[seqfan] Re: Fwd: Proof of the A186080 conjecture
Max Alekseyev
maxale at gmail.com
Mon Mar 7 08:31:43 CET 2011
Matevž,
You use expansion:
> (n*10^k + C)^4 = n^4*10^(4k) + 4*10^(3k)*n^3*C + 6*10^(2k)*n^2*C^2 + 4*10^k*n*C^3 + C^4
In my counterexample, n=5, k=1, C=9 it becomes:
= 5^4*10^4 + ... + 9^4.
Then you say :
> We want the first and the last number to be the same := n^4*10^(4k) = C^4*10^d
> (where d is the offset that we use to compare those two numbers),
You should really care about *digits* not numbers. I would rather
expect you to say something like "the first digit of n^4 must be equal
to the last digit of C^4".
But even in this form, it should be first proved that the decimal
representation of (n*10^k + C)^4 starts with the first digit of n^4.
And this is still not the case in my counterexample. 59^4=12117361
does not start with 6 (the first digit of 5^4=625).
Regards,
Max
On Mon, Mar 7, 2011 at 3:32 AM, Matevž Markovič
<matevz.markovic.v at gmail.com> wrote:
> Hy!
>
> Sorry for another reply but I thought about the flaw you mentioned and I
> came to the following conclusion: something does not fit.
>
> "..., the decimal representation of (n*10^k + C)^4 does not
>
> necessary starts with n^4 (e.g., 59^4=12117361 does not start with
> 5^4=625)."
>
> I found out, that your counter-example is not valid, because 59 can be
> written as (5*10 + 9)^4, which is, when you use the binominal coeficient,
> (5*10)^4 + ___ , not 5^4 + ___. I think that the real problem with my proof
> is not corectness itself, but making the proof understandable.
>
> I will cite my proof in full:
>
> "Proof of A186080 conjecture, Matevz Markovic, Feb 21 2011: (Start)
>
> Natural number (n*10^k + C)^4, where n,k,C are all natural numbers, is
> palindromic only when the number (n*10^k + C) is of the form 100...0001
>
> First we should proove it for the first number of the form 100..0001 (for
> 11)
>
> 11 = 1*10^k + C
> (1*10^1 + 1)^4 = 10^4 + 4*10^3*1 + 6*10^2*1^2 + 4*10^1*1^3 + 1^4
>
> 10^4 = 1^4*10^d. OK, first and the last number are the same
> 4*10^3*1 = 4*10^1*1^3 *10^d. OK, second and the fourth number are the same.
> This conjecture is valid for 11.
>
> Is it valid for every natural number X?
>
> X = n*10^k + C (we expect this number to be of the form 1*10^k + 1)
>
> (n*10^k + C)^4 = n^4*10^(4k) + 4*10^(3k)*n^3*C + 6*10^(2k)*n^2*C^2 +
> 4*10^k*n*C^3 + C^4
>
> We want the first and the last number to be the same := n^4*10^(4k) =
> C^4*10^d (where d is the offset that we use to compare those two numbers),
>
> therefore n^4 = C^4 ==> n=C //they are the same only if n=C
>
> This number is a palindrome only if (n^4) is one digit long (< 10).
> Therefore, it is true only for 1, because 2^4>10 (every natural number other
> than 1 is at least 2 digits long when quadrupaled); n = 1 ==> C=1 ==> every
> natural number X, for which X^4 is a palindrome, is of the form (1*10^k +
> 1).
>
> End of the proof"
>
> Yours,
> Matevž Markovič
>
>
>
> --
> Matevž Markovič
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>
More information about the SeqFan
mailing list