1,1

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.

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)

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

If x=28 minimal m= (1/2) (2^2+1)(7^2+1)=125

If x=32 minimal m=2(4^2+1)(2^2+1)=170

If x=96 m=2(6^2+1)(4^2+1)=1258

If x=100 m= (1/2) (5^2+1)(10^2+1)=1313

Cf. A090073 A091459 A092357.

Sequence in context: A109552 A007692 A025285 this_sequence A102803 A039473 A146170

Adjacent sequences: A092538 A092539 A092540 this_sequence A092542 A092543 A092544

nonn,uned

Robin Garcia (verob99(AT)teleline.es), Apr 08 2004