OFFSET
1,1
COMMENTS
Conjecture: The sequence has infinitely many terms.
It is easy to see that no term is divisible by 5. In the b-file we list all the 62 terms not exceeding 10^8.
Note that 2*(n^2+3)^2 - 2^4 = (n+1)^4 + (n-1)^4 with (n+1) - (n-1) = 2. This implies that any integer n > 4 can be written as x + y + 2^(z-1) with x,y,z positive integers such that x^4 + y^4 + (2^z)^4 is twice a square.
See also A343913 for a similar conjecture.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..62
EXAMPLE
a(1) = 284, and 2*284^2 - 2^4 = 20^4 + 6^4 with |20-6| > 2.
a(62) = 97077407, and 2*97077407^2 - 2^4 = 18848045899687282 = 11563^4 + 5583^4 with |11563-5583| > 2.
MATHEMATICA
QQ[n_]:=IntegerQ[n^(1/4)];
n=0; Do[Do[If[QQ[2*m^2-16-x^4]&&(2*m^2-16-x^4)^(1/4)-x>2, n=n+1; Print[n, " ", m]; Goto[aa]], {x, 0, (m^2-8)^(1/4)}]; Label[aa], {m, 3, 8041364}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 04 2021
STATUS
approved