THE COUNT OF HERONIAN TETRAHEDRA IS BOUNDED BETWEEN TWO POWER LAWS ================Warren D. Smith Dec 2011================================= In this post, I will prove that the number of primitive Heronian tetrahedra ["Heronian" means tetrahedra with all sides, face-areas, and volume all integer, and "primitive" means gcd(sidelengths)=1] with diameter<N grows with N at least proportionally to N^0.4999. (Fred Lunnon helped somewhat, but I'd say this proof is 99% due to me.) It had previously been proven that the number of Heronian tetrahedra was infinite. However, I think all the previous proofs employed integer points on elliptic curves and therefore only yield a count-growth law of order loglog(N) roughly. While we know that loglogN goes to infinity, this has never actually been observed. The present power-law count-growth is the correct kind of growth, since a power-law upper bound can be shown trivially. What is the correct power? -- that is now the question. The PROOF is divided into parts I, II, II, IV: I. Solutions of "the 4.2.2 diophantine equation" x^4+y^4=u^4+w^4: ================================================================= This is discussed here by Tito Piezas: http://sites.google.com/site/tpiezas/013 A.Gerardin (about 1910?) 1-parameter family: (a+3*a^2-2*a^3+a^5+a^7)^4 + (1+a^2-2*a^4-3*a^5+a^6)^4 = (a-3*a^2-2*a^3+a^5+a^7)^4 + (1+a^2-2*a^4+3*a^5+a^6)^4. This shows that the number of primitive solutions (x,y,u,w) with max(x,y,u,w)<N grows at least as fast as order N^(1/7). The following 2-parameter solution is mentioned by Hardy & Wright: Intro to the theory of numbers (I think it is supposed to be a cleaned up version of Euler, but not sure where it ultimately came from). [E.g. see GH Hardy, EM Wright, JH Silverman: Intro to the theory of numbers, EQ13.7.10 and thereabouts] as "the simplest... but not in any sense complete": (a+b)^4+(c-d)^4 = (a-b)^4+(c+d)^4 where a = n*(m^2+n^2)*(18*m^2*n^2-m^4-n^4), b = 2*m*(m^6+10*m^4*n^2+m^2*n^4+4*n^6), c = 2*n*(4*m^6+m^4*n^2+10*m^2*n^4+n^6), d = m*(m^2+n^2)*(18*m^2*n^2-m^4-n^4). It shows that the number of primitive solutions (x,y,u,w) with max(x,y,u,w)<N grows at least as fast as order N^(2/7). J.Stegall 2-parameter family (sort of): (a^3+b)^4 + (1+a*d)^4 = (a^3+d)^4 + (1+a*b)^4, where d=2*a*c-b and {a,b,c} must satisfy a quadratic in b, namely b^2*c - 2*a*b*c^2 - a^4+3*a*2*c + 2*a^2*c^3 - 1 = 0. Solving this in the rationals entails making its discriminant a square, hence involves the elliptic curve z^2 = (1+a^4)*c+3*a^2*c^2-a^2*c^4 If we do the substitution a = (n+1)/(n-1), then one solution is c = 9*n^4/(2*n^8-2*n^6+n^4-2*n^2+2) from which we can compute other rational points. Then we can solve the quadratic for b trivially using quadratic formula. Just using this c-from-n formula without trying to generate other rational points on the elliptic curve(s), we would obtain a 1-parameter family of solutions also showing power-law growth, but even slower than Gerardin. Euler: found several solutions. Dickson Introduction 60-62. II. Certain kinds of "imperfect pythagorean boxes": =================================================== If x=p^2*q^2-r^2*s^2, y=2*p*q*r*s, z=p^2*r^2-q^2*s^2 then x^2+y^2 = (p^2*q^2+r^2*s^2)^2 and y^2+z^2 = (p^2*r^2+q^2*s^2)^2 are squares. [But x^2+z^2, in general, is nonsquare.] If, further, p^4+s^4=q^4+r^4 (see solutions in part I), then x^2+y^2+z^2 = (p^4+s^4)*(q^4+r^4) is a square too. Indeed, more generally, this is true if (p^4+s^4)*A^2=(q^4+r^4)*B^2 which means we can take any solution of p^4+s^4=q^4+r^4 and then multiply p and s by A and q and r by B, then the above x,y,z will still be such that x^2+y^2, y^2+z^2, and x^2+y^2+z^2 all will simultaneously be squares. Using the (I)solution from Hardy & Wright we see that this shows that the number of primitive (x,y,z) with max(x,y,z)<N satisfying these three simultaneous squareness conditions, grows with N at least as fast as order N^(0.4999). Proof: from (I) the number of (p,q,r,s) with max<M grows like M^(2/7) at least; the number of (A,B) with max<J grows like J^2, and then for (x,y,z) we have N of order M^4*J^4. Pick J=N^0.24999 and M=N^0.00001 to get M^4*J^4=N and J^2*M^(2/7)=N^0.499982857... QED. III. Certain kinds of Heron tetrahedra: ======================================= A tetrahedron with these sidelengths s01=a, s23=b, s02=s03=s12=s13=c has exactly two kinds of triangle faces (both isoceles) (a,c,c) and (b,c,c). Its two face areas arise from 16*TriArea^2 = (a+2*c) * (a)^2 * (2*c-a) 16*TriArea^2 = (b+2*c) * (b)^2 * (2*c-b) and these both are integer iff T^2 = (2*c+a)*(2*c-a) = 4*c^2 - a^2 U^2 = (2*c+b)*(2*c-b) = 4*c^2 - b^2 both are squares. Its volume is given by 144*TetVol^2 = a^2*b^2*(4*c^2-b^2-a^2) Thus for the tetrahedron to be Heronian we also need that S^2 = 4*c^2 - b^2 - a^2 also be a square. Note that multiplying by 4 does not affect squaritude. Comment: Jan Fricke discovered that this kind of tetrahedron always arises from gluing together 4 congruent copies of a tetrahedron with all faces being right triangles. IV. Certain Heron tetrahedra from certain kinds of imperfect pythagorean boxes. =============================================================================== Now suppose we have an imperfect pythagorean box as in (II) having diagonal c where c^2 = x^2+y^2+z^2. We can now make a tetrahedron with sides s01=2*z, s23=2*x, s02=s03=s12=s13=c. This will be Heron of the type in (III) since T^2 = 4*c^2-4*z^2 = 4*(x^2 + y^2) = square U^2 = 4*c^2-4*x^2 = 4*(y^2 + z^2) = square S^2 = 4*c^2 - 4*x^2 - 4*y^2 = 4*z^2 = square. I have not checked the following as carefully: Related Heron tetrahedron (i): sides s01=s02=c, s13=s23=squareroot(x^2+y^2)=integer, s03=z, s12=2*x; Related Heron tetrahedron (ii): sides s01=c, s02=squareroot(y^2+z^2)=integer, s13=squareroot(x^2+y^2)=integer, s03=z, s12=x, s23=y; Related Heron tetrahedron (iii): sides s01=2*z, s02=s12=c, s03=s13=squareroot(y^2+z^2), s23=x. CONCLUSION: We have found a class of Heron tetrahedra (actually several classes, but just the first will suffice) with sides s01=a, s23=b, s02=s03=s12=s13=c such that the number of primitive tetrahedra in this class with max(a,b,c)<N grows with N at least as the same order as N^0.4999. --end. -- Warren D. Smith http://RangeVoting.org  <-- add your endorsement (by clicking "endorse" as 1st step)