A366861 Largest elements of triples (a, b, c) with a > b > c = 1 such that ab+c, ac+b and bc+a are perfect squares.

9, 12, 19, 24, 28, 33, 40, 51, 52, 57, 60, 64, 73, 84, 96, 99, 105, 112, 116, 129, 136, 144, 145, 156, 163, 177, 180, 184, 201, 204, 217, 220, 231, 232, 243, 249, 260, 264, 273, 276, 280, 288, 289, 291, 295, 312, 336, 339, 352, 355, 364, 369, 376, 385, 388, 393, 420, 424, 435, 441, 448

Offset

1,1

Comments

Equivalently, numbers a such that there exists 1 < b < a such that ab+1 and a+b are perfect squares.

When a = b+2, then ab+1 = (b+1)^2, so there remains only the condition that a+b = 2b+2 = (2m)^2 for some integer m >= 2 (since the square must be even and b > c = 1, so m > 1). Thus, all a = 2*m^2 + 1 with m >= 2, A058331(2..) = 9, 19, 33, ..., are in this sequence.

Links

David A. Corneth, Table of n, a(n) for n = 1..10000

Wendy Appleby, Find all triples..., Number Theory group on LinkedIn.com, Oct 25 2023.

Example

a(1) = 9 is in the sequence because for b = 7, we have that ab+1 = 8^2 and a+b = 4^2 both are squares (whence ab+c, ac+b and bc+a are squares for c = 1).
a(2) = 12 is in the sequence because for b = 4, we have ab+1 = 7^2 and a+b = 4^2 both squares.

Prog

(PARI) is(n)=for(b=2, n-1, issquare(n*b+1)&&issquare(n+b)&&return(1))
select(is, [1..456])
(PARI) is(n) = {
	for(i = sqrtint(n+2)+!issquare(n+2), sqrtint(2*n-1),
		if(issquare(n*(i^2-n)+1),
			return(1)
		)
	);
	0
} \\ faster than prog above \\ David A. Corneth, Jan 17 2024

Crossrefs

Cf. A000290 (the squares), A058331.

Sequence in context: A072702 A357185 A366865 * A364343 A157973 A057577

Adjacent sequences: A366858 A366859 A366860 * A366862 A366863 A366864

Keyword

nonn,easy

Author

M. F. Hasler, Oct 25 2023

Status

approved