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I. Two variables 1. {x2+axy+by2, x2+cxy+dy2} 2. {x2-ny2, x2+ny2} (Congruent numbers) 3. {x2+y, x+y2} 4. {x2+y2-1, x2-y2-1} 5. {x2+y2+1, x2-y2+1}
II. Three variables 6. {x ± y, x ± z, y ± z} 7. {x2-y2, x2-z2, y2-z2} 8. {x2+y2, x2+z2, y2+z2} (Euler Brick Problem) 9. {x2+y2+z2, x2y2+x2z2+y2z2} 10. {-x2+y2+z2, x2-y2+z2, x2+y2-z2} 11. {2x2+y2+z2, x2+2y2+z2, x2+y2+2z2} 12. {2x2+2y2-z2, 2x2-y2+2z2, -x2+2y2+2z2} 13. {x2+yz, y2+xz, z2+xy} 14. {x2+y2+xy, x2+z2+xy, y2+z2+xy} 15. {x2-xy+y2, x2-xz+z2, y2-yz+z2}
III. Four variables 16. {x2+axy+y2, x2+bxz+z2, y2+cyz+z2} 17. {a2+b2+c2, a2+b2+d2, a2+c2+d2, b2+c2+d2} 18. {a2b2+c2d2, a2d2+b2c2} 19. {a2b2+c2d2, a2c2+b2d2, a2d2+b2c2} 20. {1+abc, 1+abd, 1+acd, 1+bcd}
Most of these were solved by Euler. If you have additional polynomial solutions, pls send them. Or if you have poly solns to a set of simultaneous poly not in the list, those are also most welcome.
I. Two variables
Form 1: {x2+axy+by2, x2+cxy+dy2}
Euler
{x,y} = {(a-c)4-8(b+d)(a-c)2+16(b-d)2, 8(a-c)(a2-4b-c2+4d)}
though for the special case when {a,b} = {-c,d}, this implies discriminants are a2-4b = c2-4d, hence yields trivial y = 0. In general, for integral {a,b,c,d} there is an infinite number of non-trivial integral {x,y} such that the two polynomials are squares. To see this, consider the two eqns,
{x12+ax1+b, x22+cx2+d} = {t12, t22}
which have the soln,
{x1, x2} = {(p2-b)/(a-2p), (q2-d)/(c-2q)}
for free variables {p,q}. To make x1 = x2, it is easy to equate the two formulas to form a quadratic equation in {p,q}.
(p2-b)/(a-2p) = (q2-d)/(c-2q)
Solving for q, this has a rational root iff the discriminant, a quartic polynomial in p, is made a square and is given by,
4p4-8cp3-4(2b-ac-4d)p2+8(bc-2ad)p+4(b2-abc+a2d) = z2
Since the polynomial is monic, this is easily attained with one soln as,
4p = (a2+4b-2ac+c2-4d)/(a-c)
Treating the quartic as an “elliptic curve”, from this initial point, in general, one can then compute an infinite number of other points.
Form 2: {x2-ny2, x2+ny2}
These simultaneous equations involve what is known as congruent numbers:
1. Lucas
{x,n,y} = {p2+q2, p3q-pq3, 2}
2. A.Gerardin
{x,n,y} = {16p8+24p4q4+q8, 4p4+q4, 4pq(4p4-q4)}
{x,n,y} = {p8+6p4q4+q8, 2(p4+q4), 2pq(p4-q4)}
Q: Any other n as a polynomial with small degree?
3. J.Maurin, A. Cunningham
The ff identity proves that one soln leads to another.
{u2-nv2, u2+nv2} = {(x4-2nx2y2-n2y4)2, (x4+2nx2y2-n2y4)2},
where {u,v} = {x4+n2y4, 2xyzt}, if (x2-ny2)(x2+ny2) = t2z2.
Any poly soln to the more general concordant forms {x2+my2, x2+ny2} = {t2, z2}, esp to the case when m ≠ 1?
Form 3: {x2+y, x+y2}
Simply equate {x2+y, x+y2} = {(x+p)2, (y+q)2} and solve for x,y. (Euler)
Form 4: {x2+y2-1, x2-y2-1}
1. Bhaskara
{x,y} = {y2/2+1, (8a2-1)/(2a)}
{x,y} = {8a4+1, 8a3}
with the last one also found by E. Clere. It may be the case that this has a series of higher degree parametrizations, like the recursive one for x3+y3+z3 = 1 found by D. Lehmer. (See update below.)
2. A.Genocchi (complete rational soln in p,q,r,s)
{x,y} = {(2r2-t)/t, 4pqrs/t}, if t = r2 - (p4+4q4)s2
For integral solns, it suffices to solve the Pell equation r2 - (p4+4q4)s2 = ±1. The following identity is essentially the same then,
3. T. Pepin
If p2-(r4+4s4)q2 = ±1,
(p2+(r4+4s4)q2)2 + (4pqrs)2 - 1 = (2pq(r2+2s2))2, (p2+(r4+4s4)q2)2 - (4pqrs)2 - 1 = (2pq(r2-2s2))2,
Update: After a closer inspection, it turns out Bhaskara’s second soln is just the first in a family. Note that if r = 1 in Pepin's identity, then the Pell equation,
p2-(1+4s4)q2 = -1
has fundamental polynomial soln {p,q} = {2s2, 1} and from this we can then generate an infinite sequence of polynomial solns. The first yields,
(8s4+1)2 ± (8s3)2 = (8s4±4s2)2 + 1
which is Bhaskara’s, while the next is,
(2t2-1)2 ± (16s3t)2 = (8s4±4s2)2(2t)2 + 1, where t = 8s4+1
and so on. However, if in Pepin's identity we let s=1,
p2-(r4+4)q2 = -1
this also has a parametric soln given by {p,q} = {r2(r4+3)/2, (r4+1)/2} which is integral for odd r. Thus this gives rise to a second family with the first member,
((r4+2)(t-2)/2)2 ± (r3t)2 = (r2(r2±2)t/2)2 +1, if t = (r4+1)(r4+3)
and so on. In summary, the two families deal with the Pell equation x2-dy2 = ±1 with discriminant d of form m2+1 and n2+4 (where the latter is restricted only to odd n since it reduces to the former for even n). Interestingly, as will be seen later, a similar polynomial family exists for third powers x3+y3 = z3±1 which also depends on a parametric soln to a Pell equation, though now this involves a discriminant of form 3(4m3-1).
Form 5: {x2+y2+1, x2-y2+1}
J. Drummond
{x,y} = {2n2, 2n}
Q: Does this pair have a complete soln similar to the one found by Genocchi-Pepin? This is the only soln known so far.
Update (7/3/09): This author checked all {x,y} < 1000 and found that, other than the one given by Drummond, the only other solns were x = y = v such that v satisfies the Pell eqn u2-2v2 = 1.
II. Three variables
Form 6: {x ± y, x ± z, y ± z}
1. Petrus
{x,y,z} = {2(pr+qs)2+2(pq+rs)2, 2(pr-qs)2+2(pq-rs)2, 2(pr+qs)2-2(pq+rs)2}
where the expressions {pqrs, (p2+s2)(q2+r2)} are to be made squares. One particular soln is {p,q,r,s} = {112, 12, 35, 15}.
2. Euler
{x,y,z} = {a2d2+b2c2+2t2, 2(abcd+t2), 2(abcd-t2)}, where t2 = (a2-b2)(c2-d2)/4
It remains to make {abcd, (a2-b2)(c2-d2)} squares, a very similar condition to the one above. One particular soln is {a,b,c,d} = {9, 4, 81, 49}. By squaring variables, this reduces to the single condition (a4-b4)(c4-d4) = n2. There are in fact polynomial solns to the general problem, one of deg 16 and another of deg 20, given below:
3. Euler: (16-deg)
{x,y,z} = {(81+14n4+n8)2+z, 16n2(27+3n2+n4+n6)2-z, 32n4(81+30n4+n8)}
4. M. Rolle: (20-deg)
{x,y,z} = {1+21a4-6a8-6a12+21a16+a20, 10a2-24a6+60a10-24a14+10a18, 6a2+24a6-92a10+24a14+6a18}
Any poly soln of deg < 16? Note that if {x ± y, x ± z, y ± z} are squares, this immediately implies that {x2-y2, x2-z2, y2-z2} are also squares.
Form 7: {x2-y2, x2-z2, y2-z2}
W. Lenhart
{x,y,z} = {(p2+q2)/(p2-q2), (r2+s2)/(r2-s2), 1}
Two expressions are already squares. To make it all three,
{p,q,r,s} = {8n, n2+9, 8n(n2-9), n4+2n2+81}
Form 8: {x2+y2, x2+z2, y2+z2}
Also known as the Euler Brick Problem: Find {x,y,z,}, or the length, width, height of a brick such that the diagonals {u,v,w} are rational, given by the formulas {x2+y2, x2+z2, y2+z2} = {u2, v2, w2}. Polynomial identities are known for these, given below. A perfect cuboid would have the space diagonal a perfect square as well, or x2+y2+z2 = t2 for some rational t, but none are known with odd side less than 100 billion (1011). See Durango Bill's The Integer Brick Problem, or Oliver Knill's Hunting for the Perfect Cuboid. Olson proved that an integer Euler brick has the product uvwxyz as divisible by 344452.
Note: While there are no known "small" perfect cuboid, that certainly does not imply there are none at all, as nobody has proven it cannot exist. After all, a system of equations may simply have LARGE solns. To illustrate, the quadratic-quartic system used by Choudhry,
ak+27bk = ck+27dk, for k = 2,4
to prove that all rational numbers N is the sum of eight 7th powers has the smallest soln, among an infinity of non-trivial rational ones, with 33 digits. See Assorted Identities for more details. An even more extreme example is Archimedes' Cattle Problem, which entails solving a Pell equation with the smallest integral soln of more than 10200,000.
Polynomial identities solving the Euler Brick Problem are:
1. Euler
(x,y,z} = {z(p2-1)/(2p), z(q2-1)/(2q), z}, where {p,q} = {4t/(t2+1), (3t2-1)/(t3-3t)}
He stated that this problem can be reduced to making the expression p2(q2-1)2 + q2(p2-1)2 a square and his {p,q} satisfy this condition.
2. W. Lenhart
Used the same substitution for {x,y,z} as Euler. These values make x2+z2, y2+z2 squares and it remains to make x2+y2 one, or,
(p2-1)2(q2-1)2 + 16p2q2 = t2
This belongs to the more general problem of,
(p2-a)2(q2-a)2 + 4abp2q2 = t2
for some constants a,b which has the soln t = (p2-a)(q2-a) + 2ab, if p2+q2 = a+b, hence is reduced to this simpler problem. For Lenhart’s case, with {a,b} = {1, 4), one needs to solve p2+q2 = 5 and has soln, {p,q} = {(s2+4s-1)/(s2+1), 2(s2-s-1)/(s2+1)}.
3. C. Kunze
{x,y,z} = {2pq, pq2-p, p2q-q}
These values make the first two diagonals as squares while the third becomes,
p2q2(p2+q2-4) + p2+q2
which is a square if p2+q2 = 4, so {p,q} = {2(r2-s2)/(r2+s2), 4rs/(r2+s2)}.
4. R.Rignaux
{x,y,z} = {2pqrs, rs(p2-q2), pq(r2-s2)}, if p2q2(r2-s2)2 + r2s2(p2-q2)2 = t2
which, after equating some variables as equal to unity, is essentially the same condition given by Euler.
5. Euler, (and independently, C. Chabanel, J. Neuberg)
{x,y,z} = {2ab(p2-4q2), 8abpq, (16a2b2-p2)q},
where {p,q}= {a2+b2, a2-b2}. Note that the third diagonal, x2+z2, is a sixth power.
(Update: 10/12/09) 6. N. Saunderson
If a2+b2 = c2, then {x2+y2, x2+z2, y2+z2} = {u2, v2, w2}, where,
{x,y,z} = {a(4b2-c2), b(4a2-c2), 4abc} {u,v,w} = {c3, a(4b2+c2), b(4a2+c2)}
Q: Any others?
Form 9: {x2+y2+z2, x2y2+x2z2+y2z2}
1. Euler
{x,y,z} = {p2+q2-r2, 2pr, 2qr}
where {p,q,r} = {8n(n2-1), n4-10n2+5, n(n2+1)(n2-3)}
2. J.Euler
{x,y,z} = {(p4-6p2+1)(p2+1), 4p(p2-1)2, 8p2(p2-1)}
3. J.Euler
{x2+y2+z2, x2y2+x2z2+y2z2} = {(a2+b2)2c2, (2ab)2(a4+b4)2}
where {x,y,z} = {2a2b, 2ab2, (a2-b2)c}, if a2+b2 = c2
The next three forms can be grouped together.
Form 10: {-x2+y2+z2, x2-y2+z2, x2+y2-z2}
Euler
{x,y,z} = {a(a+b)p-2c2q, a(a+b)p+2c2q, acp-2cq2}, where {p,q} = {3a+4b, a+2b}
First two expressions are already squares. To make it all three,
{a,b,c} = {m2+2n2, 2mn, m2-2n2}
Form 11: {2x2+y2+z2, x2+2y2+z2, x2+y2+2z2}
Legendre
{x,y,z} = {u2-11v2, u2+4uv+11v2, u2-4uv+11v2}, if u4+8u2v2+121v4 = t2.
with small solns {u,v} = {2, 1} and {11, 2}. More generally,
Piezas
For constant m of form m = n2-2, for the three expressions,
{mx2+y2+z2, x2+my2+z2, x2+y2+mz2}
where {x,y,z} = {u2-bv2, u2+auv+bv2, u2-auv+bv2} and {a,b} = {2n, 5n2-9}, the last two are squares. The first is also a square if,
n2u2-2(5n4-33n2+36)u2v2+n2(5n2-9)2v4 = t2
with Legendre’s as the case n = 2.
Form 12: {2x2+2y2-z2, 2x2-y2+2z2, -x2+2y2+2z2}
E. Grebe:
{x,y,z} = {p-q, p+2q, 2p+q}
These three forms belong to the more general class {ax2+ay2+bz2, ax2+by2+az2, bx2+ay2+az2}.
Q: Any solns for other a,b?
Form 13: {x2+yz, y2+xz, z2+xy}
Euler
If z = 4(x+y) the first two become squares. It remains to make the last expression 16(x+y)2+xy a square. Two of his solns are:
{x,y} = {s2+8st, -8st+t2} {x,y} = {5s2+8st, 8st+13t2}
Form 14: {x2+y2+xy, x2+z2+xy, y2+z2+xy}
S. Ryley
{x,y,z} = {p(p2-3q2)(3p2-q2), -p(3p2-5q2)(p2+q2), q(5p4-10p2q2+q2)}
Form 15: {x2-xy+y2, x2-xz+z2, y2-yz+z2}
J. Cunliffe
{x,y,z} = {(3-5n+n2)(-1-n+3n2) /(-2+2n+n2), n(4-5n), 4-8n+3n2}
Form 16: {x2+axy+y2, x2+bxz+z2, y2+cyz+z2}
A.Martin
{x,y,z} = {n(2m+an), m2-n2, n(2m+cn)}
This makes two of the expressions squares. To make it all three, one should solve for m,n in {2m+an, 2m+cn} = {p2-q2, 2pq+bq2} for arbitrary p,q. This involves the denominator (a-c) so the method runs into a complication when a=b=c. Cunliffe’s soln is for the case a=b=c=-1. Any soln for a=b=c=1? In general, E.Turriere proved that if {ax2+bxy+cy2, dx2+exz+fz2, gy2+hyz+iz2} are all squares with rational variables then a polynomial soln involves finding rational points on a certain sextic surface.
III. Four variables
Form 17: {a2+b2+c2, a2+b2+d2, a2+c2+d2, b2+c2+d2}
Euler
{a,b,c,d} = {4n(-1+n4), 2n(1-6n2+n4), -1+7n2-7n4+n6, 4n(-1+n4)}
Note that a2+b2+c2 = (1+n2)6. Also,
{a,b,c,d} = {4pqr, (p-q)(p+3q)r, 2(p2+3q2)s, (p-3q)(p+q)s} where,
{r,s} = {(p-3q)(p+q)(p2+3q2), 2pq(p-q)(p+3q)}
Again, a2+b2+c2 is a sixth power equal to (p2+3q2)6.
Form 18: {a2b2+c2d2, a2d2+b2c2}
Euler
{a,b,c,d} = {3pq, q(2p2+q2), 2(p2-q2), p(p2+2q2)}
Form 19: {a2b2+c2d2, a2c2+b2d2, a2d2+b2c2}
Euler
{a,b,c,d} = {4pq, 2(p2+3q2), (p-q)(p+3q), (p-3q)(p+q)}
Note that if their squares are u2, v2, w2, then they can also be expressed as,
{u,v,w} = {m2+7n2, 2(m2-mn+2n2), 2(m2+mn+2n2)}
where {m,n} = {p2-3q2, 2pq}.
Form 20: {1+abc, 1+abd, 1+acd, 1+bcd}
W. Wright
Let d = 4p(ac+p)(bc+p)/(abc2-p2)2, where p = (ab-ac-bc)/2. The last three expressions become squares. One can then easily solve the first as 1+abc = q2 for any of the three free variables a,b,c.
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