This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

I found a few complex solutions too (where $j = \sqrt{-1}$ ). I was also able to find the same real-valued solutions you did. This makes sense, since there should be five solutions for a quintic equation.

My solution (wasn't going to work without changing R.H.S of equation $1$):

First rearrange the second equation to $a = 2 + b$.

Then simplify the L.H.S into $(a + b)^5$ and expand it to $32b^5 + 160b^4 + 256b^3 + 192b^2 + 96b + 32$.

Then divide both sides by $32$ and move the R.H.S to the L.H.S and you'll obtain $b^5 + 4b^4 + 5b^3 + 4b^2 + 4b$. Label this as equation $3$.

Divide both sides by $b$ to get $b (b^4 + 4b^3 + 5b^2 + 4b + 4)$. Label this equation $4$.

Then assign equation $4$ to the function $f(x)$ and perform factor theorem on it. You'll find that $f(-2) = 0$.
.
Therefore $(b + 2)$ is a factor of $b^4 + 4b^3 + 5b^2 + 4b + 4$.

Perform algebraic long division to obtain $b^3 + 2b^2 + b + 2$. Label this equation $5$ and assign it to $f(x)$.

Perform factor theorem on equation $5$ - you'll find that $f(-2) = 0$.

This is my best pair of simultaneous equations by far - I have spent the last $2$ - $3$ weeks finding a really extreme pair. I challenge anybody to get full marks for this. You have until $14^{th}$ November, $1:30$pm to get all $22$ marks.

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestFrom eq 2,

$a = b+2$

Substituting in eq 1,

$(b+b+2)^{3}(b+b+2)^{2}=b^{4}+ 3b^{3} + b^{2} - b$

$(2b+2)^{3}(2b+2)^{2}=b^{4}+ 3b^{3} + b^{2} - b$

$(2b+2)^{5}=b^{4}+ 3b^{3} + b^{2} - b$

$32(b^{5} + 5b^{4} + 10b^{3} + 10b^{2} + 5b + 1 )=b^{4}+ 3b^{3} + b^{2} - b$

$32b^{5} + 160b^{4} + 320b^{3} + 320b^{2} + 160b + 32=b^{4}+ 3b^{3} + b^{2} - b$

$32b^{5} + 159b^{4} + 317b^{3} + 319b^{2} + 161b + 32 = 0$ (I don't know how to find roots of this mammoth...)

@Yajat Shamji - This next part took a long time to calculate, mostly because I brute-forced the roots of the 5th degree term after graphing it...

One solution is A = 1 and B = -1, and other real solutions can be found when graphed -

Real Solutions -

$a = 1, b = -1$ (got it easily using Remainder Theorem)

$a = 0.464537, b = -1.53546$

$a = 1.42518, b = -0.574825$

Log in to reply

I don't get when you change $(2b + 2)^5 = b^4 + 3b^3 + b^2 - b$ into $32(b^5 + 5b^4 + 10b^3 + 10b^2 + 5b + 1) = b^4 + 3b^3 + b^2 - b$ can you explain plz.

Log in to reply

Sure.

(2b+2) = 2(b+1)

(2b+2)^5 = 2^5 x (b+1)^5 = 32 x (b+1)^5 = 32(b^5+5b^4+10b^3+10b^2+5b+1)

The last expansion is done by (b+1) x (b+1)x(b+1) x (b+1) x (b+1)

Log in to reply

I found a few complex solutions too (where $j = \sqrt{-1}$ ). I was also able to find the same real-valued solutions you did. This makes sense, since there should be five solutions for a quintic equation.

First complex solution:

$a_1 \approx 1.0708 + 0.5192 j \\ b_1 \approx -0.9292 + 0.5192 j$

Second complex solution:

$a_2 \approx 1.0708 - 0.5192 j \\ b_2 \approx -0.9292 - 0.5192 j$

Log in to reply

Nice @Steven Chase! That's all five solutions of the quintic equation. @Yajat Shamji - Challenge complete!!!

Log in to reply

My solution (wasn't going to work without changing R.H.S of equation $1$):

First rearrange the second equation to $a = 2 + b$.

Then simplify the L.H.S into $(a + b)^5$ and expand it to $32b^5 + 160b^4 + 256b^3 + 192b^2 + 96b + 32$.

Then divide both sides by $32$ and move the R.H.S to the L.H.S and you'll obtain $b^5 + 4b^4 + 5b^3 + 4b^2 + 4b$. Label this as equation $3$.

Divide both sides by $b$ to get $b (b^4 + 4b^3 + 5b^2 + 4b + 4)$. Label this equation $4$.

Then assign equation $4$ to the function $f(x)$ and perform factor theorem on it. You'll find that $f(-2) = 0$. . Therefore $(b + 2)$ is a factor of $b^4 + 4b^3 + 5b^2 + 4b + 4$.

Perform algebraic long division to obtain $b^3 + 2b^2 + b + 2$. Label this equation $5$ and assign it to $f(x)$.

Perform factor theorem on equation $5$ - you'll find that $f(-2) = 0$.

Then perform 'intelligent guessing' - you'll obtain $(b + 2) (b^2 + 1)$

$\therefore$ Factors of $b^5 + 4b^4 + 5b^3 + 4b^2 + 4b$ are:

$b(b + 2)^2(b^2 + 1) = 0$

Either:

$b = 0$

$b + 2 = 0, b = - 2$

or:

$b^2 + 1 = 0, b^2 = - 1, b = \sqrt - 1 = \pm i$

Finally, substitute the values of $b$ into $a = 2 + b$:

When $b = 0$:

$a = 2 + 0 = 2$

When $b = - 2$:

$a = 2 + - 2 = 2 - 2 = 0$

When $b = i$:

$a = 2 + i$

When $b = - i$:

$a = 2 + - i = 2 - i$

Therefore the solutions to the simultaneous equations are (in the form $(a, b)$):

$(2, 0)$ $(0, -2)$ $(i, 2 + i)$ $(-i, 2 - i)$

Log in to reply

Here are the solutions I found for your revised equation ($j = \sqrt{-1}$ ):

$(a_1, b_1) = (1,-1) \\ (a_2, b_2) \approx (1.7622,-0.2378) \\ (a_3, b_3) \approx (-0.1207,-2.1207) \\ (a_4, b_4) \approx (1.6793-1.3711 j, -0.3207-1.3711 j ) \\ (a_5, b_5) \approx (1.6793+1.3711 j, -0.3207+1.3711 j )$

Log in to reply

Algebraically, not graphically!

(Sorry if it appears I'm shouting - I'm actually just pressed for time!)

Log in to reply

@Vinayak Srivastava, @Frisk Dreemurr

Log in to reply

This is my best pair of simultaneous equations by far - I have spent the last $2$ - $3$ weeks finding a really extreme pair. I challenge anybody to get full marks for this. You have until $14^{th}$ November, $1:30$pm to get all $22$ marks.

Log in to reply

Ok!!!

Log in to reply

Me with the help of internet.

Log in to reply

$a=\sqrt[5]{b^4+3b^3+b^2-b}-b$

Log in to reply

That's not a solution.

Log in to reply

Well, @Percy Jackson, @Steven Chase, you have completed the challenge, yes!

But keep it a secret for now - I want other people to solve it...

Also, I'll post my solution as well. It is done an entirely different way to yours.

And Percy, there was a simpler way than brute force (that's all I'm telling you.).

Log in to reply

Update - I have changed equation $1$ to compensate for my solution...

@Percy Jackson, @Steven Chase, @Half pass3, @Ravi Singh

You have until tomorrow to solve it... again!

Log in to reply

bruh...

Log in to reply