login
Number of pairs (a,b), 1 <= a <= b <= n, such that (a+b)^2+n^2 is a square.
3

%I #25 Feb 15 2023 13:45:00

%S 0,0,2,1,0,3,0,4,4,0,0,11,0,0,10,8,0,7,0,17,18,0,0,28,0,0,10,16,0,19,

%T 0,15,18,0,6,33,0,0,14,42,0,35,0,16,42,0,0,77,0,0,18,19,0,19,24,53,20,

%U 0,0,120,0,0,60,29,30,34,0,25,24,12,0,114,0,0,46,28,18,27,0,103,28,0,0,140

%N Number of pairs (a,b), 1 <= a <= b <= n, such that (a+b)^2+n^2 is a square.

%C Number of cuboids of maximal side length n having integral shortest path going on the surface from one vertex to the opposite one.

%C Number of subsets {a,b} of {1..n} such that (a+b,n) form the shorter two legs of a Pythagorean triple.

%C For all primes p, p > 3: a(p)=0 (this directly follows from SierpiƄski's proof that one of the shorter sides of a Pythagorean triple must be a multiple of 3, and one must be a multiple of 4). - _Michael Turniansky_, Jul 27 2016

%H Antti Karttunen, <a href="/A143714/b143714.txt">Table of n, a(n) for n = 1..4096</a>

%H Project Euler, <a href="http://projecteuler.net/index.php?section=problems&amp;id=86">Problem 86: Cuboid route</a>

%e For n=3, we have the 3 X 3 X 1 and the 3 X 2 X 2 cuboid, for which the shortest path on the surface from one vertex to the opposite is of integral length sqrt(3^2 + (2+2)^2) = sqrt(3^2 + (3+1)^2) = 5.

%e For n=4, there is the 4 X 2 X 1 cuboid having this property.

%e For n=1,2 and 5 there is no cuboid having this property, i.e., no s >= 2, s <= 2n such that s^2 + n^2 would be a square.

%o (PARI) A143714(M)=sum(a=1,M,sum(b=a,M,issquare((a+b)^2+M^2)))

%Y Cf. A143715 (partial sums).

%K easy,nonn

%O 1,3

%A _M. F. Hasler_, Aug 29 2008