A024411 - OEIS

A024411

Short leg of more than one primitive Pythagorean triangle.

20, 28, 33, 36, 39, 44, 48, 51, 52, 57, 60, 65, 68, 69, 75, 76, 84, 85, 87, 88, 92, 93, 95, 96, 100, 104, 105, 108, 111, 115, 116, 119, 120, 123, 124, 129, 132, 133, 135, 136, 140, 141, 145, 147, 148, 152, 155, 156, 159, 160, 161, 164, 165, 168, 172, 175, 177, 180, 183, 184

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

OFFSET

1,1

COMMENTS

Every term is composite. - Clark Kimberling, Feb 04 2024

Proof by contradiction: let p prime be the short leg. Then p^2 + b^2 = c^2 i.e., p^2 = (c - b) * (c + b). Then (c - b, c + b) in {(1, p^2), (p, p)}. If (c - b, c + b) = (p, p) then c = p and b = 0 which is impossible. Hence there is at most one solution for (c - b, c + b). A contradiction. - David A. Corneth, Feb 04 2024

LINKS

Ray Chandler, Table of n, a(n) for n = 1..10000

Ron Knott, Pythagorean Triples and Online Calculators

MATHEMATICA

aa=1; s=""; For[a=1, a<=10^2, For[b=a+1, ((b+1)^2-b^2)<=a^2, c=(a^2+b^2)^0.5; If[c==Round[c]&&GCD[a, b]==1, If[a==aa, s=s<>ToString[a]<>", "]; If[a!=aa, aa=a, aa=1]]; b++ ]; a++ ]; s (* Vladimir Joseph Stephan Orlovsky, Apr 29 2008 *)

PROG

(PARI)

is(n) = {

my(d = divisors(n^2), q = 0, b, c);

for(i = 1, #d\2,

if(!bitand(d[#d + 1 - i] - d[i], 1),

c = (d[i] + d[#d + 1 - i])/2;

b = d[#d + 1 - i] - c;

if(gcd(n, b) == 1 && n < b,

q++;

if(q >= 2,

return(1)