I. Two variables 1. {x^{2}+axy+by^{2}, x^{2}+cxy+dy^{2}} 2. {x^{2}ny^{2}, x^{2}+ny^{2}} (Congruent numbers) 3. {x^{2}+y, x+y^{2}} 4. {x^{2}+y^{2}1, x^{2}y^{2}1} 5. {x^{2}+y^{2}+1, x^{2}y^{2}+1}
II. Three variables 6. {x ± y, x ± z, y ± z} (Mengoli's SixSquare Problem) 7. {x^{2}y^{2}, x^{2}z^{2}, y^{2}z^{2}} 8. {x^{2}+y^{2}, x^{2}+z^{2}, y^{2}+z^{2}} (Euler Brick Problem) 9. {x^{2}+y^{2}+z^{2}, x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}} 10. {x^{2}+y^{2}+z^{2}, x^{2}y^{2}+z^{2}, x^{2}+y^{2}z^{2}} 11. {2x^{2}+y^{2}+z^{2}, x^{2}+2y^{2}+z^{2}, x^{2}+y^{2}+2z^{2}} 12. {2x^{2}+2y^{2}z^{2}, 2x^{2}y^{2}+2z^{2}, x^{2}+2y^{2}+2z^{2}} 13. {x^{2}+yz, y^{2}+xz, z^{2}+xy} 14. {x^{2}+y^{2}+xy, x^{2}+z^{2}+xy, y^{2}+z^{2}+xy} 15. {x^{2}xy+y^{2}, x^{2}xz+z^{2}, y^{2}yz+z^{2}}
III. Four variables 16. {x^{2}+axy+y^{2}, x^{2}+bxz+z^{2}, y^{2}+cyz+z^{2}} 17. {a^{2}+b^{2}+c^{2}, a^{2}+b^{2}+d^{2}, a^{2}+c^{2}+d^{2}, b^{2}+c^{2}+d^{2}} 18. {a^{2}b^{2}+c^{2}d^{2}, a^{2}d^{2}+b^{2}c^{2}} 19. {a^{2}b^{2}+c^{2}d^{2}, a^{2}c^{2}+b^{2}d^{2}, a^{2}d^{2}+b^{2}c^{2}} 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: {x^{2}+axy+by^{2}, x^{2}+cxy+dy^{2}}
Euler
{x,y} = {(ac)^{4}8(b+d)(ac)^{2}+16(bd)^{2}, 8(ac)(a^{2}4bc^{2}+4d)}
though for the special case when {a,b} = {c,d}, this implies discriminants are a^{2}4b = c^{2}4d, hence yields trivial y = 0. In general, for integral {a,b,c,d} there is an infinite number of nontrivial integral {x,y} such that the two polynomials are squares. To see this, consider the two eqns,
{x_{1}^{2}+ax_{1}+b, x_{2}^{2}+cx_{2}+d} = {t_{1}^{2}, t_{2}^{2}}
which have the soln,
{x_{1}, x_{2}} = {(p^{2}b)/(a2p), (q^{2}d)/(c2q)}
for free variables {p,q}. To make x_{1} = x_{2}, it is easy to equate the two formulas to form a quadratic equation in {p,q}.
(p^{2}b)/(a2p) = (q^{2}d)/(c2q)
Solving for q, this has a rational root iff the discriminant, a quartic polynomial in p, is made a square and is given by,
4p^{4}8cp^{3}4(2bac4d)p^{2}+8(bc2ad)p+4(b^{2}abc+a^{2}d) = z^{2}
Since the polynomial is monic, this is easily attained with one soln as,
4p = (a^{2}+4b2ac+c^{2}4d)/(ac)
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: {x^{2}ny^{2}, x^{2}+ny^{2}}
These simultaneous equations involve what is known as congruent numbers:
1. Lucas
{x,n,y} = {p^{2}+q^{2}, p^{3}qpq^{3}, 2}
2. A.Gerardin
{x,n,y} = {16p^{8}+24p^{4}q^{4}+q^{8}, 4p^{4}+q^{4}, 4pq(4p^{4}q^{4})}
{x,n,y} = {p^{8}+6p^{4}q^{4}+q^{8}, 2(p^{4}+q^{4}), 2pq(p^{4}q^{4})}
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.
{u^{2}nv^{2}, u^{2}+nv^{2}} = {(x^{4}2nx^{2}y^{2}n^{2}y^{4})^{2}, (x^{4}+2nx^{2}y^{2}n^{2}y^{4})^{2}},
where {u,v} = {x^{4}+n^{2}y^{4}, 2xyzt}, if (x^{2}ny^{2})(x^{2}+ny^{2}) = t^{2}z^{2}.
Any poly soln to the more general concordant forms {x^{2}+my^{2}, x^{2}+ny^{2}} = {t^{2},^{ }z^{2}}, esp to the case when m ≠ 1?
Form 3: {x^{2}+y, x+y^{2}}
Simply equate {x^{2}+y, x+y^{2}} = {(x+p)^{2}, (y+q)^{2}} and solve for x,y. (Euler)
Form 4: {x^{2}+y^{2}1, x^{2}y^{2}1}
1. Bhaskara
{x,y} = {y^{2}/2+1, (8a^{2}1)/(2a)}
{x,y} = {8a^{4}+1, 8a^{3}}
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 x^{3}+y^{3}+z^{3} = 1 found by D. Lehmer. (See update below.)
2. A.Genocchi (complete rational soln in p,q,r,s)
{x,y} = {(2r^{2}t)/t, 4pqrs/t}, if t = r^{2 } (p^{4}+4q^{4})s^{2 }
For integral solns, it suffices to solve the Pell equation r^{2 } (p^{4}+4q^{4})s^{2 }= ±1. The following identity is essentially the same then,
3. T. Pepin
If p^{2}(r^{4}+4s^{4})q^{2} = ±1,
(p^{2}+(r^{4}+4s^{4})q^{2})^{2} + (4pqrs)^{2}  1 = (2pq(r^{2}+2s^{2}))^{2}, (p^{2}+(r^{4}+4s^{4})q^{2})^{2}  (4pqrs)^{2}  1 = (2pq(r^{2}2s^{2}))^{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,
p^{2}(1+4s^{4})q^{2} = 1
has fundamental polynomial soln {p,q} = {2s^{2}, 1} and from this we can then generate an infinite sequence of polynomial solns. The first yields,
(8s^{4}+1)^{2} ± (8s^{3})^{2} = (8s^{4}±4s^{2})^{2} + 1
which is Bhaskara’s, while the next is,
(2t^{2}1)^{2} ± (16s^{3}t)^{2} = (8s^{4}±4s^{2})^{2}(2t)^{2} + 1, where t = 8s^{4}+1
and so on. However, if in Pepin's identity we let s=1,
p^{2}(r^{4}+4)q^{2} = 1
this also has a parametric soln given by {p,q} = {r^{2}(r^{4}+3)/2, (r^{4}+1)/2} which is integral for odd r. Thus this gives rise to a second family with the first member,
((r^{4}+2)(t2)/2)^{2} ± (r^{3}t)^{2} = (r^{2}(r^{2}±2)t/2)^{2} +1, if t = (r^{4}+1)(r^{4}+3)
and so on. In summary, the two families deal with the Pell equation x^{2}dy^{2} = ±1 with discriminant d of form m^{2}+1 and n^{2}+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 x^{3}+y^{3} = z^{3}±1 which also depends on a parametric soln to a Pell equation, though now this involves a discriminant of form 3(4m^{3}1).
Form 5: {x^{2}+y^{2}+1, x^{2}y^{2}+1}
J. Drummond
{x,y} = {2n^{2}, 2n}
Q: Does this pair have a complete soln similar to the one found by GenocchiPepin? 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 u^{2}2v^{2} = 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(prqs)^{2}+2(pqrs)^{2}, 2(pr+qs)^{2}2(pq+rs)^{2}}
where the expressions {pqrs, (p^{2}+s^{2})(q^{2}+r^{2})} are to be made squares. One particular soln is {p,q,r,s} = {112, 12, 35, 15}.
2. Euler
{x,y,z} = {a^{2}d^{2}+b^{2}c^{2}+2t^{2}, 2(abcd+t^{2}), 2(abcdt^{2})}, where t^{2} = (a^{2}b^{2})(c^{2}d^{2})/4
It remains to make {abcd, (a^{2}b^{2})(c^{2}d^{2})} 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 (a^{4}b^{4})(c^{4}d^{4}) = n^{2}. There are in fact polynomial solns to the general problem, one of deg 16 and another of deg 20, given below:
3. Euler: (16deg)
{x,y,z} = {(81+14n^{4}+n^{8})^{2}+z, 16n^{2}(27+3n^{2}+n^{4}+n^{6})^{2}z, 32n^{4}(81+30n^{4}+n^{8})}
4. M. Rolle: (20deg)
{x,y,z} = {1+21a^{4}6a^{8}6a^{12}+21a^{16}+a^{20}, 10a^{2}24a^{6}+60a^{10}24a^{14}+10a^{18}, 6a^{2}+24a^{6}92a^{10}+24a^{14}+6a^{18}}
Any poly soln of deg < 16? Note that if {x ± y, x ± z, y ± z} are squares, this immediately implies that {x^{2}y^{2}, x^{2}z^{2}, y^{2}z^{2}} are also squares.
Form 7: {x^{2}y^{2}, x^{2}z^{2}, y^{2}z^{2}}
W. Lenhart
{x,y,z} = {(p^{2}+q^{2})/(p^{2}q^{2}), (r^{2}+s^{2})/(r^{2}s^{2}), 1}
Two expressions are already squares. To make it all three,
{p,q,r,s} = {8n, n^{2}+9, 8n(n^{2}9), n^{4}+2n^{2}+81}
Form 8: {a^{2}+b^{2}, a^{2}+c^{2}, b^{2}+c^{2}}
Also known as the Euler Brick Problem: Find {a,b,c,}, or the length, width, height of a brick such that the diagonals {x,y,z} are rational, given by,
a^{2} + b^{2} = x^{2} a^{2} + c^{2} = y^{2} b^{2} + c^{2} = z^{2}
The smallest integer soln is {44, 117, 240},
44^{2} + 117^{2} = 125^{2} 44^{2} + 240^{2} = 244^{2} 117^{2} + 240^{2} = 267^{2} Olson has proved that an integer Euler brick has the product abcxyz as divisible by 3^{4}4^{4}5^{2}. A perfect cuboid would have the space diagonal a square as well, or a^{2}+b^{2}+c^{2} = d^{2} for some rational d, but none are known with odd side less than 100 billion (10^{11}). See Durango Bill's The Integer Brick Problem, or Oliver Knill's Hunting for the Perfect Cuboid.
Theorem (A. Bremner): “Given the Euler brick {a,b,c,x,y,z}, then {ab, bc, ac, by, az, cx} is another Euler brick.”
Some polynomial solns are:
1. N. Saunderson (in 1740. Later rediscovered by Euler, C. Chabanel, J. Neuberg)
{a,b,c} = {(4p^{2}r^{2})q, 4pqr, (4q^{2}r^{2})p}
where {p,q,r} is the Pythagorean triple, {m^{2}n^{2}, 2mn, m^{2}+n^{2}}. Explicitly,
{a,b,c} = {2mn(3m^{4}10m^{2}n^{2}+3n^{4}), 8mn(m^{4}n^{4}), m^{6}15m^{4}n^{2}+15m^{2}n^{4}n^{6}}
though these are factorable polynomials. This soln has the smallest degree known, and it is unknown if there others with deg ≤ 6. (Note also that the middle diagonal a^{2}+c^{2} = y^{2} is a perfect 6th power.)
2. C. Kunze
{a,b,c} = {2pq, pq^{2}p, p^{2}qq}
The first two diagonals become squares and it remains to make b^{2}+c^{2} one,
p^{2}q^{2}(p^{2}+q^{2}4) + p^{2}+q^{2} = z^{2}
which is a square if p^{2}+q^{2} = 4, so {p,q} = {2(r^{2}s^{2})/(r^{2}+s^{2}), 4rs/(r^{2}+s^{2})}. This yields the SaundersonEuler soln given in [1].
3. Euler
{a,b,c} = {(p^{2}1)/(2p), (q^{2}1)/(2q), 1}, where {p,q} = {4t/(t^{2}+1), (3t^{2}1)/(t^{3}3t)}
The last two diagonals become squares while a^{2}+b^{2} is,
p^{2}(q^{2}1)^{2 }+ q^{2}(p^{2}1)^{2} = x^{2}
with the given {p,q} satisfying this condition. This yields an 8th deg polynomial. However, it can also be derived from Saunderson’s soln using the theorem cited above.
4. W. Lenhart
{a,b,c} = {(p^{2}1)/(2p), 2q/(q^{2}1), 1}
a very similar transformation to Euler's. Likewise, these values make the last two diagonal as squares but one still has to solve a^{2}+b^{2} = x^{2}, or,
(p^{2}1)^{2}(q^{2}1)^{2} + 16p^{2}q^{2} = x^{2}
This belongs to the more general problem of,
(p^{2}m)^{2}(q^{2}m)^{2} + 4mnp^{2}q^{2} = x^{2}
for some constants m,n which has the soln x = (p^{2}m)(q^{2}m) + 2mn, if p^{2}+q^{2} = m+n, hence is reduced to this simpler problem. For Lenhart’s case, with {m,n} = {1, 4), one needs to solve p^{2}+q^{2} = 5 which has soln, {p,q} = {(s^{2}+4s1)/(s^{2}+1), 2(s^{2}s1)/(s^{2}+1)}.
5. R. Rignaux
{a,b,c} = {2pqrs, rs(p^{2}q^{2}), pq(r^{2}s^{2})}, where
p^{2}q^{2}(r^{2}s^{2})^{2 }+ r^{2}s^{2}(p^{2}q^{2})^{2} = t^{2} (eq.1)
After equating some variables as equal to unity, eq.1 is essentially the same condition given by Euler, and can be solved using Euler’s soln.
Note: This author checked the equation a^{2}+b^{2}+c^{2} = d^{2} for all given polynomial solns and found that identities (1), (2), after minor change of variables, yield the 8deg,
v^{8}+68v^{6}122v^{4}+68v^{2}+1 = d^{2}
while (3),(4),(5) is a 16deg poly to be made a square. As the parametrizations are not complete, a negative result to this hyperelliptic curve certainly does not imply the nonexistence of a perfect Euler brick.
(Update, 7/2/10): Bremner’s paper, The rational cuboid and a quartic surface gives more parametrizations of 8th deg, and points out that evidence suggests these can be found for every even deg ≥ 6.
Form 9: {x^{2}+y^{2}+z^{2}, x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}}
1. Euler
{x,y,z} = {p^{2}+q^{2}r^{2}, 2pr, 2qr}
where {p,q,r} = {8n(n^{2}1), n^{4}10n^{2}+5, n(n^{2}+1)(n^{2}3)}
2. J.Euler
{x,y,z} = {(p^{4}6p^{2}+1)(p^{2}+1), 4p(p^{2}1)^{2}, 8p^{2}(p^{2}1)}
3. J.Euler
{x^{2}+y^{2}+z^{2}, x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2}} = {(a^{2}+b^{2})^{2}c^{2}, (2ab)^{2}(a^{4}+b^{4})^{2}}
where {x,y,z} = {2a^{2}b, 2ab^{2}, (a^{2}b^{2})c}, if a^{2}+b^{2} = c^{2}
The next three forms can be grouped together.
Form 10: {x^{2}+y^{2}+z^{2}, x^{2}y^{2}+z^{2}, x^{2}+y^{2}z^{2}}
Euler
{x,y,z} = {a(a+b)p2c^{2}q, a(a+b)p+2c^{2}q, acp+2cq^{2}}, where {p,q} = {3a+4b, a+2b}
First two expressions are already squares. To make it all three,
{a,b,c} = {m^{2}+2n^{2}, 2mn, m^{2}2n^{2}}
Form 11: {2x^{2}+y^{2}+z^{2}, x^{2}+2y^{2}+z^{2}, x^{2}+y^{2}+2z^{2}}
Legendre
{x,y,z} = {u^{2}11v^{2}, u^{2}+4uv+11v^{2}, u^{2}4uv+11v^{2}}, if u^{4}+8u^{2}v^{2}+121v^{4} = t^{2}.
with small solns {u,v} = {2, 1} and {11, 2}. More generally,
Piezas
For constant m of form m = n^{2}2, for the three expressions,
{mx^{2}+y^{2}+z^{2}, x^{2}+my^{2}+z^{2}, x^{2}+y^{2}+mz^{2}}
where {x,y,z} = {u^{2}bv^{2}, u^{2}+auv+bv^{2}, u^{2}auv+bv^{2}} and {a,b} = {2n, 5n^{2}9}, the last two are squares. The first is also a square if,
n^{2}u^{2}2(5n^{4}33n^{2}+36)u^{2}v^{2}+n^{2}(5n^{2}9)^{2}v^{4} = t^{2}
with Legendre’s as the case n = 2.
Form 12: {2x^{2}+2y^{2}z^{2}, 2x^{2}y^{2}+2z^{2}, x^{2}+2y^{2}+2z^{2}}
E. Grebe:
{x,y,z} = {pq, p+2q, 2p+q}
These three forms belong to the more general class {ax^{2}+ay^{2}+bz^{2}, ax^{2}+by^{2}+az^{2}, bx^{2}+ay^{2}+az^{2}}.
Q: Any solns for other a,b?
Form 13: {x^{2}+yz, y^{2}+xz, z^{2}+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} = {s^{2}+8st, 8st+t^{2}} {x,y} = {5s^{2}+8st, 8st+13t^{2}}
Form 14: {x^{2}+y^{2}+xy, x^{2}+z^{2}+xy, y^{2}+z^{2}+xy}
S. Ryley
{x,y,z} = {p(p^{2}3q^{2})(3p^{2}q^{2}), p(3p^{2}5q^{2})(p^{2}+q^{2}), q(5p^{4}10p^{2}q^{2}+q^{2})}
Form 15: {x^{2}xy+y^{2}, x^{2}xz+z^{2}, y^{2}yz+z^{2}}
J. Cunliffe
{x,y,z} = {(35n+n^{2})(1n+3n^{2}) /(2+2n+n^{2}), n(45n), 48n+3n^{2}}
Form 16: {x^{2}+axy+y^{2}, x^{2}+bxz+z^{2}, y^{2}+cyz+z^{2}}
A.Martin
{x,y,z} = {n(2m+an), m^{2}n^{2}, 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} = {p^{2}q^{2}, 2pq+bq^{2}} for arbitrary p,q. This involves the denominator (ac) 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 {ax^{2}+bxy+cy^{2}, dx^{2}+exz+fz^{2}, gy^{2}+hyz+iz^{2}} are all squares with rational variables then a polynomial soln involves finding rational points on a certain sextic surface.
III. Four variables
Form 17: {a^{2}+b^{2}+c^{2}, a^{2}+b^{2}+d^{2}, a^{2}+c^{2}+d^{2}, b^{2}+c^{2}+d^{2}}
Euler
{a,b,c,d} = {4n(1+n^{4}), 2n(16n^{2}+n^{4}), 1+7n^{2}7n^{4}+n^{6}, 4n(1+n^{4})}
Note that a^{2}+b^{2}+c^{2} = (1+n^{2})^{6}. Also,
{a,b,c,d} = {4pqr, (pq)(p+3q)r, 2(p^{2}+3q^{2})s, (p3q)(p+q)s} where,
{r,s} = {(p3q)(p+q)(p^{2}+3q^{2}), 2pq(pq)(p+3q)}
Again, a^{2}+b^{2}+c^{2} is a sixth power equal to (p^{2}+3q^{2})^{6}.
Form 18: {a^{2}b^{2}+c^{2}d^{2}, a^{2}d^{2}+b^{2}c^{2}}
Euler
{a,b,c,d} = {3pq, q(2p^{2}+q^{2}), 2(p^{2}q^{2}), p(p^{2}+2q^{2})}
Form 19: {a^{2}b^{2}+c^{2}d^{2}, a^{2}c^{2}+b^{2}d^{2}, a^{2}d^{2}+b^{2}c^{2}}
Euler
{a,b,c,d} = {4pq, 2(p^{2}+3q^{2}), (pq)(p+3q), (p3q)(p+q)}
Note that if their squares are u^{2}, v^{2}, w^{2}, then they can also be expressed as,
{u,v,w} = {m^{2}+7n^{2}, 2(m^{2}mn+2n^{2}), 2(m^{2}+mn+2n^{2})}
where {m,n} = {p^{2}3q^{2}, 2pq}.
Form 20: {1+abc, 1+abd, 1+acd, 1+bcd}
W. Wright
Let d = 4p(ac+p)(bc+p)/(abc^{2}p^{2})^{2}, where p = (abacbc)/2. The last three expressions become squares. One can then easily solve the first as 1+abc = q^{2} for any of the three free variables a,b,c.
