A355778 Numbers k such that both k and k^2 + 2 can be written as the sum of two nonzero squares.

40, 68, 72, 104, 148, 180, 320, 392, 468, 544, 612, 648, 720, 788, 832, 900, 936, 968, 1040, 1044, 1156, 1192, 1256, 1300, 1332, 1508, 1732, 1796, 1800, 1832, 1872, 1940, 2056, 2196, 2308, 2336, 2372, 2448, 2664, 2696, 2740, 2804, 2848, 2880, 3060, 3200, 3280

Offset: 1

Author

Angad Singh, Jul 16 2022

Keywords

• nonn

Comments

The terms in this sequence can be considered as a solution to the "near miss" problem which occurs frequently while solving Diophantine equations. It is known that if a number k can be written as the sum of two nonzero distinct squares then so can k^2 and k^2+1. Thus, finding numbers k such that k^2+2 satisfies the same property makes it quite interesting.

Examples

			40 is a term since 40 = 2^2 + 6^2 as well as 40^2 + 2 = 1602 = 9^2 + 39^2.
320 is a term since 320 = 8^2 + 16^2 as well as 320^2 + 2 = 102402 = 201^2 + 249^2.
		

Crossrefs

Cf. A000290, A000404, A000415.

Programs

PARI

is1(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)); \\ A000404
isok(k) = is1(k) && is1(k^2+2); \\ Michel Marcus, Jul 18 2022

Python

from itertools import count, islice
from sympy import factorint
def A355778_gen(startvalue=1): # generator of terms >= startvalue
    for n in count(max(startvalue, 1)):
        c = False
        for p in (f:=factorint(n)):
            if (q:= p & 3)==3 and f[p]&1:
                break
            elif q == 1:
                c = True
        else:
            if c or f.get(2, 0)&1:
                c = False
                for p in (f:=factorint(n**2+2)):
                    if (q:= p & 3)==3 and f[p]&1:
                        break
                    elif q == 1:
                        c = True
                else:
                    if c or f.get(2, 0)&1:
                        yield n
A355778_list = list(islice(A355778_gen(),30)) # Chai Wah Wu, Sep 14 2022