Algebra problem (problem from Swedish 12th grade ‘Student Exam’ from 1932) 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

 A: Since $x+y=a$,
$$10a^3=x^3+y^3=(x+y)^3-3xy(x+y)=a^3-3axy\,,$$
and $xy=-3a^2$.
$$\begin{align*}
x^4+y^4&=(x+y)^4-2xy\left(2x^2+3xy+2y^2\right)\\
&=(x+y)^4-2xy\left(2(x+y)^2-xy\right)\\
&=a^4+6a^2\left(2a^2+3a^2\right)\\
&=a^4+30a^4\\
&=31a^4\,.
\end{align*}$$
A: 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 $$f_n = \left(\frac{a}{2}\right)^n \left( (1 - \sqrt{13})^n + (1 + \sqrt{13})^n \right).$$
A: 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 $s=x+y$, $t=x^3 + y^3$, $u=x^4 + y^4$.
A: 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?
$\quad x^6 + y^6 = \,?$
