@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

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?

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.

\( 2^n + n \mid 8^n + n \Leftrightarrow 2^n + n \mid (2^n + n ) \times 4^n - n \times 4^n + n \Leftrightarrow 2^n + n \mid n \times 4^n - n \) 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?

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## Comments

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TopNewestWhat have you tried? Where are you stuck?

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I tried. cant get a clue to start

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

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

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

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

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@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

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

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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.

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

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

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

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

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

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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?

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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

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@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

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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.

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

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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.

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

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

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

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

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\( 2^n + n \mid 8^n + n \Leftrightarrow 2^n + n \mid (2^n + n ) \times 4^n - n \times 4^n + n \Leftrightarrow 2^n + n \mid n \times 4^n - n \) 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?

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

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

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