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2. Form: x2+ny2 = zk
This is just a generalization of the form x2+y2 = zk.
Euler
x2+ny2 = (p2+nq2)k
Same technique of equating factors over Ö-n,
{x+yÖ-n, x-yÖ-n} = {(p+qÖ-n)k, (p-qÖ-n)k}
and then solve for x,y. Example, for k=2,
(p2-nq2)2 + n(2pq)2 = (p2+nq2)2
which for n=1 gives the familiar and complete parametrization (after scaling) for Pythagorean triples. For k = 3,
(p3-3npq2)2 + n(3p2q-nq3)2 = (p2+nq2)3
though for k > 2 this does not generally give all solns (Pepin). For example, for the particular case c = 47, all relatively prime solns with odd z are given by the method. But a class of even z is given by,
(13u3+60u2v-168uv2-144v3)2 + 47(u3-12u2v-24uv2+16v3)2 = 23(3u2+2uv+16v2)3
by Pepin. This is also discussed in Quadratic Polynomials as a kth Power.
2b. Form: ax2+by2 = cz2
To solve this form for an infinite number of solns, all it takes is one initial point {x,y,z}. Either of two methods can be used, each with their own strengths.
Method 1: Use the identity by this author,
ax2+bxy+cy2+dz2 = (am2+bmn+cn2+dp2)(u2+cdv2)2
where {x,y,z} = {mu2+cdmv2, nu2-2pduv-d(bm+cn)v2, pu2+(bm+2cn)uv-cdpv2}
For example, given,
m2+2n2 = 2p2
we have {a,b,c,d} = {1, 0, 2, -2}, and initial soln {m,n,p} = {4, 1, 3}. Using these, we get,
x2+2y2 = 2z2
where {x,y,z} = {4(u2-4v2), u2+12uv+4v2, 3u2+4uv+12v2}.
Method 2: Use the identity,
ax12+bx22+cx32 = (ay12+by22+cy32)(az12+bz22+cz32)2
where {x1, x2, x3} = {uy1-vz1, uy2-vz2, uy3-vz3} with {u,v} = {az12+bz22+cz32, 2(ay1z1+by2z2+cy3z3)}
with initial soln {y1, y2, y3} and arbitrary {z1, z2, z3}, thus giving a three-variable parametrization. Given the same example,
y12+2y22 = 2y32
we have {a,b,c} = {1, 2, -2} and small soln {y1, y2, y3} = {4, 1, 3}. Substituting these in the formula for the xi, and letting {z1, z2, z3} = {x,y,z} for convenience, we get,
x12+2x22 = 2x32
where {x1, x2, x3} = {4(x+2y-2z)(x-y-z), x2-8xy-2y2+12yz-2z2, 3x2+6y2-8xz-4yz+6z2}. The advantage of the first method is that it yields simple forms and can be used on eqns with cross terms (when b is non-zero). However, the second method is useful in that it can be generalized into n diagonal forms:
Theorem: "In general, given one solution to a1y12+a2y22+…+ anyn2 = 0, then an infinite more can be found."
Proof: We simply generalize the soln given in Method 2. For four addends,
ax12+bx22+cx32+dx42 = (ay12+by22+cy32+dy42)(az12+bz22+cz32+dz42)2
where {x1, x2, x3, x4} = {uy1-vz1, uy2-vz2, uy3-vz3, uy4-vz4} with {u,v} = {az12+bz22+cz32+dz42, 2(ay1z1+by2z2+cy3z3+dy4z4)}
with initial {y1, y2, y3, y4} for arbitrary {z1, z2, z3, z4}, thus now giving a four-variable parametrization. And so on for any n addends as the pattern for the xi and {u,v} are easily seen. (These methods are also discussed in Section 006.)
3. Form: ad-bc = ±1
This is inserted here because it is important to the next two forms, x2+y2 = z2±1. One can solve ad-bc = ±1 in the integers using the transformation,
{a,b,c} = {q2, p+q, p-q}
such that it becomes the Pell equation,
p2-(d+1)q2 = ±1
Alternatively, let {a,b,c,d} = {p+s, -q+r, q+r, p-s} to transform it to,
p2+q2 ±1 = r2+s2
simple solns of which are,
E.Fauquembergue:
(a+1)2 + (2a)2 + 1 = (2a+1)2 + (a-1)2
(2a2-b2+1)2 + (2ab)2 + 1 = (2a2+1)2 + (b2-1)2
Piezas
(a2-c2+a+1)2 + (2c)2 + 1 = (a2-c2+a-1)2 + (2a+1)2
This form is discussed more in the section on Sums of Three Squares.
Q: Any more solns to p2+q2 ± 1 = r2+s2?
4. Form: x2+y2 = z2+1
In general, given an initial soln to x2+y2 = z2+h, one can find a quadratic parametrization (by this author) in the form,
(pn+a)2 + (pn2+2an+b)2 = (pn2+2an+c)2 + h, if a2+b2 = c2+h
where p = 2(-b+c) for arbitrary {n,h}.
(dx)2 + (d2y2-1)2 = (d2y2+d)2 + 1, where x2-2(d+1)y2 = 1
More generally, one can convert this to a Pell equation for a broader class:
Piezas
Given u2-dv2 = 1, if we define {x,y,z} = {(d-v2)/2, u, (d+v2)/2}, then x2+y2 = z2+1.
Of course, d should be appropriately chosen such that x,y,z are integers. Another way would be to use Euler’s complete soln for x2+y2 = z2+t2,
(ac+bd)2 + (ad-bc)2 = (ac-bd)2 + (ad+bc)2
and set one term, say, ad-bc = ±1, a form discussed previously. Alternatively, one can use a general identity found by this author,
(2m)2 + (2m2-n)2 = (2m2-n+1)2 + (2n-1)
and set n = {1, 0} to get,
(2m)2 + (2m2-1)2 = (2m2)2 + 1 (2m)2 + (2m2)2 = (2m2+1)2 - 1
the second of which is for the next section.
5. Form: x2+y2 = z2-1
This form is also solvable in terms of a quadratic parametrization using the identity given in the previous section. Also,
E. Grigorief
((a2-b2+c2-d2)/2)2 + (ab+cd)2 = ((a2+b2+c2+d2)/2)2 - 1, where ad-bc = ±1
Similarly for the previous section, but using the negative Pell eqn:
Piezas
Given u2-dv2 = -1, if we define {x,y,z} = {(d-v2)/2, u, (d+v2)/2}, then x2+y2 = z2-1.
One can also use Genocchi’s and Pepin’s method and solve the more specialized form,
(2pq(r2-2s2))2 + (4pqrs)2 = (p2+(r4+4s4)q2)2 - 1, if p2-(r4+4s4)q2 = ±1
which for the case r = 1 has a parametric soln to be discussed in the section on Simultaneous Polynomials. Incidentally, the complete soln such that {x2+y2-1, x2-y2-1} are both squares is given by Genocchi and Pepin and also involves solving a Pell eqn. As an overview, note that it turns out,
x2+y2 = z2±1 x3+y3 = z3±1 x4+y4 = z2+1
can be solved using Pell equations, the higher powers to be discussed in their respective sections. Q: Any more solns to x2+y2 = z2 ± 1?
Piezas
This is more for the sums of three squares but can also be discussed here.
Let, t = a2+b2+c2
(4ac-2bc+2t)2 + (2ac-4bc+2t)2 = (4ac-4bc+3t)2 - d4, if a2+b2 = c2 ± d2
which implies one soln to u2+v2 = w2-x2 leads to another. (Or equivalently, that a soln to the sum of three squares leads to another, with a simpler identity given by Monck in the section on Sums of Three Powers.) Furthermore, iterative use of the identity leads to u2+v2 = w2-xk with k a power of two. Thus, there is this relationship between the two equations x2+y2 = z2±1.
6: Form x2+y2 = z2+nt2
This generalizes the two forms above. For the special case when n = 1, this is discussed in mx2+ny2 = mz2+nt2, the last section below.
A.Gerardin
(1-ma)2 + (1-mb)2 = (mc)2 + 2, if m = 2(a+b)/(a2+b2-c2)
More generally,
Piezas
(2a+bn)2 + (2b+an)2 = (2c-cn)2 + n(2a+2b)2, if a2+b2 = c2
for some arbitrary constant n which was derived by modifying Gerardin’s. Explicitly, after some minor changes, this is,
(x2+2xy-nz)2 + (nx2+2nxy-z)2 = (n-1)2(x2+z)2 + 2n(x2-2y2)2
where z = 2xy+2y2. If we let the last term as x2-2y2 = ±1, this belongs to the special case of the Pythagorean-like triples x2+y2 = z2+h, for some constant h. It can easily be proven that for any h, there is an infinity of solns given by,
Piezas
(2m+1)2 + (2m2+2m-n)2 = (2m2+2m-n+1)2 + 2n (2m)2 + (2m2-n)2 = (2m2-n+1)2 + (2n-1)
and the general quadratic parametrization given earlier. Q: Any other simple formulas that work for all h not derived from these two?
Note: For the case h = 0, or the Pythagorean triples x2+y2 = z2, it is known that the ratio N/B where N is the number of primitive solns with z below a bound B asymptotically approaches 1/(2π) ≈ 0.159154, a result established by Lehmer. If we generalize this to x2+y2 = z2+h and define the ratio S(h) = N/B for any h, it seems for h < 0, there might be other asymptotics which involve 1/√-h. For example, for h = -1, given N with increasing bound B = 10, 102, 103, and so on gives the sequence of ratios as,
S(-1) = 0.2, 0.14, 0.126, 0.1238, 0.1251, 0.12497, 0.12499, …
which seem to be converging to 0.125 = 1/8. For small square-free h at bound B = 106 and S(h) rounded to five decimal places, paired with its conjectured asymptotic value:
S(-2) = 0.17679 1/(4√2) = 0.17677
S(-3) = 0.28868 1/(2√3) = 0.28867
S(-5) = 0.22366 1/(2√5) = 0.22360
S(-6) = 0.20396 1/(2√6) = 0.20412
S(-7) = 0.37799 1/√7 = 0.37796
S(-10) = 0.15814 1/(2√10) = 0.15811
is highly suggestive considering the simplicity of the form. Interestingly, for h > 0, the S(h) do not seem to have simple ones involving √h. (Note: Computing S(h) is time-consuming, and I’m grateful to the Mathematica code provided by Daniel Lichtblau of Wolfram Research, as well as help provided by Andrzej Kozlowski and James Waldby. If someone can prove the exact values are in fact the correct asymptotics, pls let me know.)
7. Form: x2+y2 = z2+ntk
J. Rose
(4n2)2 + (4n3)2 = (4n2(n-1))2 + (2n)5
Mehmed-Nadir
x2 + y2 = z2 + (a2+b2)5
{x,y,z} = {(a2+b2)(a2-b2)b, ((a2+1)(a2+b2)2+4b4)/2, ((a2-1)(a2+b2)2-4b4)/2}
E. Barisien
((n+2)(n2-2n-2))2 + (4n(n+1))2 = (2(n+1)(n+2))2 + n6
These can be generalized by the identity given previously by this author for near-Pythagorean triples,
(2m+1)2 + (2m2+2m-n)2 = (2m2+2m-n+1)2 + 2n (2m)2 + (2m2-n)2 = (2m2-n+1)2 + (2n-1)
where 2n or 2n-1 can then be equated to any kth power pk or (a2+b2)k and n easily solved for.
Q: Can Mehmed-Nadir’s identity be generalized to x2 + y2 = z2 + (a2+b2)k for other k not using the identity found by this author? See also sum of three squares x2 + y2 + z2 = (a2+b2)k, one of which is also by Mehmed-Nadir.
8. Form: x2+y2 = mz2+nt2
For the special case, x2+y2 = 2(z2+t2):
Euler
{x,y,z,t} = {2(pr-qs), 2(ps+qr), (p+q)r-(p-q)s, (p-q)r+(p+q)s}
M. Gruber
{x,y,z,t} = {4(p-q)q-2s, 4(p-q)r, s, 4p(p-q)-s}, where s = 2p2-q2-r2
More simply, for this case, it suffices to solve the form x2+y2 = p2+q2 by using the transformation of {2z, 2t} = {p+q, p-q}. Alternatively, if n is a constant that is the sum of two squares n = a2+b2,
x2+y2 = n(z2+t2)
this has the soln {x,y} = {az+bt, bz-at} for constant a,b and arbitrary z,t since this yields,
x2+y2 = (a2+b2)(z2+t2)
For example, x2+y2 = 5(z2+t2) has soln {x,y} = {2z+t, z-2t}, and so on.
Q: Any soln to the more general form x2+y2 = mz2+nt2?
9. Form: c1(x2+ny2) = c2(z2+nt2)
In a manner, this generalizes the previous section. Given an equation of form with constants m1, m2
m1(a2+b2) = (r2+s2)(c2+d2), or m1(a2+b2) = m2(c2+d2),
these can be solved if m1, m2 are sums of two squares using the general identity,
(a2+b2)(p2+q2) = (c2+d2)(r2+s2)
where {a,b,c,d} = {ru+sv, su-rv, pu+qv, qu-pv}. For the former case, it is constant p,q with arbitrary {r,s,u,v}; for the latter, constant p,q,r,s with arbitrary {u,v}. More generally,
(a2+nb2)(p2+nq2) = (c2+nd2)(r2+ns2)
where {a,b,c,d} = {ru+nsv, su-rv, pu+nqv, qu-pv}.
10. Form: mx2+ny2 = mz2+nt2
Theorem: “There is a complete integral solution to ap2+bpq+cq2 = ar2+brs+cs2.”
First. we transform this to diagonal form by the transformation {p,q,r,s} = {u-bv, 2av, x-by, 2ay} to get,
u2-Dv2 = x2-Dy2
where D = b2-4ac is the discriminant. This can be solved in an even more general form.
Theorem: “The equation mu2+nv2 = mx2+ny2 has the complete solution,
m(ac+bdn)2 + n(bc-adm)2 = m(ac-bdn)2 + n(bc+adm)2, for arbitrary a,b,c,d.”
The identity is by Euler. Note that either side is equal to (ma2+nb2)(mc2+nd2). Proof (Piezas): Given any non-trivial soln u,v,x,y, one can always find rational a,b,c,d using the formulas,
{a,b,c} = {(y-v)/(2dm), (u-x)/(2dn), dn(v+y)/(u-x)}
where it can be set d=1 without loss of generality. For example, substituting these into u = ac+bdn one finds it is true only if mu2+nv2 = mx2+ny2, and similarly for the other variables. The idea was to solve for {a,b,c,d} and express them in terms of u,v,x,y. These were found using Mathematica’s solve command,
Solve[{u,v,x} = {ac+bdn, bc-adm, ac-bdn}, {a,b,c}]
which finds a,b,c and where the resulting radical solns can then be simplified using the relation mu2+nv2 = mx2+ny2. Incidentally, if we set the fourth variable y = bc+adm = 0, Euler’s identity reduces to,
m(ma2-nb2)2 + n(2abm)2 = m(ma2+nb2)2
which for m=n=1, not surprisingly, is the well-known formula for Pythagorean triples. There are also other versions.
S. Realis (complete)
u2+nv2 = x2+ny2
{u,v,x,y} = {a2-n(a-b)2+n(a-c)2, b2-(a-b)2+n(b-c)2, a2+nb2-nc2, c2-(a-c)2-n(b-c)2}
V. Lebesgue
(p+q+r-s)2 + (p+q-r+s)2 = (p-q+r+s)2 + (p-q-r-s)2, if pq = rs
or, alternatively,
(p+q)2 + (r-s)2 = (r+s)2 + (p-q)2, if pq = rs
A.Desboves
x2+y2 = (m2+n2)z2
{x,y,z} = {(u2-dv2)m, nu2-2duv+dnv2, u2-2nuv+dv2}, where d = m2+n2
Q: Any soln to the more general form au2+bv2 = cx2+dy2?
11. Form: x2+y2 = zk+tk
(Update, 11/9/09): W. Lewis emailed and asked if results were known for the eqn x2+y2 = z5+t5. Since I didn't know of any relevant identity, my first thought was to use Mathematica's search function to look for small solns and see if a parametrization could be found. One can use two approaches:
Piezas
Method 1: Use the Bramagupta-Fibonacci Two-Square Identity for the problem in its general case. This is given by,
(pr+qs)2 + (ps-qr)2 = (p2+q2)(r2+s2) (eq.1) There are 4 free variables {p,q,r,s} to dispose of. Select {p,q} = {xk, yk}. The RHS, after distributing, becomes,
(pr+qs)2 + (ps-qr)2 = x2k(r2+s2) + y2k(r2+s2)
Then solve for r2+s2 = tk which was discussed in the previous page and is easy to do. Eq. 1 then becomes,
(pr+qs)2 + (ps-qr)2 = (tx2)k + (ty2)k
For k = 3, this is satistied by {p,q,r,s,t} = {x5, y5, u3-3uv2, 3u2v-v3, u2+v2} with 4 free variables. For ex, let {u,v,x,y} = {2,1,3,4}, then,
7582 + 1692 = (5*9)3 + (5*16)3
Method 2: For co-prime terms, the small solns given by Mathematica were enough to provide a clue for a parametrization to the eqn,
x2+y2 = zk+1
for all ODD k in terms of powers and near-powers of 2. This is given by,
k = 1; {x,y,z} = {21m+1, 22m+1-1, 24m+2} k = 3; {x,y,z} = {23m+2, 26m+3-1, 24m+2} k = 5; {x,y,z} = {25m+3, 210m+5-1, 24m+2}
and so on for higher k. The pattern is easily seen and notice how z has a constant form regardless of k. (End update.) |