# [seqfan] Re: A006451 = n such that n*(n+1)/2+1 is square?

franktaw at netscape.net franktaw at netscape.net
Sat Oct 10 13:54:07 CEST 2009

```The answer is 1/8.

I.e., there are no solutions for k == 1/8 (mod p), except k == 1/8 (mod
p^2), where p is any prime == 3 or 5 mod 8.  But my conjecture is not
quite correct; one must extend to higher powers of p; for such p, there
are no solutions for k == 1/8 (mod p^3), unless k == 1/8 mod (p^4),
etc.  (The first counter-example to the original conjecture is k = 422,
where 8k-1 = 15^3.)

This suggests the following algorithm: factor 8k-1.  Discard any square
factors (cf. A007913).  If any prime == 3 or 5 (mod 8) remains, there
are no solutions.  (This is certain.)  If no such prime factor remains,
there are solutions.  (This is conjecture).

-----Original Message-----
From: franktaw at netscape.net

There are no solutions for k == 2 (mod 3), except k == 8 (mod 9).
There are no solutions for k == 2 (mod 5), except k == 22 (mod 25).
There are no solutions for k == 7 (mod 11), except k == 106 (mod 121).
There are no solutions for k == 5 (mod 13), except k == 148 (mod 169).
There are no solutions for k == 12 (mod 19), except k == 316 (mod 361).
There are no solutions for k == 11 (mod 29), except k == 736 (mod 841).

This covers all k up to 100.

The primes in this list, to this point, are
http://research.att.com/~njas/sequences/A001122, primes with primitive
root 2.  I think the correct sequence is primes for which 2 is a
quadratic non-residue, which is to say primes congruent to 3 or 5 mod
8.  This makes sense; the factor of 2 in n(n+1)/2 takes the squares to
non-squares (we can complete the square n(n+1) = (n+1/2)^2-1/4, which
can evaluated mod p for odd p), excluding the multiples of p; there
will be one solution mod p^2 so that multiples of p^2 appear in the
list of sums.  The first prime with 2 as a non-residue but not a
primitive root is 43, and I would expect there to be an expression of
this form for 43.

I'm too tired right now to try to determine what the critical residues
are mod p and p^2.

I will conjecture that these conditions are sufficient to determine
which k have solutions.  How large a prime you have to check to ensure
that k does not fail any of these tests is not obvious to me.

Definitely not trivial, and definitely worth including.

-----Original Message-----
From: Joerg Arndt <arndt at jjj.de>
...

Some Diophantinista may want to enter seq
k such that n*(n+1)+k can be square.
(might be unworthy if trivial for some reason).
Quick & dirty:
?
for(k=0,20,q=0;for(n=1,10000,t=n*(n+1)/2+k;if(issquare(t),q=1;break()));

if(q,
print1(k,", ")))
0, 1, 3, 4, 6, 8, 9, 10, 13, 15, 16, 19,
seq. not in OEIS, neither complement,
neither for negative k.

```