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How many positive integers n satisfy the following equation:
n 5 − 8 0 n 3 = 1 0 n 4 + 2 6 0 n 2 + 4 0 0 n + 2 4 2
Difficulty − Sharper than most † † † † †
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5 solutions
Amazing method.
The whole equation, after some adding and subtracting, can be re-written as: ( n + 3 ) 5 = n 5 + ( n + 1 ) 5 By Fermat's Last Theorem , there are no solutions for positive integer n .
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You are finding real solutions, not integer.
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Oh! I didn't read the question, just the equation LOL
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@A Former Brilliant Member – How is the problem?
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@Vinayak Srivastava – Great! Especially the name :)
Btw, did you and Deprave Tripartite delete your comments in your pirate question @Vinayak Srivastava , I got notifications, but I couldn't see the comments.
Edit : Pradeep Tripathi got autocorrected to Deprave Tripartite LOL
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@A Former Brilliant Member – I deleted that thread, it wasn't related to the problem.
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@Vinayak Srivastava – Oh! Ok, I wanted to see his reaction when he realized you were stalking him on Brilliant for some science problems LOL XD @Vinayak Srivastava
@Vinayak Srivastava , please don't make the problem question too fanciful. The main thing is too learn math. Need not use LaTex for text. It is difficult and not a standard here in Brilliant. Don't not use software app to to solve algebra problem. We only learn how to punch in data and not learning math.
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Ok, I'll remember it next time. Sorry. But I didn't use software app, @Percy Jackson has used. I'll remove latex text from my solution also.
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OK. Yes, I notice it was @Percy Jackson . Nice solution, I have upvoted
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@Chew-Seong Cheong – Thank you! :)
@Chew-Seong Cheong – Sir, do you know any good source of Biology problems? I am very weak in Biology, and can't find any Biology problems on Brilliant.
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@Vinayak Srivastava – There were. They were trying to start that field but did not succeed.
@Chew-Seong Cheong – @Chew-Seong Cheong - I was only verifying sir, I had solved the problem, but didn't notice the positive integer part, so I got real number roots :)
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@A Former Brilliant Member – Please share your method, how did you solve a 5th degree polynomial?????
@Vinayak Srivastava , the Biology problems are still there.
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Can you please share some profile pages of people who posted them?
@Vinayak Srivastava - Did Aryan Sanghi teach you the daggers?
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No, I saw your note.
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Oh! You can use \large for text and \LARGE for the daggers, to make it look more like the original daggers, if you want :)
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@A Former Brilliant Member – I used it, but Sir said that I should not make the problem too fanciful. I won't edit it, as Sir has edited, so I don't think it's good to edit again.
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@Vinayak Srivastava – Ok (some text)
I noticed 1 thing, that even if we had to solve it in a different way, we had to check only the integers ending with 2 where n is not equal to 2, i.e n = 1 2 . 2 2 , 3 2 , 4 2 , . . because 1 0 n 4 + 2 6 0 n 2 + 4 0 0 n have at least common factor of 10, leaving n and thus no matter what integer n will be when this terms will be added the last digit will be 0 and then we add 242 after that, so the last digit would be obviously 2 on the right side. Similarly, on the left side, n 5 > 8 0 n 3 will be possible only if n 2 is greater than 80, otherwise the left-hand side would have a negative answer and the right side a positive one. Now as the right side will have a 2 in its last digit, the left side should also have that. The term 8 0 n 3 will also have 0 as the last digit due to 10 as a factor, so we need to now depend on n 5 now only when 2 is multiplied to itself 5 times, the last digit is 2, thus n would have the last digit as 2, and the minimum it would be 12 as 2^2<80. But sadly we find n doesn't exist.
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Yes, good notice! I think if we follow this, we only need to check some cases, and later the difference b/w both sides will increase, so we can leave after that.
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Mention somebody who could help in coding this, because I see some other result also looking at this, which I can tell if it is true with some calculations, but for now, let's see if somebody can do the coding.
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@Siddharth Chakravarty – @Pi Han Goh , please help!
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@Vinayak Srivastava – I cannot mention the famous person who writes good coding solution in the Daily challenges, I don't remember the name either.
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@Siddharth Chakravarty – Carsten Kaminski, I mentioned him @Siddharth Chakravarty :)
@Siddharth Chakravarty – @Aryan Sanghi and @Páll Márton are good with Python :)
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@A Former Brilliant Member – Okay, I think it will be an easy code, I will see if I can program if I have the time.
@Siddharth Chakravarty – @Carsten Kaminski - Siddharth and Vinayak need help in coding this solution.
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@A Former Brilliant Member – Yes, Carsten. I could not spell his last name. Thanks @Percy Jackson
@Siddharth Chakravarty – The solutions here already demonstrated that no solution exists. Yes, should such an integer n exist, its last digit must be 2.
So you don't need to check for so many cases when applying rational root theorem.
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@Pi Han Goh – No, I asked for a code which would show the difference between the RHS and LHS for the first 100 terms ending with 2 starting from n=12 and then 22,32,42 and so on. :) Because I think it will help in another problem for which I noticed a pattern.
FUC* ME
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I'll be the first one to say..............Ewwwww................. @Lin Le
n 5 − 1 0 n 4 − 8 0 n 3 − 2 6 0 n 2 − 4 0 0 n = 2 4 2
- n divides 2 4 2 = 2 × 1 1 × 1 1
- odd n makes LHS is odd.
- If n = 2 , 2 2 , 2 4 2 , divide both sides by n , LHS is still even while RHS is odd.
∴ no solution.
There are no solutions for a n + b n = c n , when n ≥ 3
Yes, same thing :)
⇒
n
5
−
1
0
n
4
−
8
0
n
3
−
2
6
0
n
2
−
4
0
0
n
−
2
4
2
=
0
B y a p p l y i n g i n t e g e r c o r o l l a r y o f Rational root theorem
We can see there is no integral solution.
This solution is easier to understand, because it uses the Rational Root Theorem. +1 Brilliant, Helpful :)
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Thanks! I can never think about the method used by vinayak in 1.5 minute.But,mathematically that one is creative and this one is calculative. However,if it is mentioned that i have to prove it. Then, i will surely try to use fermat's theorem or trival inequality. I have learned it to use trickily.Such as there will no positive choices and -2 can't be the option. I have just calculated 10 11^4 and 80 11^3 and it was clear that there will be no solutions.
@Vinayak Srivastava
,your solution is far better than mine but,not always you will get question of no solution.You can use rational root
theorem to solve many prmo questions.
It is a bit calculative but you can easily learn it. :-)
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I don't know this theorem.
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Go to wiki, it will be very useful to solve prmo questions.
In fact, I would have never been able to factorize it correctly if I were not the creator of the problem :)
The polynomial f ( X ) = X 5 − 1 0 X 4 − 3 4 0 X 3 − 4 0 0 X − 2 4 2 is irreducible over the integers, using Eisenstein's Irreducibility Criterion (the prime 2 divides 1 0 , 3 4 0 , 4 0 0 , 2 4 2 , but 2 2 does not divide 2 4 2 ), and so f ( X ) has no rational roots, let alone positive integer ones.