A091459 -id:A091459

Search: a091459 -id:a091459

Displaying 1-2 of 2 results found.

page 1

A092357

Smallest value of x=a+b+c+d (a,b,c,d positive integers) such that there are n different values of m=a^2+b^2=c^2+d^2, or 0 if no such x exists.

+10
2

18, 30, 36, 42, 54, 66, 78, 60, 80, 102, 72, 84, 138, 112, 90, 184, 154, 186, 452, 170, 126, 162, 196, 160, 120, 150, 652, 144, 692, 344, 318, 376, 266, 192, 200, 168, 272, 228, 304, 220, 472, 426, 234, 1052, 1076, 180, 474, 260, 368, 722, 584, 418, 534, 434

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

This is an infinite sequence because if x=4*p (p=any prime), the number of different n values of m is n=k for p=6k+/-1. I do not know if there is an x for every natural number n.

LINKS

Table of n, a(n) for n=1..54.

EXAMPLE

We denote m=a^2+b^2=c^2+d^2 by writing (a,b,c,d). Then:

x=18->(1,7,5,5)=50 for n=1

x=30->(1,12,8,9)=145 (3,11,7,9)=130 for n=2

x=36->(2,14,10,10)=200 (3,14,6,13)=205 (4,13,8,11)=185 for n=3

CROSSREFS

Cf. A091459 A090073.

KEYWORD

nonn

AUTHOR

Robin Garcia, Mar 18 2004

EXTENSIONS

More terms from Ray Chandler, Mar 26 2004

STATUS

approved

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).

+10
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.

KEYWORD

nonn,uned

AUTHOR

Robin Garcia, Apr 08 2004

STATUS

approved

page 1

Search completed in 0.004 seconds