Determine $ax^4 + by^4$ for system of equations

Question

I found the following recreational problem without further specification for $a,b$.

Let $x,y$ be real numbers s.t.

$a + b = 6$, $ax + by = 10$, $ax^2 + by^2 = 24$, $ax^3 + by^3 = 62$.

Determine $ax^4 + by^4$.

I am new to problem solving exercises like this and therefore appreciate diverse approaches to this problem as well as comments on how to tackle those types of exercises.

Answer 1

$$ \color{red}{3 \cdot 62 - 24 = 162}. $$ $$ 3 \cdot 10 - 6 = 24. $$ $$ 3 \cdot 24 - 10 = 62. $$

This is the observation of @Ross, $$ (xy-1)(x-y)^2 = 0. $$ Ross points out in comment (below) that an ingredient was $44=abxy(x-y)^2,$ so that $$ a \neq 0, b \neq 0, x \neq 0, y \neq 0, x \neq y. $$ Finally, $$ xy = 1. $$

The simple fact I got was $$ a (x^2 - 3x+1) + b ( y^2 - 3 y + 1) = 0. $$

We have $xy = 1.$ If $x^2 - 3 x + 1 \neq 0$ and $y^2 - 3 y + 1 \neq 0,$ $$ y^2 - 3 y + 1 = (x^2 - 3 x + 1) / x^2,$$ $$ x^2 ( y^2 - 3 y + 1 ) = x^2 - 3 x + 1. $$ $$ a x^2 (y^2 - 3 y + 1) + b (y^2 - 3 y + 1) = 0,$$ $$ (a x^2 + b)(y^2 - 3 y + 1) = 0. $$ Switch the letters, we get $$ (a + b y^2)(x^2 - 3 x + 1) = 0. $$

Alright, $$ a x^2 + b y^2 = 24, a x^2 + b = 0. $$ $$ b(y^2 - 1) = 24. $$ Also $$ a ( x^2 - 1) = 24. $$ Add, $$ a(x^2 - 1) + b ( y^2 - 1) = 48.$$ However, we know $$ a x^2 + b y^2 - a - b = 24 - 6 = 18. $$ This contradicts the assumption $ x^2 - 3 x + 1 \neq 0. $

$$ \color{red}{ x^2 - 3 x + 1 = 0} $$ $$ \color{red}{ y^2 - 3 y + 1 = 0} $$

Answer 2

We can start with $$(ax+by)(ax^3+by^3)-(ax^2+by^2)^2=620-24^2=44=abxy(x-y)^2\\ (a+b)(ax^2+by^2)-(ax+by)^2=44=ab(x-y)^2$$ Which shows $xy=1$ Now, if nothing else, you can rewrite the middle two in terms of $a,x$ and solve them.

Answer 3

This is not a nice solution.

Considering the equations $$a+b-6=0\tag 1$$ $$ax+by-10=0\tag 2$$ $$ax^2+by^2-24=0\tag 3$$ $$ax^3+by^3-62=0\tag 4$$ Let us eliminate the variables one at the time and express them as a function of $a$.

From $(1)$, $b=6-a$. Replacing in $(2)$ and assuming $a\neq6$, we have $$a x+(6-a) y-10=0 \implies y=\frac{a x-10}{a-6}\tag 5$$ Replacing $b$ and $y$ in $(3)$, we have $$\frac{a \left(-6 x^2+20 x-24\right)+44}{a-6}=0\implies x_{\pm}=\frac{5a\pm\sqrt{11} \sqrt{6 a-a^2}}{3 a}\tag 6$$ (assuming $a\neq 0$).

Let us use $x_+$ and replace in $(4)$ to end with $$\frac{22 \left(\sqrt{(6-a) a}-2 \sqrt{11} a+6 \sqrt{11}\right)}{9 \sqrt{(6-a) a}}=0 \tag 7$$ So, we need to solve $$\sqrt{(6-a) a}=2 \sqrt{11} a-6 \sqrt{11} \tag 8$$ Squaring both sides $$-a^2+6a=44 a^2-264 a+396\implies 45 a^2-270 a+396=0 \implies a_{\pm}=\frac{1}{5} \left(15\pm\sqrt{5}\right)$$ Considering $a_+$ and going backwards, we should get $$a=3+\frac{1}{\sqrt{5}}\qquad b=3-\frac{1}{\sqrt{5}}\qquad x=\frac{1}{2} \left(3+\sqrt{5}\right)\qquad y=\frac{1}{2} \left(3-\sqrt{5}\right)$$ and then the value of $(a x^n+b y^n)$ for any $n$.

Quite laborious, isn't it ?

Thanks for the fun.

Answer 4

I prefer to add another answer for the most general case.

Consider the four equations which I shall write $$ax^{i-1}+b y^{i-1}=c_i \qquad (i=1,2,3,4)$$ Manipulating them, we can show that $x y=\lambda$ and $a b=\mu$ using $$\lambda=\frac{c_3^2-c_2 c_4}{c_2^2-c_1 c_3}$$ $$\mu=-\frac{\left(c_2^2-c_1 c_3\right)^3}{(4 c_4 c_2^3-3 c_3)^2 c_2^2-6 c_1 c_2 c_3 c_4+c_1 \left(4 c_3^3+c_1 c_4^2\right)}$$ This makes $$a_{\pm}=\frac{1}{2} \left(c_1\pm\sqrt{c_1^2-4 \mu }\right)$$ Using $a_+$ , this gives $$b=c_1-a\qquad x=\frac{c_2+\sqrt{c_2^2-4 \lambda \mu }}{2 a}\qquad y=\frac \lambda x$$ Using the given values for the $c_i$'s, this leads to $$\lambda=1\qquad \mu=\frac{44}{5}$$ $$a=3+\frac{1}{\sqrt{5}}\qquad b=3-\frac{1}{\sqrt{5}}\qquad x=\frac{1}{2} \left(3+\sqrt{5}\right)\qquad y=\frac{1}{2} \left(3-\sqrt{5}\right)$$

Now, play with the numbers of your choice.