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A299068 Number of pairs of factors of n^2*(n^2-1) which differ by n.

%I A299068 #25 Feb 05 2018 02:55:21
%S A299068 3,4,8,7,11,6,10,12,11,9,9,9,13,22,12,7,7,11,21,28,9,7,17,14,13,14,13,
%T A299068 13,11,9,10,12,17,33,28,8,7,20,19,15,9,10,21,29,10,7,14,19,18,21,11,9,
%U A299068 16,44,46,14,7,9,15,9,9,18,40,24,18,8,9,30,18,17,11
%N A299068 Number of pairs of factors of n^2*(n^2-1) which differ by n.
%C A299068 The question arose when seeking triples of numbers for which the sum of the squares of any two is congruent to 1 modulo the third.
%C A299068 From _Robert Israel_, Feb 04 2018: (Start)
%C A299068 For n > 7, a(n)>= 7, as there are at least the following pairs:
%C A299068 (1,n+1), (n,2*n), (2*n,3*n), ((n^2-n)/2,(n^2+n)/2), (n^2-n,n^2), (n^2,n^2+n), and (3*n, 4*n) (if n is odd) or (n/2,3*n/2) (if n is even).
%C A299068 If k in A299159 is sufficiently large, then a(12*k-2)=7.  Dickson's conjecture implies there are infinitely many such k, and thus infinitely many n with a(n)=7. (End)
%H A299068 Robert Israel, <a href="/A299068/b299068.txt">Table of n, a(n) for n = 2..10000</a>
%p A299068 a:= n-> (s-> add(`if`(i+n in s, 1, 0), i=s))(
%p A299068         numtheory[divisors](n^2*(n^2-1))):
%p A299068 seq(a(n), n=2..100);  # _Alois P. Heinz_, Feb 01 2018
%t A299068 Array[With[{d = Divisors[# (# - 1)] &[#^2]}, Count[d + #, _?(MemberQ[d, #] &)]] &, 71, 2] (* _Michael De Vlieger_, Feb 01 2018 *)
%Y A299068 Cf. A299159.
%K A299068 nonn
%O A299068 2,1
%A A299068 _John H Mason_, Feb 01 2018