OFFSET
1,1
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.
LINKS
Orlov Nikita, Pascal program.
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:
PROG
(Pascal) (* see link *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Orlov Nikita and Nikolay Osipov, Mar 07 2024
EXTENSIONS
Edited by Nikolay Osipov, Jun 11 2024
STATUS
approved