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A050795
Numbers n such that n^2 - 1 is expressible as the sum of two nonzero squares in at least one way.
10
3, 9, 17, 19, 33, 35, 51, 73, 81, 99, 105, 129, 145, 147, 161, 163, 179, 195, 201, 233, 243, 273, 289, 291, 297, 339, 361, 387, 393, 451, 465, 467, 483, 489, 513, 521, 577, 579, 585, 611, 627, 649, 675, 721, 723, 739, 777, 801, 809, 819, 849, 883, 899, 915
OFFSET
1,1
COMMENTS
Analogous solutions exist for the sum of two identical squares z^2-1 = 2.r^2 (e.g. 99^2-1 = 2.70^2). Values of 'z' are the terms in sequence A001541, values of 'r' are the terms in sequence A001542.
Looking at a^2 + b^2 = c^2 - 1 modulo 4, we must have a and b even and c odd. Taking a = 2u, b = 2v and c = 2w - 1 and simplifying, we get u^2 + v^2 = w(w+1). - Franklin T. Adams-Watters, May 19 2008
If n is in this sequence, then so is n^(2^k), for all k >= 0. - Altug Alkan, Apr 13 2016
LINKS
E.-B. Escott, Query 2521, L'Intermédiaire des Mathématiciens, 10 (1903), 285. [Contains errors]
FORMULA
a(n) = 2*A140612(n) + 1. - Franklin T. Adams-Watters, May 19 2008
{k : A025426(k^2-1)>0}. - R. J. Mathar, Mar 07 2022
EXAMPLE
E.g. 51^2 - 1 = 10^2 + 50^2 = 22^2 + 46^2 = 34^2 + 38^2.
MATHEMATICA
t={}; Do[i=c=1; While[i<n&&c!=0, If[IntegerQ[Sqrt[n^2-1-i^2]], c=0; AppendTo[t, n]]; i++], {n, 3, 920}]; t (* Jayanta Basu, Jun 01 2013 *)
Select[Range@ 1000, Length[PowersRepresentations[#^2 - 1, 2, 2] /. {0, _} -> Nothing] > 0 &] (* Michael De Vlieger, Apr 13 2016 *)
PROG
(Python)
from itertools import islice, count
from sympy import factorint
def A050795_gen(startvalue=2): # generator of terms >= startvalue
for k in count(max(startvalue, 2)):
if all(map(lambda d: d[0] % 4 != 3 or d[1] % 2 == 0, factorint(k**2-1).items())):
yield k
A050795_list = list(islice(A050795_gen(), 20)) # Chai Wah Wu, Mar 07 2022
(PARI) select( {is_A050795(n)=#qfbsolve(Qfb(1, 0, 1), n^2-1, 2)}, [1..999]) \\ M. F. Hasler, Mar 07 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Patrick De Geest, Sep 15 1999
STATUS
approved