A104461 Number of instances of nonprimes m in Pythagorean triples x,y,z such that x^2 + y^2 = z^2. Except for 1, the number of instances of composite numbers m in Pythagorean triples.
0, 1, 1, 2, 2, 2, 4, 1, 5, 3, 2, 5, 4, 1, 7, 4, 2, 3, 4, 5, 4, 4, 2, 5, 7, 1, 5, 8, 4, 4, 8, 1, 10, 2, 4, 5, 5, 3, 5, 7, 4, 2, 14, 1, 7, 5, 8, 4, 5, 4, 5, 12, 2, 9, 4, 4, 5, 11, 4, 2, 13, 8, 1, 5, 7, 8, 5, 4, 4, 1, 5, 13, 2, 7, 9, 5, 8, 14, 2, 10, 5, 5, 10, 4, 5, 5, 8, 1, 5, 23, 2, 2, 5, 4, 6, 7, 6, 4, 8, 13
Offset: 1
Examples
For m=30 there are 5 Pythagorean triples that have a 30: 30, 224, 226 30, 72, 78 30, 40, 50 30, 16, 34 18, 24, 30
Programs
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PARI
\\ instances of m in Pythagorean triples using a direct method for x,y pythm3(m) = { local(m2,ln,j,j2=0,d,d2,q2,q,a,b,x,x1,x2,xx,y,y2,z,c,c2,r,f,str,stp); d=divisors(m^2); /* get the divisors of m^2 */ ln=length(d)-1; d2=q2=vector(ln); m2=m^2; if(m%2,r=1,r=0); for(j=1,ln, /* save only the both even r=0, both odd r=1 */ if(d[j]%2==r, if(m2/d[j]%2==r, j2++; d2[j2]=d[j]; q2[j2]=m2/d[j]; /* save m/factor to solve (z-y)(z+y) = m^2 */ ) ) ); x2=y2 = vector(20); for(j=1,j2, z=(d2[j] + q2[j])/2; y= z - d2[j]; if(y>0, c++; ) ); if(m%2==0,start=2;step=2,start=1;step=1); forstep(y=start,m-1,step, /* esolve when z is m */ x1 = (m2-y^2); if(issquare(x1), c2++; x2[c2]=floor(sqrt(x1)); /* save to later mask dupes */ y2[c2]=y; ) ); for(x=1,c2, /* mask the dupes routine */ for(y=x,c2, if(x2[x]==y2[y], ) ) ); return(c+c2/2) /* print total */} for(k=1,400,if(isprime(k)==0,print1(pythm3(k)", ")))
Formula
Consider Pythagorean triples x^2 + y^2 = z^2. We seek to find the total number of instances of an integer m being x or y or z. The solution for x or y is straightforward by considering appropriate lesser and greater pairwise factors, L, G of m^2 in z^2 - y^2 = (z-y)(z+y) = m^2. Then solve for z and y with the relations, z-y = L z+y = G 2z = L+G, z = (L+G)/2 where L and G are both even if m is even or both odd if m is odd. The number of L factors < m is the number of instances of x or y. The count of instances z=m is solved by trial on x^2 = m^2 - y^2.
Comments