login
Numbers k such that 1 + (k-1)^2 and ((k-1)/2)^2 + ((k+1)/2)^2 = (1/2)*(k^2+1) are primes.
1

%I #25 Feb 16 2025 08:32:56

%S 3,5,11,15,25,85,95,121,131,171,181,205,231,261,271,315,441,445,471,

%T 545,571,595,715,751,781,861,921,951,1011,1055,1081,1095,1125,1151,

%U 1185,1315,1411,1421,1495,1615,1661,1701,2035,2051,2055,2065,2175,2261,2315

%N Numbers k such that 1 + (k-1)^2 and ((k-1)/2)^2 + ((k+1)/2)^2 = (1/2)*(k^2+1) are primes.

%C Conjectured to be infinite.

%D G. H. Hardy and W. M. Wright, Unsolved Problems Concerning Primes, Section 2.8 and Appendix 3 in An Introduction to the Theory of Numbers, 5th ed. Oxford, England: Oxford University Press, p. 19.

%D P. Ribenboim, The New Book of Prime Number Records, 3rd ed. New York: Springer-Verlag, pp. 206-208, 1996.

%H Robert Israel, <a href="/A102751/b102751.txt">Table of n, a(n) for n = 1..10000</a>

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/LandausProblems.html">Landau's Problems</a>.

%e 11 is a term because 10^2+1=101 and 5^2+6^2=(1/2)*(11^2+1)=61 are primes.

%p a:=proc(n) if isprime(1+(n-1)^2)=true and type((n^2+1)/2,integer)=true and isprime((n^2+1)/2)=true then n else fi end: seq(a(n),n=1..3000); # _Emeric Deutsch_, May 31 2005

%t Select[Range[2,2500], PrimeQ[1+(#-1)^2]&&PrimeQ[(1/2)*(#^2+1)]&] (* _James C. McMahon_, Jan 10 2024 *)

%Y Cf. A036468, A002496, A005574.

%K nonn,changed

%O 1,1

%A _Robin Garcia_, Feb 09 2005

%E More terms from _Emeric Deutsch_, May 31 2005