cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A370721 Positive integers k == 2 (mod 4) such that the parametric Pell-type equation x^2 - m*x*y + y^2 = m^2 + k has no integer solutions (x,y) for all integer m >= 1.

Original entry on oeis.org

14, 94, 114, 118, 154, 158, 214, 238, 254, 294, 358, 414, 478, 574, 594, 598, 614, 654, 658, 694, 718, 758, 790, 814, 834, 862, 874, 878, 934, 958, 994, 1014, 1054, 1106, 1174, 1198, 1294, 1414, 1434, 1454, 1486, 1494, 1498, 1558, 1634, 1678, 1738, 1774, 1794, 1834, 1894, 1918, 1978
Offset: 1

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Author

Orlov Nikita and Nikolay Osipov, Mar 07 2024

Keywords

Comments

For a positive integer k == 2 (mod 4), it suffice to check that the equation x^2-m*x*y+y^2 = m^2+k (*) has no integer solutions (x,y) for all integer m with 1 <= m <= k/2 (see references for the proof of some similar assertions). This condition can be verified by an algorithm similar to brute force search for the general Pell equation x^2-Dy^2 = N (see, for example, sect. 4.4.5 in: Andreescu T., Andrica D. Quadratic Diophantine Equations. New York: Springer, 2015).
Also, the equation (*) has no integer solutions (x,y) for all integer m >= 1 when k = 1 or k = 4. For any other positive integer k, the equation (*) has integer solutions (x,y) for infinitely many integers m >= 1.

References

  • N. Osipov, A Pell-Type Diophantine Equation, Amer. Math. Monthly, 128 (2021), p. 858-860.
  • N. Osipov, A Pell-type Equation in Disguise, Amer. Math. Monthly, 129 (2022), p. 389-390.

Crossrefs

Cf. A371957 (for the equation x^2-m*x*y+y^2=-m^2-k).

Programs

  • Maple
    check:=proc(k) local flag,y,m,yy,mm; flag:=0;
    for y from 0 to evalf(2*sqrt((k+1)/3)+1) while flag=0 do
    if issqr(-3*y^2+4*k+4)=true then flag:=1; mm:=1; yy:=y; fi; od;
    for m from 3 to k/2 while flag=0 do
    if m mod 4<>2 then for y from 0 to evalf(sqrt((m^2+k)/(m+2)))+1 while flag=0 do
    if issqr((m^2-4)*y^2+4*(m^2+k))=true then flag:=1; mm:=m; yy:=y; fi; od; fi; od;
    if flag=0 then return 0 else return [mm,yy]; fi; end proc:
    for k from 1 to 2000 do if k mod 4=2 and check(k)=0 then print(k); fi; od:
  • Pascal
    (* see link *)

Extensions

Edited by Nikolay Osipov, Jun 11 2024