A064716 Smallest member of three consecutive numbers each of which is the sum of two nonzero squares (not necessarily different).

72, 232, 288, 520, 584, 800, 808, 1096, 1152, 1224, 1312, 1600, 1664, 1744, 1800, 1872, 1960, 2248, 2312, 2384, 2592, 2600, 2824, 3328, 3392, 3528, 3600, 4112, 4176, 4328, 4624, 5120, 5328, 5408, 5904, 6056, 6120, 6272, 6352, 6408, 6568, 6920, 8080

Offset: 1

Author

Robert G. Wilson v, Oct 13 2001

Keywords

• nonn

Comments

a(n) == 0 (modulo 4) since no integer == 3 (modulo 4) can be represented as the sum of two squares.

This sequence has as a subsequence 72, 288, 800, 1800, ... which is 8 * (triangular numbers)^2. Proof: If x = 8*(n(n+1)/2)^2 then x = (n(n+1))^2 + (n(n+1))^2, x+1 = ((n-1)(n+1))^2 + (n(n+2))^2 and x+2 = (n^2+n-1)^2 + (n^2+n+1)^2. See A254371 - Joshua Zucker, Nov 01 2002

From Altug Alkan, Apr 13 2016: (Start)

If n is in this sequence, so is n*(n+2). Proof:

If n is in this sequence, then n = a^2 + b^2, n+1 = c^2 + d^2, n+2 = e^2 + f^2 for a, b, c, d, e, f being nonzero integers.

So, n*(n+2) = (a^2 + b^2)*(e^2 + f^2) = (a*e + b*f)^2 + (a*f - b*e)^2. Note that a*f cannot be equal to b*e because of their definitions.

n*(n+2) + 1 = n^2 + 2*n + 1 = (n+1)^2. Since we know that n mod 4 = 0, then n+1 cannot be of the form 2*k^2, that is, c and d must be different. So (n+1)^2 is the sum of two nonzero squares because n+1 = c^2 + d^2.

n*(n+2) + 2 = (n+1)^2 + 1, that is obviously the sum of two nonzero squares.

So if n is in this sequence, then n*(n+2), n*(n+2) + 1 and n*(n+2) + 2 are the sums of two nonzero squares, that is n*(n+2) must also be member of this sequence.

Note that it can be produced by repeating of this result and n*(n+2)*(n*(n+2)+2)*(n*(n+2)*(n*(n+2)+2)+2)... is always a member, if n is a member. (End)

For k > 0, 25*k^2*(10*k+2)^2 and 8*A001080(k)^2 are terms. - Jinyuan Wang, Feb 23 2019

Examples

			72 = 6^2 + 6^2, 73 = 3^2 + 8^2, 74 = 5^2 + 7^2.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• W. Allen Whitworth, Problem 356, The Mathematical Gazette, Vol. 1, No. 20 (Mar. 1900), p. 338.

Crossrefs

Cf. A000404, A001080.

Cf. A254371 \ {0, 8} (a subsequence).

Programs

Maple

N:= 10000: # to get all terms <= N
S:= {seq(seq(a^2+b^2, b=1..floor(sqrt(N+2-a^2))),a=1..floor(sqrt(N+2)))}:
sort(convert(S intersect map(`-`,S,1) intersect map(`-`,S,2),list)); # Robert Israel, Apr 14 2016

Mathematica

a = Table[n^2, {n, 1, 100}]; c = {}; Do[ c = Append[c, a[[i]] + a[[j]]], {i, 1, 100}, {j, 1, i} ]; c = Union[c]; c[[ Select[ Range[ Length[c] - 2], c[[ # ]] + 2 == c[[ # + 2 ]] & ]]]
Select[Range@ 8080, AllTrue[# + {0, 1, 2}, Length[ PowersRepresentations[#, 2, 2] /. {0, } -> Nothing] > 0 &] &] (* _Michael De Vlieger, Apr 13 2016, Version 10 *)

PARI

is(n)= for( i=1, #n=factor(n)~%4, n[1, i]==3 && n[2, i]%2 && return); n && ( vecmin(n[1, ])==1 || (n[1, 1]==2 && n[2, 1]%2));
lista(nn) = {for(n=1,nn,if(is(n)==1&&is(n+1)==1&&is(n+2)==1,print1(n,", ")))}; \\ Jinyuan Wang, Feb 23 2019