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From: bhaskar@cvl.UUCP
Newsgroups: net.math
Subject: Re: Nifty math problem
Message-ID: <1305@cvl.UUCP>
Date: Sat, 15-Mar-86 15:37:24 EST
Article-I.D.: cvl.1305
Posted: Sat Mar 15 15:37:24 1986
Date-Received: Mon, 17-Mar-86 03:27:32 EST
References: <2222@jhunix.UUCP>
Distribution: net
Organization: Center for Automation Research, Univ. of Md.
Lines: 45
Summary: Solution to "Nifty math problem".

In article <2222@jhunix.UUCP>, ins_adsf@jhunix.UUCP (David S Fry) writes:
> 
> 	Here's a nifty yet simple number theory problem.  Prove that there
> is a unique palindromic prime number with an even number of digits.  Generalize
> to other bases.
> 


Let b ( > 1 )  be the base under consideration and let the notation

S{p <= i <= q, a[i]} represent the sum a[p] + a[p+1] +...+ a[q] and x^y

represent x_to_power_y.

Let the number sought, x, be represented by 

   a[n-1] a[n-2] ...a[1] a[0] a[0] a[1]... a[n-2] a[n-1] in a b-ary base. 

Assume that n > 1. 

Then value(x)  

= S{ 0 <= i <= n-1 , a[i] * ( b^(n-i-1) + b^(n+i)) }

= S{ 0 <= i <= n-1 , a[i] * b^(n-i-1) * (1 + b^(2i+1)) }

= S{ 0 <= i <= n-1 , a[i] * b^(n-i-1) * (1 + b) * (1-b+b^2-... +b^2i) }.

Thus x cannot be prime for it has b+1 as factor.

Thus we must have n = 1  and the number sought is of the form a[0] a[0].

Then, value(x) = a[0] * ( 1 + b ) .

Since x is prime, we must have a[0] = 1 and 1+b = x.

Thus the only "even length palindromic primes" are of form 11 in a base

b, where b+1 is itself prime. For example, eleven in base 10 or three in

base 2.

In particular, we cannot have such primes as described, in any odd-valued 

base b ( b > 1 ).