|
| |
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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(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 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 * A335233 A180103 A102803
Adjacent sequences: A092538 A092539 A092540 * A092542 A092543 A092544
|
|
|
KEYWORD
|
nonn,uned
|
|
|
AUTHOR
|
Robin Garcia, Apr 08 2004
|
|
|
STATUS
|
approved
|
| |
|
|
