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

[c.zizka at email.cz]

> ------------ Původní zpráva ------------
> Od:  <franktaw at netscape.net>
> Předmět: [seqfan] Re: A006451 = n such that n*(n+1)/2+1 is square?
> Datum: 10.10.2009 13:05:35
> ----------------------------------------
> 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.
> 
> Franklin T. Adams-Watters
> 
> 
> -----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.
> 
> 
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