$\sqrt{4} = 2$

Algebra Level 4

Consider the function $f: \mathbb{R} \to \mathbb{R}$ such that, for two defined constants $a,b$ , $f(x) = \dfrac{a^x - b^x}{a-b}.$

We know that $f(0) = 0, \; f(1) = 1, \; f(2) = \sqrt{2}, \; f(3) = \sqrt{3}$ .

If $f(4) = w_1\sqrt{w_2-\sqrt{w_3}}$ , where $w_1, w_2, w_3$ are all integers with $w_3$ is square-free, find $w_1 + w_2 + w_3$ .

The answer is 9.

1 solution

Guilherme Dela Corte
Jan 5, 2017

The values for $f(0)$ and $f(1)$ are tautological. From the others, we get that $f(2)=\frac{a^2-b^2}{a-b}=a+b=\sqrt{2}, \; f(3) = \frac{a^3-b^3}{a-b} = a^2+ab+b^2=\sqrt{3}.$

By doing $f(2)^2 - f(3)$ , we find out that $ab = 2 - \sqrt{3}$ . This way we can evaluate $f(4) = \frac{a^4-b^4}{a-b} = a^3+a^2b+ab^2+b^3$ .

Note that $a^3+a^2b+ab^2+b^3 = (a+b)^3 - 2ab(a+b)$ , so $f(4) = 2(\sqrt{6} - \sqrt{2})$ . Squaring this, we get $f(4)^2 = 4(6+2-4\sqrt{3}) = 16(2 - \sqrt{3})$ .

Applying the square root, we get it back as $f(4) = 4\sqrt{2-\sqrt{3}}$ . Thus $a+b+c = \boxed{9.}$

Tricky but good problem! I already have $f(4) = 2(\sqrt{6} -\sqrt{2})$ , but then i got tricked!

Fidel Simanjuntak - 4 years, 5 months ago

Another solution using linear recurrences:

It can be proved that the sequence $f(n)=\frac{a^n-b^n}{a-b}$ satisfies the recurrence relation $f(n+2)=(a+b)f(n+1)-abf(n),\quad\quad\quad (*)$ for all non-negative integer number $n.$ Using that $f(2)=\sqrt 2$ and $f(3)=\sqrt 3$ and the recurrence equation $(*),$ It can be obtained that $a+b=\sqrt 2$ and $(a+b)\sqrt 2-ab=\sqrt 3.$ Then, $a+b=\sqrt 2$ and $ab=2-\sqrt 3.$ So, the recurrence relation $(*)$ can be written in the form $f(n+2)=\sqrt 2f(n+1)-(2-\sqrt 3) f(n),$ and making $n=2,$ it follows that $f(4)=\sqrt 2 f(3)-(2-\sqrt 3)f(2)=\sqrt 2 \sqrt 3 -(2-\sqrt 3) \sqrt 2= 2\sqrt 6 -2\sqrt 2=4\sqrt{2-\sqrt 3.}$

Arturo Presa - 1 year, 10 months ago

4 = 2 \sqrt{4} = 2 4 ​ = 2

The answer is 9.

1 solution

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$\sqrt{4} = 2$