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%I #6 Mar 09 2013 13:21:06
%S 50,65,85,125,130,170,185,221,250,305,325,338,425,410,425,481,578,610,
%T 725,650,697,905,850,845,925,1037,1066,1325,1258,1250,1313,1450,1445,
%U 1517,1586,1625,1810,2105,1885,2405,2050,2210,2210,2257,2465,2650,2525,2665
%N 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 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 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 infinitely 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 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 If x=28 minimal m= (1/2) (2^2+1)(7^2+1)=125
%e If x=32 minimal m=2(4^2+1)(2^2+1)=170
%e If x=96 m=2(6^2+1)(4^2+1)=1258
%e If x=100 m= (1/2) (5^2+1)(10^2+1)=1313
%Y Cf. A090073, A091459, A092357.
%K nonn,uned
%O 1,1
%A _Robin Garcia_, Apr 08 2004
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