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A355778
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Numbers k such that both k and k^2 + 2 can be written as the sum of two nonzero squares.
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0
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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PROG
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(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
(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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