007: Simultaneous Polynomials made squares


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}  (Mengoli's Six-Square Problem)

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}




{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: {a2+b2, a2+c2, b2+c2}


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,


a2 + b2 = x2

a2 + c2 = y2

b2 + c2 = z2


The smallest integer soln is {44, 117, 240},


442 + 1172 = 1252

442 + 2402 = 2442

1172 + 2402 = 2672


Olson has proved that an integer Euler brick has the product abcxyz as divisible by 344452A perfect cuboid would have the space diagonal a square as well, or a2+b2+c2 = d2 for some rational d, 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


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 re-discovered by Euler, C. Chabanel, J. Neuberg)


{a,b,c} = {(4p2-r2)q,  4pqr,  (4q2-r2)p}


where {p,q,r} is the Pythagorean triple, {m2-n2, 2mn, m2+n2}.  Explicitly,


{a,b,c} = {2mn(3m4-10m2n2+3n4),  8mn(m4-n4),  m6-15m4n2+15m2n4-n6}


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 a2+c2 = y2 is a perfect 6th power.)


2. C. Kunze


{a,b,c} = {2pq,  pq2-p,  p2q-q} 


The first two diagonals become squares and it remains to make b2+c2 one,


p2q2(p2+q2-4) + p2+q2 = z2


which is a square if p2+q2 = 4, so {p,q} = {2(r2-s2)/(r2+s2), 4rs/(r2+s2)}.  This yields the Saunderson-Euler soln given in [1].


3. Euler


{a,b,c} = {(p2-1)/(2p), (q2-1)/(2q), 1},  where {p,q} = {4t/(t2+1), (3t2-1)/(t3-3t)}


The last two diagonals become squares while a2+b2 is,


p2(q2-1)2 + q2(p2-1)2  = x2


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} = {(p2-1)/(2p), 2q/(q2-1), 1}


a very similar transformation to Euler's. Likewise, these values make the last two diagonal as squares but one still has to solve a2+b2 = x2, or,


(p2-1)2(q2-1)2 + 16p2q2 = x2


This belongs to the more general problem of,


(p2-m)2(q2-m)2 + 4mnp2q2 = x2


for some constants m,n which has the soln x = (p2-m)(q2-m) + 2mn, if p2+q2 = m+n, hence is reduced to this simpler problem.  For Lenhart’s case, with {m,n} = {1, 4), one needs to solve p2+q2 = 5 which has soln, {p,q} = {(s2+4s-1)/(s2+1),  2(s2-s-1)/(s2+1)}.


5. R. Rignaux


{a,b,c} = {2pqrs,  rs(p2-q2),  pq(r2-s2)}, where 


p2q2(r2-s2)2 + r2s2(p2-q2)2 = t2      (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 a2+b2+c2 = d2 for all given polynomial solns and found that identities (1), (2), after minor change of variables, yield the 8-deg,


v8+68v6-122v4+68v2+1 = d2


while (3),(4),(5) is a 16-deg poly to be made a square.  As the parametrizations are not complete, a negative result to this hyper-elliptic curve certainly does not imply the non-existence 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: {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}




{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}




{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,




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}




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}




{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}




{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}




{a,b,c,d} = {3pq,  q(2p2+q2),  2(p2-q2),  p(p2+2q2)}



Form 19: {a2b2+c2d2, a2c2+b2d2, a2d2+b2c2}




{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.



Previous Page        Next Page