Infinity is lovely

Algebra Level 4

$K= \large \sqrt{5 + \sqrt{5-\sqrt{5+\sqrt{5-\sqrt{5+\ldots}}}}}$

If $K$ can be expressed in simplest form as $\dfrac{a+\sqrt{b}}{c}$ , where $a,b,c$ are positive integers, find the value of $a^2+b^2+c^2$ .

The answer is 294.

4 solutions

Alan Enrique Ontiveros Salazar
Aug 14, 2015

Another approach:

$K=\sqrt{5+\sqrt{5-\sqrt{5+\cdots}}} \\ L=\sqrt{5-\sqrt{5+\sqrt{5-\cdots}}}$

Then $K=\sqrt{5+L}$ and $L=\sqrt{5-K}$ . Square both equations: $K^2=5+L$ and $L^2=5-K$ and subtract them: $K^2-L^2=K+L \implies (K+L)(K-L)=K+L$ . Since $K+L \neq0$ : $K-L=1 \implies L=K-1$ .

Finally substitute that in the first equation:

$K^2=5+K-1 \implies K^2-K-4=0$

Using the quadratic formula we obtain, and with the fact that $K>0$ :

$K=\dfrac{1+\sqrt{17}}{2}$

Comparing we get $a=1$ , $b=17$ and $c=2$ . Hence the answer is $1^2+17^2+2^2=\boxed{294}$ .

Nice solution,it avoids something like x^4...

Dave Day - 5 years, 10 months ago

True, but if you use the "x^4 approach" you get to practice your skills in factorizing polynomials. Still, i love this solution

Michele Franzoni - 2 years, 1 month ago

I saw this method stack exchange and used it to great effect on two nested radical problems. Thank you, sir. :)

Kevin Mano - 5 years, 9 months ago

Nice solution buddy up upvoted it

Department 8 - 5 years, 10 months ago

Thank you :D

Alan Enrique Ontiveros Salazar - 5 years, 10 months ago

I've definitely seen it before...

Satvik Golechha - 5 years, 9 months ago

Matías Bruna
Aug 14, 2015

Solution:

Umm......

See the answer comes as $\frac{a+\sqrt{b}}{c}$ and you are asking $a^{2}+b^{2}+c^{2}$ but in your solution , it says $b=\sqrt{17}$ but according to $\frac{a+\sqrt{b}}{c}$ , $b=17$ .

Please look over my problem.

Akshat Sharda - 5 years, 10 months ago

Im really sorry, the real problem said "is at the form (a+b)/c" but brilliant edited this.

Matías Bruna - 5 years, 10 months ago

Ok.........

Akshat Sharda - 5 years, 10 months ago

Ya i also confirmed it by wolfram alpha

Mohit Gupta - 5 years, 10 months ago

Now i fixed this!, brilliant edited this $\dfrac{a + b}c$ to this! $\dfrac{a + \sqrt{b}}c$ im really sorry for the trouble, but brilliant did a bad-edition.

Matías Bruna - 5 years, 10 months ago

ok,we don't talk about the problem ,I wonder how u recognized 「(x²-x-4)(x²+x-5)」?I finally ask Wolfram.

Dave Day - 5 years, 10 months ago

This is my method.

$x^{4} - 10x^{2} + x + 20 = 0$

$4x^{4} - 40x^{2} + 4x + 80 = 0$

$4x^{4} - 36x^{2} + 81 = 4x^{2} - 4x + 1$

$(2x^{2} - 9)^{2} = (2x - 1)^{2}$

$(2x^{2} - 9)^{2} - (2x - 1)^{2} = 0$

$(2x^{2} + 2x - 10)(2x^{2} - 2x - 8) = 0$

$(x^{2} + x - 5)(x^{2} - x - 4) = 0$

汶良林 - 5 years, 10 months ago

This how i recognized the factors:- $factors\quad of\quad the\quad above\quad polynomial\quad can\quad be\quad of\quad the\quad form:-\\ (a{ k }^{ 2 }+{ bk }^{ 1 }+c)({ a }^{ ' }{ k }^{ 2 }+{ b }^{ ' }{ k }^{ 1 }+{ c }^{ ' })\_ \_ \_ \_ (1)\\ Now\quad coefficient\quad of\quad { k }^{ 4 }\quad is\quad 1\\ so\quad a=a^{ ' }=1\quad also\quad coefficeint\\ of\quad { k }^{ 3 }\quad is\quad 0\quad so\quad b=1\quad and\quad b^{ ' }=-1\\ Now\quad substituting\quad a,b,a^{ ' },b'\quad and\quad expanding\quad (1)\\ we\quad get\\ { k }^{ 4 }-{ k }^{ 2 }(c+c'+1)+k(c-c')+cc'\\ comparing\quad it\quad with\quad the\quad original\quad polynomial\\ we\quad get\quad c=5\quad and\quad c'=4\\ so\quad the\quad factors\quad are\\ ({ k }^{ 2 }+k-5)({ k }^{ 2 }-k-4)$

Rishi Sharma - 5 years, 10 months ago

How do you know that b,b' = ±1 not ±2,±3,±4,... ???

Anuchit Thanasrisurbwong - 5 years, 10 months ago

@Anuchit Thanasrisurbwong – Rishi's got the right idea, but it can be a bit more involved.

First, the possible rational zeros don't produce any zeros, so we can see if there are 2 quadratics that fit the bill. The coefficient of the $k^{4}$ term is 1, so that makes things a tad easier.

$(k^{2}+ak+b)(k^{2}+ck+d)=k^{4}+(a+c)k^{3}+(ac+b+d)k^{2}+(ad+bc)k+bd$

From this, we can match coefficients and get 4 equations. If they can be solved, then this is possible.

1) $a+c=0$

2) $ac+b+d=-10$

3) $ad+bc=1$

4) $bd=20$

Trying to solve them (at least for me) yields:

1) $\Rightarrow c=-a$

3) $\Rightarrow a(d-b)=1 \Rightarrow d-b=\frac{1}{a}$

For this to work $\frac{1}{a}$ must be an integer, so a must be 1 or -1. Choosing either works (I chose 1)

1) $a=1 \Rightarrow c=-1$

3) $\Rightarrow d=b+1$

4) $\Rightarrow b(b+1)=20 \Rightarrow b^{2}+b-20=0 \Rightarrow b=4, -5$

If $b=4$ then $d=5$ , but then equation 2 doesn't work. So:

4) $b=-5$

3) $\Rightarrow d=-4$

2) $1(-1)-4-5=-10$ Check

Therefore, the assumption that $a=1$ works and you are left with $(k^{2}+k-5)(k^{2}-k-4)$ .

Louis W - 5 years, 10 months ago

@Anuchit Thanasrisurbwong – see @louis w's solution.. you can easily notice that two get the simplest possible valuse of the unknowns it is important to keep b and b' unity. even if you change the values you will get the same values but a little more complex.

Rishi Sharma - 5 years, 10 months ago

Lu Chee Ket
Oct 18, 2015

(K^2 - 5)^2 = 5 - K

K^4 - 10 K^2 + K + 20 = 0

K^4 + (u - 10) K^2 - u K^2 + K + 20 = 0

u^3 - 20 u^2 + 20 u - 1 = 0 is having 3 real roots but 1 simplest which is 1.

Therefore (K^2 - K - 4)(K^2 + K - 5) = 0

As we know that K = 2.56 approximately from the original expression,

K^2 - K - 4 = 0 is the one contains 2.56,

(K - 1/ 2)^2 = (1/ 2)^2 + 4 = 17/ 4

K = 1/ 2 +/- Sqrt (17) / 2 initially but not negative,

K = [1 + Sqrt (17)]/ 2 = 2.561552812808830274910704927987

1^2 + 2^2 + 17^2 = 294

William Isoroku
Aug 20, 2015

${ K }^{ 2 }=5+\sqrt { 5-\sqrt { 5+\sqrt { 5-\sqrt { 5+.... } } } }$

${ K }^{ 2 }-5=\sqrt { 5-\sqrt { 5+\sqrt { 5-\sqrt { 5+.... } } } }$

${ ({ K }^{ 2 } }-5)^{ 2 }=5-\sqrt { 5+\sqrt { 5-\sqrt { 5+.... } } }=5-K$

${ K }^{ 4 }-10{ K }^{ 2 }+25=5-K\rightarrow { K }^{ 4 }-{ 10K }^{ 2 }+K+20=0$

${ K }^{ 4 }-{ 10K }^{ 2 }+K+20=0\rightarrow ({ K }^{ 2 }-K-4)({ K }^{ 2 }+K-5)=0$

Using the quadratic formula, we get the roots:

$\frac { 1\pm \sqrt { 17 } }{ 2 }$

$\frac { -1\pm \sqrt { 21 } }{ 2 }$

We choose the root with all positive numbers.

Moderator note:

Both $\dfrac{-1+\sqrt{17}}2$ and $\dfrac{-1+\sqrt{21}}2$ are positive numbers, how do you know which one is the right answer?

Infinity is lovely

The answer is 294.

4 solutions

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