%I A092541
%S A092541 50,65,85,125,130,170,185,221,250,305,325,338,425,410,425,481,578,610,
               725,
%T A092541 650,697,905,850,845,925,1037,1066,1325,1258,1250,1313,1450,1445,1517,
%U A092541 1586,1625,1810,2105,1885,2405,2050,2210,2210,2257,2465,2650,2525,2665
%N A092541 Minimal values of m=a^2+b^2=c^2+d^2 for each x=a+b+c+d (a,b,c,d positive 
               integers).
%C A092541 A general solution to m=a^2+b^2=c^2+d^2 for a known x=a+b+c+d is: c=(x(r-1)/
               2r)-a, d=(x+a(r-1))/(r+1) where r is a divisor of x/2. Thus x is 
               always even.
%C A092541 Theorem: a natural number p is prime if and only if there is never any 
               m=a^2+b^2=c^2+d^2 for x=a+b+c+d=2p. Proof: Then r=p and d=(2p+a(p-1))/
               (p+1) which is impossible. x is even,x>=18 and x is never 2p (p=any 
               prime). There are no other restrictions for the values of x. Thus 
               this is an infinite sequence and is another proof that there are 
               infinitely many primes of the form 4k+1. Proving that there are infinetely 
               many values of x with minimal m being sum of 2 squares in less than 
               4 ways would be a proof that there are infinitely many primes of 
               the form n^2+1 or 1/2(n^2*1)
%F A092541 minimal m= (1/2) (t^2+1)((x/2t)^2+1) if t is the greatest factor of x/
               2 <=floor(sqrt(x/2)) and t or x/2t are odd. Or minimal m=2(t^2+1)((x/
               4t)^2+1) if t is the greatest factor of x/2 <=floor(sqrt(x/2)) and 
               t and x/4t are even. Note that all minimal values are of the form 
               2^n(u^2+1)(v^2+1) n=-1 or 1
%e A092541 If x=28 minimal m= (1/2) (2^2+1)(7^2+1)=125
%e A092541 If x=32 minimal m=2(4^2+1)(2^2+1)=170
%e A092541 If x=96 m=2(6^2+1)(4^2+1)=1258
%e A092541 If x=100 m= (1/2) (5^2+1)(10^2+1)=1313
%Y A092541 Cf. A090073 A091459 A092357.
%Y A092541 Sequence in context: A109552 A007692 A025285 this_sequence A102803 A039473 
               A146170
%Y A092541 Adjacent sequences: A092538 A092539 A092540 this_sequence A092542 A092543 
               A092544
%K A092541 nonn,uned
%O A092541 1,1
%A A092541 Robin Garcia (verob99(AT)teleline.es), Apr 08 2004