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Find $ax^5 + by^5$ if the real numbers $a$, $b$, $x$, and $y$ satisfy the equations $$ \left\{ \begin{aligned} ax+by&=3 \\ ax^{2}+by^{2}&=7 \\ ax^{3}+by^{3}&=16 \\ ax^{4}+by^{4}&=42 \end{aligned} \right. $$ with matrix form!

Interestingly, when I googled the equations, the problem comes from 1990 AIME Problems/Problem 15. The link gives two methods for the solution.

Is there a fashioned way to find the solution in matrix form?

làntèrn
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  • This can be solved in the same way as this MSE question. For the solution in matrix form see Jack D'Aurizio's answer. – Dietrich Burde Aug 12 '15 at 13:55
  • @DietrichBurde: I think that my approach makes better use of the fact that we are given a longer segment of the sequence than what was available in the question you linked to. – Jyrki Lahtonen Aug 12 '15 at 14:10
  • @JyrkiLahtonen Yes, this is even better, of course. – Dietrich Burde Aug 12 '15 at 14:12
  • @Dietrich: Admittedly only because we were only asked to calculate the first missing entry. Otherwise we would need to figure out the coefficients of the recurrence relation, and then apply one of the techniques from that other thread. Which, BTW, has some very nice solutions! – Jyrki Lahtonen Aug 12 '15 at 14:15

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The secuence $c_n=ax^n+by^n$ satisfies a linear two step recurrence relation of the form $$c_{n+2}=Ac_{n+1}+Bc_n$$ for some constants $A,B$. This means that if we perform the row operations coming from that recurrence relation (subtract the second row multiplied by $A$ and the first row multiplied by $B$ from the third row) to the matrix $$ \left(\begin{array}{ccc}c_1&c_2&c_3\\c_2&c_3&c_4\\c_3&c_4&c_5\end{array}\right) $$ its last row becomes all zeros. This means that its determinant vanishes, i.e. $$ 0=\left\vert\begin{array}{ccc}c_1&c_2&c_3\\c_2&c_3&c_4\\c_3&c_4&c_5\end{array}\right\vert=\left\vert\begin{array}{ccc}3&7&16\\7&16&42\\16&42&c_5\end{array}\right\vert=20-c_5. $$ Therefore $c_5=20$.

Jyrki Lahtonen
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