

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).


1



50, 65, 85, 125, 130, 170, 185, 221, 250, 305, 325, 338, 425, 410, 425, 481, 578, 610, 725, 650, 697, 905, 850, 845, 925, 1037, 1066, 1325, 1258, 1250, 1313, 1450, 1445, 1517, 1586, 1625, 1810, 2105, 1885, 2405, 2050, 2210, 2210, 2257, 2465, 2650, 2525, 2665
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OFFSET

1,1


COMMENTS

A general solution to m=a^2+b^2=c^2+d^2 for a known x=a+b+c+d is: c=(x(r1)/2r)a, d=(x+a(r1))/(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(p1))/(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)


LINKS

Table of n, a(n) for n=1..48.


FORMULA

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


EXAMPLE

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


CROSSREFS

Cf. A090073, A091459, A092357.
Sequence in context: A206263 A007692 A025285 * A180103 A102803 A224559
Adjacent sequences: A092538 A092539 A092540 * A092542 A092543 A092544


KEYWORD

nonn,uned


AUTHOR

Robin Garcia, Apr 08 2004


STATUS

approved



