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

Since $x+y=a$,

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

and $xy=-3a^2$.

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

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 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 $s=x+y$, $t=x^3 + y^3$, $u=x^4 + y^4$.