1. Form: ax4+by4 = cz2
Theorem: "If the form ax4-y4 = z2 has a solution, then so does the form x4-ay4 = z2."
Proof:
1. If ax4-y4 = z2 has a soln, then so does x4+4ay4 = z2, since,
z4 + 4a(xy)4 = (ax4+y4)2, if ax4-y4 = z2.
2. And if in turn x4+4ay4 = z2 has a soln, then so does x4-ay4 = z2, since,
z4 - a(2xy)4 = (x4-4ay4)2, if x4+4ay4 = z2.
Combining the two,
(ax4+y4)4 - a(2xyz)4 = (4ax4y4-z4)2, if ax4-y4 = z2.
For example, the eqn 2x4-y4 = z2 has smallest soln {x,y,z} = {13, 1, 239}. The theorem implies that p4-2q4 = r2 has solns as well, yielding {p,q} = {57123, 6214}, which is not necessarily the smallest. Furthermore, given a non-trivial integral soln to ax4+bx2y2+cy4 = dz2, one can generally find an infinite more as proven by an identity of Desboves. For the special case when a = c = 1, we have
Lagrange
x4+by4 = z2
{x,y,z} = {p4-bq4, 2pqr, 4bp4q4+r4}, if p4+bq4 = r2
Desboves
More generally, if ap4+bq4 = cr2, then,
ax4+by4 = cz2
{x,y,z} = {p(4m2-3(m+n)2), q(4n2-3(m+n)2), r(4(m+n)4-3(m-n)4)}, where {m,n} = {ap4, bq4}
re-arranged by this author to make it more aesthetic. (Note how the algebraic form 4u2-3v2 appears). This can be generalized even more to cover the form ax4+bx2y2+cy4 = dz2. The identity involves polynomials of high degree, 9th deg in the {p,q}, but special cases like a+b = c need only cubics,
Desboves
If ap4+bq4 = (a+b)r2, then,
ax4+by4 = (a+b)z2
where {x,y,z} = {-d2p2+4bcq2, (2c2-d2)pq±2cdr, 4bdpq(4acp2+d2q2)±(2c2-d2)(d2p2+4bcq2)r}, where {c,d} = {a+b, a-b}.
with z as a cubic. And since this involves the ± sign, for this particular form there are solns that share the same x variable. For ex, the eqn 2x4-y4 = z2 has small solns {x,y} = {13, 1} and {1525, 1343}. Using Desboves’ identity, the second yields a pair of {x,y} as,
{28145221, 2165017}
{28145221, 30838067}
Euler, Lagrange
Let {u,v} be the legs of a triangle. The problem of making the sum of the legs and its squares as a square and a fourth power, respectively, or
u+v = y2,
u2+v2 = x4
can be reduced to a single condition. These simultaneous equations are true for,
{u,v} = {(y2+z)/2, (y2-z)/2}, if 2x4-y4 = z2
A soln given by Euler is,
{x,y,z} = {p3+2pq2-qr, p3-4pq2+qr, p6+p4q2-6p3qr+24p2q4-8q6}, if p4+8q4 = r2
Piezas
2x4-y4 = z2
{x,y} = {12(p+q)3-(3p+2q)v, 12(p+q)3-(4p+3q)v}, if 2p4-q4 = r2
where v = 2p2+8pq+5q2+r, and z is a sixth deg polynomial. Note that with the sign of p held constant, the other two variables q,r can come in four combinations: {+,+}, {-,+}, {+,-}, {-,-} generally giving four distinct values {x,y}.
S.Realis, A.Gerardin
In general, one soln to ax4+by4 = cz2 can lead to subsequent ones. Let n,x be unknowns,
a(u+px)4 + b(v+qx)4 = c(w-nx+rx2)2
Collecting powers of x,
(ap4+bq4-cr2)x4 + (Poly1)x3 + (Poly2)x2 + (Poly3)x + (au4+bv4-cw2) = 0
Assume that (ap4+bq4-cr2) = (au4+bv4-cw2) = 0. Since n is only linear in Poly1 (or Poly3), equate Poly1 = 0, then solve for n. The whole equation then reduces to the form,
(Poly2)x2 + (Poly3)x = 0
which, after factoring, is simply a linear eqn in x so we have both our unknowns {n,x} in terms of {p,q,r} and {u,v,w}. These generally lead to addends that are cubic polynomials in terms of p,q,r as in the explicit example given by this author.
2. Form: ax4+bx2y2+cy4 = dz2
Theorem 1: Given an initial non-trivial solution to ap4+bp2q2+cq4 = dr2, one can generally find an infinite number of solutions. (Desboves) (See also no. 4 below)
Proof:
ax4+bx2y2+cy4+dz2 = (ap4+bp2q2+cq4+dr2)(z/r)2
{x,y,z} = {p(u2-4cq4w), q(u2-4ap4w), r((p4q4v-w2)2-4p4q4u2v)}, where {u,v,w} = {ap4-cq4, b2-4ac, ap4+bp2q2+cq4}
Alternatively, to show its affinity to the diagonal form (when b = 0),
{x,y,z} = {p(4m2-3(m+n)2-nt), q(4n2-3(m+n)2-mt), r(4(m+n)4-3(m-n)4+mnt(4m+4n+t)+t(m+n)3)}, where {m,t,n} = {ap4, 4bp2q2, cq4}.
Note how when b = t = 0, the identity reduces to the one given in the previous section. Less general solns are also known,
1. J. Lagrange
x4+cy4 = z2
{x,y,z} = {p4-cq4, 2pqr, 4cp4q4+r4}, if p4+cq4 = r2
This was already given, but is a special case of the next one,
2. V. Lebesgue
x4+bx2y2+cy4 = z2
{x,y,z} = {p4-cq4, 2pqr, (b2-4c)p4q4-r4}, if p4+bp2q2+cq4 = r2
For b = 0, this reduces to the previous. This, in turn, in generalized by,
3. A. Desboves
x4+bdx2y2+acd2y4 = z2
{x,y,z} = {ap4-cq4, 2pqr, (b2-4ac)p4q4-d2r4}, if ap4+bp2q2+cq4 = dr2
which for a=d=1 again reduces to the previous.
4. A. Desboves
ax4+bx2y2+cy4 = dz2
{x,y,z} = {p(u2-4cdq4r2), q(u2-4adp4r2), r(4p4q4u2v-(p4q4v-d2r4)2)}
where {u,v} = {ap4-cq4, b2-4ac}, if ap4+bp2q2+cq4 = dr2
This is basically the identity discussed by Theorem 1 in its original form but takes too long to factor symbolically on a computer. Can the degree of the polynomials be reduced yet still apply to general {a,b,c,d}?
Euler
Let b = nx2+2x,
x4+bx2y2+y4 = (x3+y2)2, if x2-ny2 = 1
One can then use solns of Pell equations to solve the above form for certain b. More generally, let b = nx2+2v,
x4+bx2y2+y4 = (vx2+y2)2, if v2-ny2 = 1
3. Form: au4+bu2v2+cv4 = ax4+bx2y2+cy4
Piezas
(√p+√q)k + (√p-√q)k = (√r+√s)k + (√r-√s)k (eq.1)
Poly solns to eq.1 have been found this author for k=5,6,8, with k=7 found by D.Rusin. For k=8, this becomes,
p4+28p3q+70p2q2+28pq3+q4 = r4+28r3s+70r2s2+28rs3+s4
with one soln,
{p,q,r,s} = {(n-1)(n2-n-1), (n+1)(n2+n-1), n3-n-1, n3-n+1}
Since the above is a symmetric polynomial, a second transformation {p,q,r,s} = {u/2+v, u/2-v, x/2+y, x/2-y} gives it the simpler form,
u4-4u2v2+2v4 = x4-4x2y2+2y4
Any other soln for this as well as for the general case? Also, any non-trivial soln to eq.1 for k>8?
(Update, 11/10/09): In "Parametric Solutions of the Quartic Diophantine Equation f(x,y) = f(u,v)", Choudhry discusses how, if an initial soln is known, then one can find another soln in general. He also provides an explicit identity:
Choudhry
x14+2nx12x22+x24 = y14+2ny12y22+y24
{x1, x2} = {1+n2t+st2-rt3+qt4+3t5+pt6+nt7, -n+pt-3t2+qt3+rt4+st5-n2t6+t7}
{y1, y2} = {1-n2t+st2+rt3+qt4-3t5+pt6-nt7, n+pt+3t2+qt3-rt4+st5+n2t6+t7}
where {p,q,r,s} = {1-n+n2, -2+4n+n2, -4n+n3, 1+n2+n3}.
Note how, for {x1,y1} the signs are different with odd powers t, while for {x2,y2}, they are different for even powers t. For the case n = 0, this reduces to the formula found by Gerardin, while n = 1 is trivial. (End update.)
4. Form: ax4+bx3y+cx2y2+dxy3+ey4 = z2
For the 4th power univariate case to be made a square, Fermat's method still applies. If it has an initial soln, we can use the form with the square constant term,
ax4+bx3+cx2+dx+e2 = z2
and assume it as equal to,
ax4+bx3+cx2+dx+e2 = (px2+qx+e)2
Using the same approach, the unknowns {p,q} are enough to knock out enough terms such that one can solve for x, given by,
x = (be-dp)/(e(p2-a)), where p = (4ce2-d2)/(2e)3
Unfortunately, for the quintic function f(x) = z2, not enough terms can be removed so the method stops here. Interestingly, this is similar to the situation of solving f(x) = 0 where x is in radicals since there are general formulas only for degree 2,3,4 but none for higher. Has it been explicitly proven there is no general formula to find rational and non-zero x that solves f(x) = z2 where f(x) is a quintic or higher degree with rational coefficients and a square constant term?
ax4+by4 = cz2
ax4+bx2y2+cy4 = dz2
au4+bu2v2+cv4 = ax4+bx2y2+cy4
ax4+bx3y+cx2y2+dxy3+ey4 = z2