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

Offset: 1

Author

David W. Wilson

Keywords

• nonn

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

Crossrefs

Cf. A020884, A024352.

Programs

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 *)

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)
				)
			)
		)
	); 0
} \\ David A. Corneth, Feb 04 2024