[seqfan] Re: A (new) problem

[Max Alekseyev]

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