[seqfan] Re: A (new) problem
Vladimir Shevelev
shevelev at bgu.ac.il
Mon Apr 26 18:19:52 CEST 2010
Very thanks, Max!
Your proof is simple and excellent!
This an illustration of a known principle: "A good formulation of a problem is a half-solution".
Since you obtained a proof of my problem and many terms of the sequence, then I ask you to enter it into OEIS (together with your proof as a main comment).
Vladimir
----- Original Message -----
From: Max Alekseyev <maxale at gmail.com>
Date: Monday, April 26, 2010 18:39
Subject: [seqfan] Re: A (new) problem
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> On Sun, Apr 25, 2010 at 12:40 PM, Charles Greathouse
> <charles.greathouse at case.edu> wrote:
> > Interesting question. Essentially: is there some k for which all
> > positive integers can be written as the sum of an a-polygonal
> and a
> > b-polygonal number, for 3 <= a <= b <= k?
>
> There is no such k. In other words, Vladimir's sequence is infinite.
>
> Proof.
>
> Assume that such k exists. Then any positive integer q can be written
> as the sum of the m-th a-gonal and the n-th b-gonal number for some
> nonnegative integers m,n and 3 <= a <= b <= k. That is,
>
> q = ((a-2) m^2 - (a-4) m) / 2 + ((b-2) n^2 - (b-4) n) / 2
>
> implying that
>
> 8(a-2)(b-2)q + (b-2)(a-4)^2 + (a-2)(b-4)^2 = (b-2) (2(a-2)m - (a-4))^2
> + (a-2) (2(b-2)n - (b-4))^2.
>
> For every pair (a,b), 3 <= a <= b <= k, let us fix a
> prime number
> p_{a,b} > k such that the following three conditions hold:
> (i) -(a-2)/(b-2) is not a square modulo p_{a,b};
> (ii) all p_{a,b} are pairwise distinct;
> and
> (iii) the sum of reciprocals 1/p_{a,b} is less than 1.
>
> By Chinese Remainder Theorem, (ii) implies that the system of
> congruences:
> 8(a-2)(b-2)q + (b-2)(a-4)^2 + (a-2)(b-4)^2 == 0 (mod p_{a,b})
>
> where a,b run over all pairs 3 <= a <= b <= k, has a
> solution w.r.t.
> q. Moreover, (iii) implies that there exists such solution (positive
> integer) q that the above congruences do not hold modulo any
> p_{a,b}^2.
>
> By our assumption, such q is representable as the sum of
> the m-th
> a-gonal and the n-th b-gonal number for some nonnegative
> integers m,n
> and 3 <= a <= b <= k, implying that
>
> (b-2) (2(a-2)m - (a-4))^2 + (a-2) (2(b-2)n - (b-4))^2 == 0 (mod
> p_{a,b}).
> The condition (i) implies that both (2(a-2)m - (a-4)) and (2(b-
> 2)n -
> (b-4)) are divisible by p_{a,b} and thus
>
> 8(a-2)(b-2)q + (b-2)(a-4)^2 + (a-2)(b-4)^2 == 0 (mod p_{a,b}^2).
>
> This contradiction proves that our assumption on the existence
> of k
> does not hold.
>
> Max
>
>
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>
Shevelev Vladimir
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