^2$.

$$\begin{align*} x^4+y^4&=(x+y)^4-2xy\left(2x^2+3xy+2y^2\right)\\ &=(x+y)^4-2xy\left(2(x+y)^2-xy\right)\\ &=a^4+6a^2\left(2a^2+3a^2\right)\\ &=a^4+30a^4\\ &=31a^4\,. \end{align*}$$

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Yes, there is a shorter solution that relies on the identity $$(x+y)(x^n + y^n) = (x^{n+1} + y^{n+1}) + xy(x^{n-1} + y^{n-1}),$$ which is easily verified by multiplication. Then if we let $f_n = x^n + y^n$, this may be written $$f_{n+1} = (x+y)f_n - xy f_{n-1} = f_1 f_n - xy f_{n-1}.$$

Next, observe that $f_0 = 2$ for any nonzero choice of $x, y$. In addition, we are given $f_1 = a$. Then $$f_2 = f_1^2 - xy f_0 = a^2 - 2xy,$$ and $$f_3 = f_1 f_2 - xy f_1 = a(a^2 - 2xy - xy) = a(a^2 - 3xy).$$ Since we are also given $f_3 = 10a^3$, it follows that $10a^3 = a(a^2 - 3xy)$ and if $a \ne 0$, we obtain $$xy = -3a^2.$$ This gives us the information we need to compute $$f_4 = f_1 f_3 - xy f_2 = a (10a^3) - (-3a^2) (a^2 - 2(-3a^2)) = 31a^4$$ and we are done. If $a = 0$, then $f_n$ is trivially $0$ for all $n \ge 1$.

Note this method furnishes the more general recursion $$f_{n+1} = a f_n + 3a^2 f_{n-1}, \quad f_0 = 2, \quad f_1 = a,$$ for which the non-recursive solution is $$f_n = \left(\frac{a}{2}\right)^n \left( (1 - \sqrt{13})^n + (1 + \sqrt{13})^n \right).$$

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A very nice question.

You can think in this way: if we denote $$x+y = s\\ x^3 + y^3= t\\ x^4 + y^4 = u$$ $s$, $t$, $u$ depend on only two parameters $x$, $y$, so they have only two degrees of freedom. Therefore, there must be some relation between $s$, $t$, $u$. It can be obtained by eliminating $x$, $y$ from the above equation. The relation obtained will be a polynomial one. I used WA and got this relation $$s^6 - 8 s^3 t - 2 t^2 + 9 s^2 u=0$$ You can check it directly by substituting $s=x+y$, $t=x^3 + y^3$, $u=x^4 + y^4$.

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  • $\begingroup$ Thanks for pointing out this method. I learnt something new in Mathematica. $\endgroup$
    – mf67
    Dec 27, 2020 at 2:51
  • $\begingroup$ @mf67: You are very welcome! The Swedish exam problems are very interesting. $\endgroup$
    – orangeskid
    Dec 27, 2020 at 2:59
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    $\begingroup$ And I have literally loads(!) of them. I hope to return with more question of this sort if this is liked by the S.E. math community. There are some very tricky problems from time to time, especially regarding classical geometric constructions (using a non-graded ruler and compass), an area which I'm, sad to say, not very experienced in. $\endgroup$
    – mf67
    Dec 27, 2020 at 3:30
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    $\begingroup$ @mf67: All right! Looking forward to more postings. $\endgroup$
    – orangeskid
    Dec 27, 2020 at 3:43
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This answer provides some motivation for a high school student learning the binomial theorem and problem solving techniques; we also get a 'feeling' for homogeneous polynomials.

Employing the binomial theorem we naturally write out (including the quadratic),

$\tag 1 x + y \color\red{ = a} $

$\tag 2 x^2 + y^2 = (x+y)^2- (2xy) $

$\tag 3 x^3 + y^3 = (x+y)^3 - (3 x^2 y + 3 x y^2) \color\red {= 10a^3}$

$\tag 4 x^4 + y^4 = (x+y)^4 - (4 x^3 y + 6 x^2 y^2 + 4 x y^3) $

Using $\text{(1)}$,

$\tag 1 x + y \color\red{ = a} $

$\tag 2 x^2 + y^2 = \color\red {(a)^2} - (2xy) $

$\tag 3 x^3 + y^3 = \color\red {(a)^3} - (3 x^2 y + 3 x y^2) \color\red {= 10a^3}$

$\tag 4 x^4 + y^4 = \color\red {(a)^4} - (4 x^3 y + 6 x^2 y^2 + 4 x y^3) $

It looks likes we have a good shot at writing $xy$ in terms of $a^2$ and from $\text{(3)}$,

$\quad \text{(3)} \implies -3xy(x+y) = 9a^3 \land \text{(1)} \implies xy = -3a^2$

Checkpoint,

$\tag 1 x + y \color\red{ = a} $

$\tag 2 x^2 + y^2 = \color\red {7a^2}$

$\tag 2 xy = \color\red{ -3\, a^2}$

$\tag 3 x^3 + y^3 \color\red {= 10a^3}$

$\tag 4 x^4 + y^4 = \color\red {a^4} - (4 x^3 y + 6 x^2 y^2 + 4 x y^3) $

Always hopeful, we tackle the remaining part,

$\quad - (4 x^3 y + 6 x^2 y^2 + 4 x y^3) =$
$\quad -2xy(2x^2+3xy+2y^2) =$
$\quad -2xy\bigr(2(x^2+y^2) + 3xy\bigr) =$
$\quad 6a^2\bigr(14a^2 -9a^2\bigr) = 30a^4$

and so

$\tag{ANS} x^4 + y^4 = 31a^4$


Bonus Section

Keep going?

$\quad x^5 + y^5 = (x+y)^5 -(5 x^4y + 10 x^3 y^2 + 10 x^2 y^3 + 5 x y^4 ) = $
$\quad \quad \quad \quad \quad \quad a^5 -xy(5 x^3 + 10 x^2 y^1 + 10 x^1 y^2 + 5 y^3 ) =$
$\quad \quad \quad \quad \quad \quad a^5 +3a^2(50 a^3 + 10 x^2 y^1 + 10 x^1 y^2 ) =$
$\quad \quad \quad \quad \quad \quad a^5 +3a^2\bigr(50 a^3 + 10 xy(x + y )\bigr) =$
$\quad \quad \quad \quad \quad \quad a^5 +3a^2(50 a^3 -30a^3) =$
$\quad \quad \quad \quad \quad \quad 61a^5$

More?

$\quad x^6 + y^6 = \,?$

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  • $\begingroup$ A very nice pedagogical presentation. (And I learnt that LaTeX color codes works as well!) $\endgroup$
    – mf67
    Dec 27, 2020 at 3:36

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