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A024411 Short leg of more than one primitive Pythagorean triangle.

%I A024411 #29 Feb 04 2024 18:29:09
%S A024411 20,28,33,36,39,44,48,51,52,57,60,65,68,69,75,76,84,85,87,88,92,93,95,
%T A024411 96,100,104,105,108,111,115,116,119,120,123,124,129,132,133,135,136,
%U A024411 140,141,145,147,148,152,155,156,159,160,161,164,165,168,172,175,177,180,183,184
%N A024411 Short leg of more than one primitive Pythagorean triangle.
%C A024411 Every term is composite. - _Clark Kimberling_, Feb 04 2024
%C A024411 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
%H A024411 Ray Chandler, <a href="/A024411/b024411.txt">Table of n, a(n) for n = 1..10000</a>
%H A024411 Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%t A024411 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 *)
%o A024411 (PARI)
%o A024411 is(n) = {
%o A024411 	my(d = divisors(n^2), q = 0, b, c);
%o A024411 	for(i = 1, #d\2,
%o A024411 		if(!bitand(d[#d + 1 - i] - d[i], 1),
%o A024411 			c = (d[i] + d[#d + 1 - i])/2;
%o A024411 			b = d[#d + 1 - i] - c;
%o A024411 			if(gcd(n, b) == 1 && n < b,
%o A024411 				q++;
%o A024411 				if(q >= 2,
%o A024411 					return(1)
%o A024411 				)
%o A024411 			)
%o A024411 		)
%o A024411 	); 0
%o A024411 } \\ _David A. Corneth_, Feb 04 2024
%Y A024411 Cf. A020884, A024352.
%K A024411 nonn
%O A024411 1,1
%A A024411 _David W. Wilson_