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Squarefree numbers m congruent to 1 modulo 4 such that the fundamental unit of the field Q(sqrt(m)) has the form x+y*sqrt(m) with x, y integers.
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%I #12 Mar 09 2017 10:11:49

%S 17,33,37,41,57,65,73,89,97,101,105,113,129,137,141,145,161,177,185,

%T 193,197,201,209,217,233,241,249,257,265,269,273,281,305,313,321,329,

%U 337,345,349,353,373,377,381,385,389,393,401,409,417,433,449,457,465,473,481,485,489,497,505,521,537,545,553,557,561,569,573

%N Squarefree numbers m congruent to 1 modulo 4 such that the fundamental unit of the field Q(sqrt(m)) has the form x+y*sqrt(m) with x, y integers.

%C Squarefree integers m congruent to 1 modulo 4 such that the minimal solution of the Pell equation x^2 - d*y^2 = +-4 has both x and y even.

%C The sequence contains the squarefree numbers congruent to 5 modulo 8 that are not in A107997.

%C This sequence union A107997 = A039955.

%C This sequence contains all numbers of the form 4*k^2+1 (k > 1) that are squarefree.

%D Z. I. Borevich and I. R. Shafarevich. Number Theory. Academic Press. 1966.

%H Keith Matthews, <a href="http://www.numbertheory.org/php/unit.html">Finding the fundamental unit of a real quadratic field</a>

%e 33 is in the sequence since the fundamental unit of the field Q(sqrt(33)) is 23+4*sqrt(33).

%e 53 is not in the sequence since the fundamental unit of the field Q(sqrt(53)) is 3+omega, where omega = (1+sqrt(53))/2.

%Y Cf. A107997, A039955, A107998, A197170.

%K nonn

%O 1,1

%A _Emmanuel Vantieghem_, Mar 07 2017