II. Quartic Polynomials as kth Powers
1. Form: ax^{4}+by^{4} = cz^{2}
Theorem: "If the form ax^{4}y^{4} = z^{2} has a solution, then so does the form x^{4}ay^{4} = z^{2}."
Proof:
1. If ax^{4}y^{4} = z^{2} has a soln, then so does x^{4}+4ay^{4} = z^{2}, since,
z^{4} + 4a(xy)^{4} = (ax^{4}+y^{4})^{2}, if ax^{4}y^{4} = z^{2}.
2. And if in turn x^{4}+4ay^{4} = z^{2} has a soln, then so does x^{4}ay^{4} = z^{2}, since,
z^{4}  a(2xy)^{4} = (x^{4}4ay^{4})^{2}, if x^{4}+4ay^{4} = z^{2}.
Combining the two,
(ax^{4}+y^{4})^{4}  a(2xyz)^{4} = (4ax^{4}y^{4}z^{4})^{2}, if ax^{4}y^{4} = z^{2}.
For example, the eqn 2x^{4}y^{4} = z^{2} has smallest soln {x,y,z} = {13, 1, 239}. The theorem implies that p^{4}2q^{4} = r^{2} has solns as well, yielding {p,q} = {57123, 6214}, which is not necessarily the smallest. Furthermore, given a nontrivial integral soln to ax^{4}+bx^{2}y^{2}+cy^{4} = dz^{2}, 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
x^{4}+by^{4} = z^{2}
{x,y,z} = {p^{4}bq^{4}, 2pqr, 4bp^{4}q^{4}+r^{4}}, if p^{4}+bq^{4} = r^{2}
Desboves
More generally, if ap^{4}+bq^{4} = cr^{2}, then,
ax^{4}+by^{4} = cz^{2}
{x,y,z} = {p(4m^{2}3(m+n)^{2}), q(4n^{2}3(m+n)^{2}), r(4(m+n)^{4}3(mn)^{4})}, where {m,n} = {ap^{4}, bq^{4}}
rearranged by this author to make it more aesthetic. (Note how the algebraic form 4u^{2}3v^{2} appears). This can be generalized even more to cover the form ax^{4}+bx^{2}y^{2}+cy^{4} = dz^{2}. 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 ap^{4}+bq^{4} = (a+b)r^{2}, then, ax^{4}+by^{4} = (a+b)z^{2}
where {x,y,z} = {d^{2}p^{2}+4bcq^{2}, (2c^{2}d^{2})pq±2cdr, 4bdpq(4acp^{2}+d^{2}q^{2})±(2c^{2}d^{2})(d^{2}p^{2}+4bcq^{2})r}, where {c,d} = {a+b, ab}.
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 2x^{4}y^{4} = z^{2} 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 = y^{2}, u^{2}+v^{2} = x^{4}
can be reduced to a single condition. These simultaneous equations are true for,
{u,v} = {(y^{2}+z)/2, (y^{2}z)/2}, if 2x^{4}y^{4} = z^{2}
A soln given by Euler is,
{x,y,z} = {p^{3}+2pq^{2}qr, p^{3}4pq^{2}+qr, p^{6}+p^{4}q^{2}6p^{3}qr+24p^{2}q^{4}8q^{6}}, if p^{4}+8q^{4} = r^{2}
Piezas
2x^{4}y^{4} = z^{2}
{x,y} = {12(p+q)^{3}(3p+2q)v, 12(p+q)^{3}(4p+3q)v}, if 2p^{4}q^{4} = r^{2}
where v = 2p^{2}+8pq+5q^{2}+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 ax^{4}+by^{4} = cz^{2} can lead to subsequent ones. Let n,x be unknowns,
a(u+px)^{4} + b(v+qx)^{4} = c(wnx+rx^{2})^{2}
Collecting powers of x,
(ap^{4}+bq^{4}cr^{2})x^{4} + (Poly1)x^{3 }+ (Poly2)x^{2} + (Poly3)x + (au^{4}+bv^{4}cw^{2}) = 0
Assume that (ap^{4}+bq^{4}cr^{2}) = (au^{4}+bv^{4}cw^{2}) = 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)x^{2} + (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: ax^{4}+bx^{2}y^{2}+cy^{4} = dz^{2}
Theorem 1: Given an initial nontrivial solution to ap^{4}+bp^{2}q^{2}+cq^{4} = dr^{2}, one can generally find an infinite number of solutions. (Desboves) (See also no. 4 below)
Proof:
ax^{4}+bx^{2}y^{2}+cy^{4}+dz^{2} = (ap^{4}+bp^{2}q^{2}+cq^{4}+dr^{2})(z/r)^{2}
{x,y,z} = {p(u^{2}4cq^{4}w), q(u^{2}4ap^{4}w), r((p^{4}q^{4}vw^{2})^{2}4p^{4}q^{4}u^{2}v)}, where {u,v,w} = {ap^{4}cq^{4}, b^{2}4ac, ap^{4}+bp^{2}q^{2}+cq^{4}}
Alternatively, to show its affinity to the diagonal form (when b = 0),
{x,y,z} = {p(4m^{2}3(m+n)^{2}nt), q(4n^{2}3(m+n)^{2}mt), r(4(m+n)^{4}3(mn)^{4}+mnt(4m+4n+t)+t(m+n)^{3})}, where {m,t,n} = {ap^{4}, 4bp^{2}q^{2}, cq^{4}}.
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
x^{4}+cy^{4} = z^{2}
{x,y,z} = {p^{4}cq^{4}, 2pqr, 4cp^{4}q^{4}+r^{4}}, if p^{4}+cq^{4} = r^{2}
This was already given, but is a special case of the next one,
2. V. Lebesgue
x^{4}+bx^{2}y^{2}+cy^{4} = z^{2}
{x,y,z} = {p^{4}cq^{4}, 2pqr, (b^{2}4c)p^{4}q^{4}r^{4}}, if p^{4}+bp^{2}q^{2}+cq^{4} = r^{2}
For b = 0, this reduces to the previous. This, in turn, in generalized by,
3. A. Desboves
x^{4}+bdx^{2}y^{2}+acd^{2}y^{4} = z^{2}
{x,y,z} = {ap^{4}cq^{4}, 2pqr, (b^{2}4ac)p^{4}q^{4}d^{2}r^{4}}, if ap^{4}+bp^{2}q^{2}+cq^{4} = dr^{2}
which for a=d=1 again reduces to the previous.
4. A. Desboves
ax^{4}+bx^{2}y^{2}+cy^{4} = dz^{2}
{x,y,z} = {p(u^{2}4cdq^{4}r^{2}), q(u^{2}4adp^{4}r^{2}), r(4p^{4}q^{4}u^{2}v(p^{4}q^{4}vd^{2}r^{4})^{2})}
where {u,v} = {ap^{4}cq^{4}, b^{2}4ac}, if ap^{4}+bp^{2}q^{2}+cq^{4} = dr^{2}
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 = nx^{2}+2x,
x^{4}+bx^{2}y^{2}+y^{4} = (x^{3}+y^{2})^{2}, if x^{2}ny^{2} = 1
One can then use solns of Pell equations to solve the above form for certain b. More generally, let b = nx^{2}+2v,
x^{4}+bx^{2}y^{2}+y^{4} = (vx^{2}+y^{2})^{2}, if v^{2}ny^{2} = 1
3. Form: au^{4}+bu^{2}v^{2}+cv^{4} = ax^{4}+bx^{2}y^{2}+cy^{4}
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,
p^{4}+28p^{3}q+70p^{2}q^{2}+28pq^{3}+q^{4} = r^{4}+28r^{3}s+70r^{2}s^{2}+28rs^{3}+s^{4}
with one soln,
{p,q,r,s} = {(n1)(n^{2}n1), (n+1)(n^{2}+n1), n^{3}n1, n^{3}n+1}
Since the above is a symmetric polynomial, a second transformation {p,q,r,s} = {u/2+v, u/2v, x/2+y, x/2y} gives it the simpler form,
u^{4}4u^{2}v^{2}+2v^{4} = x^{4}4x^{2}y^{2}+2y^{4}
Any other soln for this as well as for the general case? Also, any nontrivial 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
x_{1}^{4}+2nx_{1}^{2}x_{2}^{2}+x_{2}^{4} = y_{1}^{4}+2ny_{1}^{2}y_{2}^{2}+y_{2}^{4}
{x_{1}, x_{2}} = {1+n^{2}t+st^{2}rt^{3}+qt^{4}+3t^{5}+pt^{6}+nt^{7}, n+pt3t^{2}+qt^{3}+rt^{4}+st^{5}n^{2}t^{6}+t^{7}} {y_{1}, y_{2}} = {1n^{2}t+st^{2}+rt^{3}+qt^{4}3t^{5}+pt^{6}nt^{7}, n+pt+3t^{2}+qt^{3}rt^{4}+st^{5}+n^{2}t^{6}+t^{7}} where {p,q,r,s} = {1n+n^{2}, 2+4n+n^{2}, 4n+n^{3}, 1+n^{2}+n^{3}}.
Note how, for {x_{1},y_{1}} the signs are different with odd powers t, while for {x_{2},y_{2}}, 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: ax^{4}+bx^{3}y+cx^{2}y^{2}+dxy^{3}+ey^{4} = z^{2}
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,
ax^{4}+bx^{3}+cx^{2}+dx+e^{2} = z^{2}
and assume it as equal to,
ax^{4}+bx^{3}+cx^{2}+dx+e^{2} = (px^{2}+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 = (bedp)/(e(p^{2}a)), where p = (4ce^{2}d^{2})/(2e)^{3}
Unfortunately, for the quintic function f(x) = z^{2}, 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 nonzero x that solves f(x) = z^{2} where f(x) is a quintic or higher degree with rational coefficients and a square constant term?
