A343913 Positive integers m such that 2*m^2 - 1 = x^4 + y^4 for some nonnegative integers x and y with |x - y| > 1.

71, 347, 1193, 2139, 2709, 17823, 18337, 26057, 32847, 34037, 65793, 87519, 159541, 245573, 383037, 421957, 489731, 520547, 574841, 800589, 1291333, 2010341, 2113003, 2990187, 4528667, 7430553, 8284063, 8402417, 8520567, 9220519, 9865989, 10621507, 11961043, 12335203, 16405581, 17648561, 22224647, 22918853, 24171273

Offset: 1

Author

Zhi-Wei Sun, May 03 2021

Keywords

• nonn

Comments

Conjecture: The sequence has infinitely many terms.

Clearly all the terms must be odd and not divisible by 5. Note also that 2*(n^2+n+1)^2 - 1 = n^4 + (n+1)^4.

See also A343917 for a similar conjecture.

Examples

			a(1) = 71, and 2*71^2 - 1 = 10^4 + 3^4 with |10 - 3| > 1.
a(53) = 99532937, and 2*99532937^2 - 1 = 19813611095691937 = 11337^4 + 7576^4 with |11337 - 7576| > 1.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..112 (all terms < 10^10; first 53 terms from Zhi-Wei Sun)

Crossrefs

Cf. A000290, A000583, A003336, A056220, A343917.

Programs

Maple

N:= 10^18: # for all terms <= sqrt(N)
R:= {}: count:= 0:
for x from 1 while 2*x^4 < 2*N-1 do
  for y from x+3 by 2 do
    v:= (x^4 + y^4 + 1)/2;
    if v > N then break fi;
    if issqr(v) then
      m:= sqrt(v);
      if not member(m,R) then
         count:= count+1; R:= R union {m};
      fi fi
od od:
sort(convert(R,list)); # Robert Israel, May 04 2021

Mathematica

QQ[n_]:=IntegerQ[n^(1/4)];
n=0;Do[Do[If[QQ[2*m^2-1-(2x)^4]&&Abs[2x-(2*m^2-1-(2x)^4)^(1/4)]>1,n=n+1;Print[n," ",m];Goto[aa]],{x,0,((2m^2-1)^(1/4))/2}];Label[aa],{m,1,25000000}]