[seqfan] Fwd: Proof of the A186080 conjecture

[Matevž Markovič]

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