OFFSET
1,1
COMMENTS
Pythagorean primes, i.e., primes of the form p = 4k+1 = A002144(n), have exactly one representation as sum of two squares: A002144(n) = x^2+y^2 = A002330(n+1)^2+A002331(n+1)^2. The corresponding (primitive) integer-sided right triangle with sides { 2xy, |x^2-y^2| } = { A002365(n), A002366(n) } has area xy|x^2-y^2| = a(n). For n>1 this is a(n) = 30*A068386(n).
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
FORMULA
EXAMPLE
The following table shows the relationship between several closely related sequences:
Here p = A002144 = primes == 1 (mod 4), p = a^2+b^2 with a < b;
with {c,d} = {t_2, t_3}, t_4 = cd/2 = ab(b^2-a^2).
---------------------------------
p a b t_1 c d t_2 t_3 t_4
---------------------------------
5 1 2 1 3 4 4 3 6
13 2 3 3 5 12 12 5 30
17 1 4 2 8 15 8 15 60
29 2 5 5 20 21 20 21 210
37 1 6 3 12 35 12 35 210
41 4 5 10 9 40 40 9 180
53 2 7 7 28 45 28 45 630
MATHEMATICA
Reap[For[p = 2, p < 500, p = NextPrime[p], If[Mod[p, 4] == 1, area = x*y/2 /. ToRules[Reduce[0 < x <= y && p^2 == x^2 + y^2, {x, y}, Integers]]; Sow[area]]]][[2, 1]] (* Jean-François Alcover, Feb 04 2015 *)
PROG
(PARI) forprime(p=1, 499, p%4==1 | next; t=[p, lift(-sqrt(Mod(-1, p)))]; while(t[1]^2>p, t=[t[2], t[1]%t[2]]); print1(t[1]*t[2]*(t[1]^2-t[2]^2)", "))
(PARI) {Q=Qfb(1, 0, 1); forprime(p=1, 499, p%4==1|next; t=qfbsolve(Q, p); print1(t[1]*t[2]*(t[1]^2-t[2]^2)", "))} \\ David Broadhurst
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, Feb 24 2009
STATUS
approved