\( 2^n + n \mid 8^n + n \Leftrightarrow 2^n + n \mid (2^n + n ) \times 4^n - n \times 4^n + n \Leftrightarrow 2^n + n \mid n \times 4^n - n \) was what I thought you said you were doing earlier.

So we were able to remove the \( 8^n \) and replaced it with a \( n \times 4 ^n \). How can we remove the \( 4^n
\) now to further simplify the statement?

Can you recheck? You are saying \( A \mid BA \) where \( A = 2^n + n \).

This doesn't quite look right to me. I now do not know what you are doing. Please list out your steps carefully, and do not just jump to the final step.

You already have the ideas, but just are not executing on them. You saw how to "simplify" \( 8^n \) into other terms, and just need to "simplify" these other terms into polynomials.

IE We have a \( 2^{2n} ( \times n) \) term. How can we get rid of it?

@CalvinLin-this was my approach. If 2^n+n devides 8^n+n then it devides 8^n+n-2^n+n.I wanted to eliminate n. Now i arrive at finding 2^n+n devides (2^n)(2^n+1)(2n-1). Here i am stuck to find the answer

Easy Math Editor

`*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:

TopNewest@Calvin Lin- could you please give an entire solution to this?

Log in to reply

i did it already. please post the entire solution.i am quite...disturbed by this probelem

Log in to reply

@Calvin Lin-i really have no idea how to go further. Could you please post a solution to this?

Log in to reply

\( 2^n + n \mid 8^n + n \Leftrightarrow 2^n + n \mid (2^n + n ) \times 4^n - n \times 4^n + n \Leftrightarrow 2^n + n \mid n \times 4^n - n \) was what I thought you said you were doing earlier.

So we were able to remove the \( 8^n \) and replaced it with a \( n \times 4 ^n \). How can we remove the \( 4^n \) now to further simplify the statement?

Log in to reply

@Calvin Lin-so sorry! it is 2^n+n| 2^n(2^n+1)(2^n-1)

Log in to reply

That's a good start. Actually, \( 8^n\) is "more complicated" than \(n\). How can we "eliminate exponents" to only obtain polynomials?

Log in to reply

@Calvin Lin-sir, please wait because i need to get the answer today. I dnt have anyone here else to help me out.

Log in to reply

@Calvin Lin-2^n+n|(2^n-1)(2^n +n )

Log in to reply

Can you recheck? You are saying \( A \mid BA \) where \( A = 2^n + n \).

This doesn't quite look right to me. I now do not know what you are doing. Please list out your steps carefully, and do not just jump to the final step.

Log in to reply

@Calvin Lin-i did it already, multiplied and subtracted to to get (2^n-1)(2^n +n )|2^n+n. all i dnt know is how to get a polynomial out of it

Log in to reply

Are you sure the RHS is 2^n+n? That seems to be the LHS. At this point in time, I'm not certain what you're doing.

Log in to reply

i did it already, multiplied and subtracted to to get (2^n-1)(2^n+n )|2^n+n. all i dnt know is how to get a polynomial out of it

Log in to reply

@Calvin Lin-can you make me understand how to get a polynomial from this?

Log in to reply

You already have the ideas, but just are not executing on them. You saw how to "simplify" \( 8^n \) into other terms, and just need to "simplify" these other terms into polynomials.

IE We have a \( 2^{2n} ( \times n) \) term. How can we get rid of it?

Log in to reply

(2^n-1)(2^n-1)|2^n+n

Log in to reply

(2^2n)n-n|2^n+n

Log in to reply

Great. So once again, " \( 2^{2n} \times n \) is more complicated than \( 2^n \)". How can we simplify further?

Log in to reply

i tried that too. I multiplied 2^2n to 2^n+n and subtracted 8^n+n from that and proceeded as above. What i dont get is what to do next.

Log in to reply

What is the equation that you get? How can we simplify it further?

Log in to reply

@CalvinLin-this was my approach. If 2^n+n devides 8^n+n then it devides 8^n+n-2^n+n.I wanted to eliminate n. Now i arrive at finding 2^n+n devides (2^n)(2^n+1)(2n-1). Here i am stuck to find the answer

Log in to reply

That's a good start. Actually, \( 8^n\) is "more complicated" than \(n\). How can we "eliminate exponents" to only obtain polynomials?

Log in to reply

i tried dat way too. I believe i am leaving a minute detail. Can ya please send me a solution?

Log in to reply

You can write up your solution and obtain feedback from the community on how to improve it.

Log in to reply

through brute force 1,2 come. but what canbe a logical way to do this?

Log in to reply

I tried. cant get a clue to start

Log in to reply

Since we have a Diophantine equation, which approach can we use?

Log in to reply

What have you tried? Where are you stuck?

Log in to reply