Such power? Much scare! Many wow!

Algebra Level 5

The roots of the cubic equation

$x^3 - 4x^2 + 6x - 5 = 0$

are $p,q,r$ . If

$\large \frac {p^3}{q^4+r^4} + \frac {q^3}{p^4+r^4} + \frac {r^3}{p^4+q^4} = - \frac {a}{b}$

for coprime positive integers $a,b$ . What is the value of $a-2b$ ?

The answer is 844.

3 solutions

Pi Han Goh
Dec 24, 2013

By Vieta's formula, we have $p+q+r = 4, pq+pr+qr=6, pqr=5$

The combination of the 3 fractions give

$\large \frac {p^3}{q^4+r^4} + \frac {q^3}{p^4+r^4} + \frac {r^3}{p^4+q^4}$

$\large = \frac { ( p^{11} + q^{11} + r^{11} ) + p^4 q^4 (p^3 + q^3) + p^4 r^4 (p^3 + r^3) + q^4 r^4 (q^3 + r^3) + (pqr)^3 (p+q+r) } { (p^4 + q^4)(p^4+r^4)(q^4+r^4) }$

Because there's sum of powers of roots ( $11,4, \text{and} \space 3$ ), this motivates us to use Newton's Sum

However before that, we prefer to find the cubic equation with roots $p^4,q^4,r^4$

Let $f(x) = x^3-4x^2+6x-5 = 0$ , then $f(\sqrt{x}) = 0$ has roots $p^2,q^2,r^2$

$\begin{aligned} (\sqrt x)^3 - 4(\sqrt x)^2 + 6(\sqrt x) - 5 & = & 0 \\ x \sqrt{x} - 4x + 6\sqrt x - 5 & = & 0 \\ \sqrt{x} (x+6) & = & 4x+5 \\ x (x+6)^2 & = & (4x+5)^2 \\ x^3-4x^2-4x-25 & = & 0 \\ \end{aligned}$

Similar, let $g(x) = x^3-4x^2-4x-25$ , then $g(\sqrt{x}) = 0$ has roots $p^4,q^4,r^4$

$\begin{aligned} (\sqrt x)^3 - 4(\sqrt x)^2 - 4(\sqrt x) - 25 & = & 0 \\ x \sqrt{x} - 4x - 4 \sqrt x - 25 & = & 0 \\ \sqrt{x} (x-4) & = & 4x+25 \\ x (x-4)^2 & = & (4x+25)^2 \\ x^3 - 24x^2 - 184x - 625 & = & 0 \\ \end{aligned}$

Then by Vieta's formula, $p^4 + q^4 + r^4 = 24, p^4 q^4 + p^4 r^4 + q^4 r^4 = -184, p^4 q^4 r^4 = 625$

Now we apply Newton's Sum

$P_1 - 4 = 0 \Rightarrow P_1 = 4$

$P_2 - 4P_1 + 2 \cdot 6 = 0 \Rightarrow P_2 = 4$

$P_3 - 4 P_2 + 6 P_1 + 3 \cdot (-5) = 0 \Rightarrow P_3 = 7$

From above, $P_4 = 24$

$P_5 - 4 P_4 + 6 P_3 - 5 P_2 = 0 \Rightarrow P_5 = 74$

Apply Recursive Relation

$P_n = 4P_{n-1} - 6 P_{n-2} + 5 P_{n-3}$

$\Rightarrow P_6 = 187, P_7 = 424, P_8 = 944, P_9 = 2167, P_{10} = 5124, P_{11} = 12214$

So the large fraction simplifies to

$\large \frac {12214 + p^4 q^4 (7 - r^3) + p^4 r^4 (7 - q^3) + q^4 r^4 (7 - p^3) + 5^3 (4) } { (24-r^4)(24-q^4)(24-p^4) }$

$\large = \frac {12714 + 7( p^4 q^4 + p^4 r^4 + q^4 r^4 ) - (pqr)^3 (p+q+r) } { -(p^4-24)(q^4-24)(r^4-24) }$

$\large = \frac {12714 + 7( -184 ) - (5)^3 (4) } { -(p^4-24)(q^4-24)(r^4-24) }$

$\large = - \frac {10926 } { (p^4-24)(q^4-24)(r^4-24) }$

To evaluate the denominator, we consider $h(x) = x^3 - 24x^2 - 184x - 625 = 0$ , then $h(x+24) = 0$ has roots $(p^4-24),(q^4-24),(r^4-24)$

$(x+24)^3 - 24(x+24)^2-184(x+24) - 625 = 0$ , the constant term of the equation is the negative of the product of roots, which yields $24^3 - 24(24^2) - 184(24) - 625 = -5041$ , so $(p^4-24)(q^4-24)(r^4-24) = 5041$

Hence, the fraction equals to $-\frac {10926}{5041}$ , and because $5041 = 71^2$ with $71 \nmid 10926$ , then the fraction is in its simplest form. So $a = 10926, b = 5041 \Rightarrow a-2b = \boxed{844}$

This is how I did it.

Define $P_n = p^n+q^n+r^n, S_n = (pq)^n+(qr)^n+(rp)^n$ and $Z_n=(pqr)^n$ for $n\ge 0$ . Then we have $P_0=3, P_1=4, S_1=6, Z_1=5$ . $P_2 = P_1^2 - 2S_1 = 4$ .

Using the recursive relation $P_{n+3}=4P_{n+2} - 6 P_{n+1} + 5 P_n$ for $n\ge 0$ , it's easy to find $P_3=7, P_4=24, P_{11}=12214$ .

We also have $S_2 = S_1^2 - 2Z_1P_1 = -4$ and $S_4 = S_2^2 - 2 Z_2 P_2 = -184$ .

The numerator of that monster is $P_{11}+S_4P_3 = 10926$ , and the denominator is $S_4P_4 - Z_4 = -5041$ .

George G - 7 years, 5 months ago

Splendid!

Pi Han Goh - 7 years, 5 months ago

What monstrosity is this?

Was what I thought after seeing your solution lol.

Abdur Rehman Zahid - 4 years, 11 months ago

Interesting way to get the value of the denominator.

Hung Woei Neoh - 4 years, 11 months ago

Yes, that is call transforming roots of polynomial .

You might also enjoy my solution here .

Glad that you enjoyed this sadistic question!

Pi Han Goh - 4 years, 11 months ago

What a coincidence, I have a solution there too

Hung Woei Neoh - 4 years, 11 months ago

just Great ! Thanks for nice question and awesome solution.

Piyushkumar Palan - 7 years, 5 months ago

can u plz help me with questions "fine with five" and "abcdefg" ?

Piyushkumar Palan - 7 years, 5 months ago

Nice method! I used the basic method: found out the roots and substituted it

Aditya Kumar - 5 years, 10 months ago

Hung Woei Neoh
Jul 1, 2016

Relevant wiki: Vieta's Formula Problem Solving - Advanced

This is really, really sadistic. I simply must give this problem a solution.

From Vieta's formula, we know that:

$\color{#3D99F6}{p+q+r=4}\\ \color{#D61F06}{pq+pr+qr=6}\\ \color{#EC7300}{pqr=5}$

Using Newton's sums, we have

$\color{#20A900}{p^2+q^2+r^2} = (\color{#3D99F6}{p+q+r})^2 - 2(\color{#D61F06}{pq+pr+qr}) = \color{#3D99F6}{4}^2-2(\color{#D61F06}{6}) =\color{#20A900}{ 4}\\ \color{#69047E}{p^3+q^3+r^3} = (\color{#3D99F6}{p+q+r})(\color{#20A900}{p^2+q^2+r^2}) - (\color{#D61F06}{pq+pr+qr}) (\color{#3D99F6}{p+q+r})+3(\color{#EC7300}{pqr}) = (\color{#3D99F6}{4})(\color{#20A900}{4}) - (\color{#D61F06}{6}) (\color{#3D99F6}{4})+3(\color{#EC7300}{5})=\color{#69047E}{7}\\ \color{magenta}{p^4+q^4+r^4} = (\color{#3D99F6}{p+q+r})(\color{#69047E}{p^3+q^3+r^3}) - (\color{#D61F06}{pq+pr+qr}) (\color{#20A900}{p^2+q^2+r^2})+(\color{#EC7300}{pqr})(\color{#3D99F6}{p+q+r}) = (\color{#3D99F6}{4})(\color{#69047E}{7}) - (\color{#D61F06}{6}) (\color{#20A900}{4})+(\color{#EC7300}{5})(\color{#3D99F6}{4})=\color{magenta}{24}$

The sum we're looking for is

$\dfrac{p^3}{\color{magenta}{q^4+r^4}}+\dfrac{q^3}{\color{magenta}{p^4+r^4}}+\dfrac{r^3}{\color{magenta}{p^4+q^4}}\\ =\dfrac{p^3}{\color{magenta}{24-p^4}}+\dfrac{q^3}{\color{magenta}{24-q^4}}+\dfrac{r^3}{\color{magenta}{24-r^4}}\\ =\dfrac{p^3(24-q^4)(24-r^4)+q^3(24-p^4)(24-r^4)+r^3(24-p^4)(24-q^4)}{(24-p^4)(24-q^4)(24-r^4)}\\ =\dfrac{p^3(24^2-24(\color{magenta}{q^4+r^4})+q^4r^4)+q^3(24^2-24(\color{magenta}{p^4+r^4})+p^4r^4)+r^3(24^2-24(\color{magenta}{p^4+q^4})+p^4q^4)}{24^3-24^2(\color{magenta}{p^4+q^4+r^4})+24(\color{#27D2E7}{p^4q^4+p^4r^4+q^4r^4})-p^4q^4r^4}\\ =\dfrac{p^3(24^2-24(\color{magenta}{24-p^4})+q^4r^4)+q^3(24^2-24(\color{magenta}{24-q^4})+p^4r^4)+r^3(24^2-24(\color{magenta}{24-r^4})+p^4q^4)}{24^3-24^2(\color{magenta}{24})+24(\color{#27D2E7}{p^4q^4+p^4r^4+q^4r^4})-(\color{#EC7300}{pqr})^4}\\ =\dfrac{24(\color{olive}{p^7+q^7+r^7})+p^3q^3r^3(\color{#D61F06}{pq+pr+qr})}{24(\color{#27D2E7}{p^4q^4+p^4r^4+q^4r^4})-(\color{#EC7300}{pqr})^4}$

Now, we extend Newton's sums to the $7$ th power:

$\color{teal}{p^5+q^5+r^5} = (\color{#3D99F6}{p+q+r})(\color{magenta}{p^4+q^4+r^4} ) - (\color{#D61F06}{pq+pr+qr}) (\color{#69047E}{p^3+q^3+r^3})+(\color{#EC7300}{pqr})(\color{#20A900}{p^2+q^2+r^2}) = (\color{#3D99F6}{4})(\color{magenta}{24}) - (\color{#D61F06}{6}) (\color{#69047E}{7})+(\color{#EC7300}{5})(\color{#20A900}{4})=\color{teal}{74}\\ \color{#624F41}{p^6+q^6+r^6}= (\color{#3D99F6}{p+q+r})(\color{teal}{p^5+q^5+r^5} ) - (\color{#D61F06}{pq+pr+qr}) (\color{magenta}{p^4+q^4+r^4})+(\color{#EC7300}{pqr})(\color{#69047E}{p^3+q^3+r^3}) = (\color{#3D99F6}{4})(\color{teal}{74} ) - (\color{#D61F06}{6}) (\color{magenta}{24})+(\color{#EC7300}{5})(\color{#69047E}{7}) =\color{#624F41}{187}\\ \color{olive}{p^7+q^7+r^7}= (\color{#3D99F6}{p+q+r})(\color{#624F41}{p^6+q^6+r^6}) - (\color{#D61F06}{pq+pr+qr}) (\color{teal}{p^5+q^5+r^5} )+(\color{#EC7300}{pqr})(\color{magenta}{p^4+q^4+r^4}) =(\color{#3D99F6}{4})(\color{#624F41}{187}) - (\color{#D61F06}{6}) (\color{teal}{74} )+(\color{#EC7300}{5})(\color{magenta}{24})=\color{olive}{424}$

And now, we arrive at the tricky part: $\color{#27D2E7}{p^4q^4+p^4r^4+q^4r^4}$

$(\color{#D61F06}{pq+pr+qr})^2 = \color{#BA33D6}{p^2q^2+p^2r^2+q^2r^2} + 2(p^2qr+pq^2r+pqr^2)\\ \color{#BA33D6}{p^2q^2+p^2r^2+q^2r^2} =(\color{#D61F06}{pq+pr+qr})^2-2\color{#EC7300}{pqr}(\color{#3D99F6}{p+q+r}) = (\color{#D61F06}{6})^2-2(\color{#EC7300}{5})(\color{#3D99F6}{4}) = \color{#BA33D6}{-4}\\ (\color{#BA33D6}{p^2q^2+p^2r^2+q^2r^2})^2 = \color{#27D2E7}{p^4q^4+p^4r^4+q^4r^4} + 2(p^4q^2r^2+p^2q^4r^2+p^2q^2r^4)\\ \color{#27D2E7}{p^4q^4+p^4r^4+q^4r^4} = (\color{#BA33D6}{p^2q^2+p^2r^2+q^2r^2})^2 -2(\color{#EC7300}{pqr})^2(\color{#20A900}{p^2+q^2+r^2})=(\color{#BA33D6}{-4})^2 -2(\color{#EC7300}{5})^2(\color{#20A900}{4})=\color{#27D2E7}{-184}$

With everything prepared, we can now substitute in the values (finally!):

$=\dfrac{24(\color{olive}{p^7+q^7+r^7})+(\color{#EC7300}{pqr})^3(\color{#D61F06}{pq+pr+qr})}{24(\color{#27D2E7}{p^4q^4+p^4r^4+q^4r^4})-(\color{#EC7300}{pqr})^4}\\ =\dfrac{24(\color{olive}{424})+(\color{#EC7300}{5})^3(\color{#D61F06}{6})}{24(\color{#27D2E7}{-184})-(\color{#EC7300}{5})^4}\\ =\dfrac{10176+750}{-4416-625}\\ =-\dfrac{10926}{5041}$

(Use your calculator, WolframAlpha or whatever to verify that this is the simplest form of the fraction. Don't lie, you're probably already using your calculator at this point)

Therefore, $a=10926,\;b=5041,\;a-2b = 10926-2(5041)=10926-10082 = \boxed{844}$

MOST COLORFUL SOLUTION EVER!!!! +1

Pi Han Goh - 4 years, 11 months ago

Thanks! I almost ran out of colors to use!!!!!!!!! Hopefully, the colors help to make this solution easier to read

Hung Woei Neoh - 4 years, 11 months ago

Are you on Slack?

Pi Han Goh - 4 years, 11 months ago

@Pi Han Goh – Nope, I'm driving

Hung Woei Neoh - 4 years, 11 months ago

@Hung Woei Neoh – You typed a colorful solution while driving? WOAHHHHHHH

Pi Han Goh - 4 years, 11 months ago

@Pi Han Goh – =.= The idea of typing this solution while driving sounds so ridiculous and insane... even with a laptop, this solution took me 45 minutes to type out (surprisingly, it took less time than the solution to your Newton's sums problem a few weeks ago)

Nope, I typed this solution a few hours ago. I am using my phone now, and I don't type solutions with my phone

Hung Woei Neoh - 4 years, 11 months ago

@Hung Woei Neoh – Haha I was being sarcastic!!

By the way, I really appreciate your write up and I'm glad you enjoyed this question! =D =D =D =D

Pi Han Goh - 4 years, 11 months ago

Lol!!! The colours!! A bright solution!

Ashish Menon - 4 years, 11 months ago

Colors!!!!!!!!!!

Hung Woei Neoh - 4 years, 11 months ago

Dont you ever get tired of plugging in so many many many colours together? :P

Ashish Menon - 4 years, 11 months ago

@Ashish Menon – Not at all. This question was especially thrilling, I had fun solving this problem and writing this colorful solution

Hung Woei Neoh - 4 years, 11 months ago

@Hung Woei Neoh – Genius in disguise :P

Ashish Menon - 4 years, 11 months ago

@Ashish Menon – Not really a genius. Just a guy who loves math and has the free time to type colorful solutions

Hung Woei Neoh - 4 years, 11 months ago

@Hung Woei Neoh – Nice, I have no free time niw :( So much homeworks

Ashish Menon - 4 years, 11 months ago

@Ashish Menon – It's okay, you can spend more time here during the holidays

Hung Woei Neoh - 4 years, 11 months ago

A very cleanly presented and colourful solution!!!

Abdur Rehman Zahid - 4 years, 11 months ago

Thanks $\ddot\smile$

Hung Woei Neoh - 4 years, 11 months ago

Same way,the only thing I did different was strategically not substituting $24-p⁴$ for $q^4+r^4$ (and similar terms) in the numerator and so on.Therefore I didnot get that nasty $p^7+q^7+r^7$ term.

Abdur Rehman Zahid - 4 years, 11 months ago

@Abdur Rehman Zahid – Google "LaTeX color list". The first result will show a list of basic, predefined LaTeX colors. I don't recommend yellow or pink, too bright.

Ehh? Then how did you calculate $p^3q^4 +p^3r^4+\ldots$ ???

Hung Woei Neoh - 4 years, 11 months ago

@Hung Woei Neoh – Thanks :)

I currently don't remember what I did :P I solved this problem about more than a week ago.I'll get back to you soon.

Abdur Rehman Zahid - 4 years, 11 months ago

Also,how do you know so many colours? Do you know a specific page of a list of colours for that?

Abdur Rehman Zahid - 4 years, 11 months ago

Thanks :) I currently don't remember what I did :P I solved this problem about more than a week ago.I'll get back to you soon.

Abdur Rehman Zahid - 4 years, 11 months ago

Loved this!!..I seriously wants to know how much time you spend on this problem to get it correct?

A Former Brilliant Member - 4 years, 11 months ago

Mmmm...I don't remember....about 20-30 minutes? Maybe?

Hung Woei Neoh - 4 years, 11 months ago

Niranjan Khanderia
Jun 22, 2017

$Let~P_k=p^k+q^k+r^k,~~~~and~~~~Q_k=(p*q*r)^k\\ \color{#EC7300}{\dfrac{p^3}{q^4+r^4}+\dfrac{p^3}{q^4+r^4}+\dfrac{p^3}{q^4+r^4}\\ =\dfrac{p^3( p^4+q^4)(p^4+r^4)+q^3( q^4+r^4)(q^4+p^4)+r^3( r^4+p^4)(r^4+q^4)}{( p^4+q^4)( q^4+r^4) ( r^4+p^4)}\\ But~~p^3( p^4+q^4)(p^4+r^4)=p^3( P_4-r^4)(P_4-q^4)=p^3*(P_4^2-P_4(q^4+r^4)+ q^4*r^4 )\\ = p^3*(P_4^2-P_4(P_4-p^4)+ Q_4/p^4 ) =p^3(P_4^2-P_4^2+P_4*p^4+ Q_4/p^4)\\ \therefore~~\color{#3D99F6}{p^3( p^4+q^4)(p^4+r^4)=P_4*p^7+Q_4/p. ..............(1) } \\ \ \ \ \ \ \color{teal}{teal}~~\color{#27D2E7}{aqua}~~\color{olive}{olive}~~\color{#BA33D6}{violet}\\ ( p^4+q^4)( q^4+r^4) ( r^4+p^4)=( P_4-r^4)( P_4-p^4) ( P_4-q^4) \\ =P_4^3-P_4^2(p^4+q^4+r^4)+P_4( p^4*q^4+q^4*r^4 + r^4*p^4)-p^4*q^4*r^4 \\ =P_4^3-P_4^2(P_4)+P_4* Q_4*(1/p^4+1/q^4 +1/r^4)-Q_4\\ =0- P_4* Q_4*P_{-4}-Q_4. } \\ \therefore~~\color{#3D99F6}{( p^4+q^4)( q^4+r^4) ( r^4+p^4)= - P_4* Q_4*P_{\!-4}-Q_4.............(2)} \\ \ \ \ \ \ \ \\ \color{#BA33D6}{(A)~ X^3-4X^2+6X-5=0,~~~dividing~~by~-5*X^3~~~~~ we~get~~~~(B) ~X^{-3}-\frac 6 5X^{-2}+\frac 4 5X-\frac 1 5 =0 \\ We~use~Newton~Sum~for~equation.~~~~~~~~ a_3*X^3+ a_2X^2+a_1*X+a_o \\ P_1=-a_2/a_3,.........P_2=-a_2/a_3*P_1 - 2*a_1/a_3.........P_3=-a_2/a_3*P_2-a_1/a_3*P_1-3*a_0/a_3\\ (A)~~\therefore~P_1=4,....................P_2=4*4-2*6=4,....................P_3=4*4-6*4+3*5=7\\ (B)~~{\color{#20A900}{P_{-1}=6/5,}.}........P_{-2}=6/5*6/5 - 2*4/5= - 4/25.........P_{-3}=6/5*(-4/25)-4/5*6/5+3*5 =-69/125 \\ \ \ \ \ \ \ \ \\ Using~Recursive ~Relation ~for~ both~ A ~ and~ B:-\\ (A)~~P_k=4P_{k-1} - 6P_{k-2}+5P_{k-3} ~~~~~~~~~~~~~~~ \\ P_4=4P_3 - 6P_2+5P_1=4*7-6*4+5*4=24 . ~~~~~~ P_5=4P_4 - 6P_3+5P_2=4*24-6*7+5*4=74 .\\ P_6=4P_5 - 6P_4+5P_3=4*74-6*24+5*7=187 .}\\ \color{#20A900}{P_7=4P_6 - 6P_5+5P_4=4*187-6*74+5*24=424 }.\\ \ \ \ \ \ \ \ \\ (B)~~\color{#20A900}{P_{-4}=\frac 6 5*P_{-3}+\frac 4 5*P_{-2} - \frac 1 5*P_{-1} \\ =\frac 6 5*(-69)/125 -\frac 4 5*(- 4/25) +\frac 1 5*\frac 6 5=184/625 } \\ \color{teal}{From~(1)~and~(2):-\\ \dfrac{p^3}{q^4+r^4}+\dfrac{q^3}{r^4+p^4}+\dfrac{r^3}{p^4+q^4}=\displaystyle \dfrac{ \sum_{cyc} P_4*p^7~~+~~Q_4/p.} {- P_4* Q_4*P_{\!-4}~~-~~Q_4} \\ =\dfrac{P_4*(p^7+q^7+r^7)~~+~~Q_4*(p^{-1}~~ +~~q^{-1}~~ +~~r^{-1} )} {~~ -~~ P_4* Q_4*P_{\!-4}~~-~~Q_4} \\ =\dfrac{P_4*P_7~~+~~Q_4*P_{-1}} {~~-~P_4* Q_4*P_{\!-4}~~-~~Q_4} \\ = \dfrac{24*424~~+~~625*6/5}{ ~-~ 24*625*184/625~~-~~625}=\dfrac{10926}{-~5041} =-\dfrac a b.\\ \therefore~~a-2b= 10926-2*5041 }= \Large \color{#D61F06}{844}.$

Are you sure P(k) = x^k +y^k + z^k also works for negative integer k?

Note that Newton's sum only applies for positive powers. But why does it appear to work for negative integers k too?

Pi Han Goh - 3 years, 11 months ago

I think this is only a representation. It is not doing any calculation. That is what I think.

Niranjan Khanderia - 3 years, 11 months ago

You are doing a lot of calculations, I don't know what you meant by "representation"...

Pi Han Goh - 3 years, 11 months ago

@Pi Han Goh – representation was not a correct word. My second comment explanations my thinking.

Niranjan Khanderia - 3 years, 11 months ago

I am an engineer. I love maths. We often use math without knowing the theory.
I used Newton's Sum, but do not know the theory. For polynomial with + powers, if we know $P_n$ for lower powers, we can calculate for higher ones. My logic. For - Powers, if we know $P_n$ for higher powers, we can calculate for lower ones. If we know for n=-1,-2, -3,-4, we can calculate for -5,-6 ... .

Niranjan Khanderia - 3 years, 11 months ago

That's a hasty generalization. Math is not "oh, because it works for higher ones, so I can reverse it and go the other way too!"

Otherwise, mathematical induction will not make sense.

Pi Han Goh - 3 years, 11 months ago

@Pi Han Goh – I agree. What I meant to say is how I got it and what I got was correct. A request. What I got was a coincident? Or there is a theoretical explanation?

Niranjan Khanderia - 3 years, 11 months ago

@Niranjan Khanderia – Hint: If P(x) is a polynomial with non-zero roots a,b,c, then what are the roots of x^3 ( P(1/x) ) ?

Pi Han Goh - 3 years, 11 months ago

@Pi Han Goh – Thank you.

Niranjan Khanderia - 3 years, 11 months ago

Such power? Much scare! Many wow!

The answer is 844.

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