2. Form: x^{2}+ny^{2} = z^{k}
This is just a generalization of the form x^{2}+y^{2} = z^{k}.
Euler
x^{2}+ny^{2} = (p^{2}+nq^{2})^{k}
Same technique of equating factors over Ön,
{x+yÖn, xyÖn} = {(p+qÖn)^{k}, (pqÖn)^{k}}
and then solve for x,y. Example, for k=2,
(p^{2}nq^{2})^{2} + n(2pq)^{2} = (p^{2}+nq^{2})^{2}
which for n=1 gives the familiar and complete parametrization (after scaling) for Pythagorean triples. For k = 3, we get,
(p^{3}3npq^{2})^{2} + n(3p^{2}qnq^{3})^{2} = (p^{2}+nq^{2})^{3}
However, for k > 2 this method does not generally give all solns (Pepin). For example, for the particular case k = 3 and n = 47, a class of solns not given by the above is,
(13u^{3}+60u^{2}v168uv^{2}144v^{3})^{2} + 47(u^{3}12u^{2}v24uv^{2}+16v^{3})^{2} = 2^{3}(3u^{2}+2uv+16v^{2})^{3}
by Pepin. This is also discussed in Quadratic Polynomials as a kth Power.
2b. Form: ax^{2}+by^{2} = cz^{2}
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,
ax^{2}+bxy+cy^{2}+dz^{2} = (am^{2}+bmn+cn^{2}+dp^{2})(u^{2}+cdv^{2})^{2}
where {x,y,z} = {mu^{2}+cdmv^{2}, nu^{2}2pduvd(bm+cn)v^{2}, pu^{2}+(bm+2cn)uvcdpv^{2}}
For example, given m^{2}+2n^{2} = 2p^{2} we have {a,b,c,d} = {1, 0, 2, 2}, and initial soln {m,n,p} = {4, 1, 3}. Using these, we get the parametrization,
x^{2}+2y^{2} = 2z^{2}
where {x,y,z} = {4(u^{2}4v^{2}), u^{2}+12uv+4v^{2}, 3u^{2}+4uv+12v^{2}}. Similarly for,
x^{2}+y^{2} = 2z^{2}
where {x,y,z} = {u^{2}2v^{2}, u^{2}+4uv+2v^{2}, u^{2}+2uv+2v^{2}}, and so on.
Method 2: Use the identity,
ax_{1}^{2}+bx_{2}^{2}+cx_{3}^{2} = (ay_{1}^{2}+by_{2}^{2}+cy_{3}^{2})(az_{1}^{2}+bz_{2}^{2}+cz_{3}^{2})^{2}
where {x_{1},^{ }x_{2}, x_{3}} = {uy_{1}vz_{1}, uy_{2}vz_{2}, uy_{3}vz_{3}} with {u,v} = {az_{1}^{2}+bz_{2}^{2}+cz_{3}^{2}, 2(ay_{1}z_{1}+by_{2}z_{2}+cy_{3}z_{3})}
with initial soln {y_{1}, y_{2}, y_{3}} and arbitrary {z_{1}, z_{2}, z_{3}}, thus giving a threevariable parametrization. Given the same ex, y_{1}^{2}+2y_{2}^{2} = 2y_{3}^{2}, we have {a,b,c} = {1, 2, 2} and small soln {y_{1},^{ }y_{2}, y_{3}} = {4, 1, 3}. Substituting these in the formula for the x_{i}, and letting {z_{1}, z_{2}, z_{3}} = {x,y,z} for convenience, we get the alternative identity,
x_{1}^{2}+2x_{2}^{2} = 2x_{3}^{2}
where {x_{1},^{ }x_{2}, x_{3}} = {4(x+2y2z)(xyz), x^{2}8xy2y^{2}+12yz2z^{2}, 3x^{2}+6y^{2}8xz4yz+6z^{2}}. The advantage of the first method is that it yields simple forms and can be used on eqns with cross terms (i.e. where b is nonzero). However, the second method is useful in that it can be generalized into n diagonal forms:
Theorem: "In general, given one solution to a_{1}y_{1}^{2}+a_{2}y_{2}^{2}+…+ a_{n}y_{n}^{2} = 0, then an infinite more can be found."
Proof: We simply generalize the soln given in Method 2. For four addends,
ax_{1}^{2}+bx_{2}^{2}+cx_{3}^{2}+dx_{4}^{2} = (ay_{1}^{2}+by_{2}^{2}+cy_{3}^{2}+dy_{4}^{2})(az_{1}^{2}+bz_{2}^{2}+cz_{3}^{2}+dz_{4}^{2})^{2}
where {x_{1},^{ }x_{2}, x_{3}, x_{4}} = {uy_{1}vz_{1}, uy_{2}vz_{2}, uy_{3}vz_{3}, uy_{4}vz_{4}} with {u,v} = {az_{1}^{2}+bz_{2}^{2}+cz_{3}^{2}+dz_{4}^{2}, 2(ay_{1}z_{1}+by_{2}z_{2}+cy_{3}z_{3}+dy_{4}z_{4})}
with initial {y_{1},^{ }y_{2}, y_{3}, y_{4}} for arbitrary {z_{1}, z_{2}, z_{3}, z_{4}}, thus now giving a fourvariable parametrization. And so on for any n addends as the pattern for the x_{i} and {u,v} are easily seen. (These methods are also discussed in Section 006.)
3. Form: adbc = ±1
This is inserted here because it is important to the next two forms, x^{2}+y^{2} = z^{2}±1. One can solve adbc = ±1 in the integers using the transformation,
{a,b,c} = {q^{2}, p+q, pq}
such that it becomes the Pell equation,
p^{2}(d+1)q^{2} = ±1
Alternatively, let {a,b,c,d} = {p+s, q+r, q+r, ps} to transform it to,
p^{2}+q^{2 }±1 = r^{2}+s^{2}
simple solns of which are,
E.Fauquembergue:
(a+1)^{2} + (2a)^{2} + 1 = (2a+1)^{2} + (a1)^{2}
(2a^{2}b^{2}+1)^{2} + (2ab)^{2} + 1 = (2a^{2}+1)^{2} + (b^{2}1)^{2}
Piezas
(a^{2}c^{2}+a+1)^{2} + (2c)^{2} + 1 = (a^{2}c^{2}+a1)^{2} + (2a+1)^{2}
This form is discussed more in the section on Sums of Three Squares.
Q: Any more solns to p^{2}+q^{2 }± 1 = r^{2}+s^{2}?
4. Form: x^{2}+y^{2} = z^{2}+1
In general, given an initial soln to x^{2}+y^{2} = z^{2}+h, one can find a quadratic parametrization (by this author) in the form,
(pn+a)^{2} + (pn^{2}+2an+b)^{2} = (pn^{2}+2an+c)^{2} + h, if a^{2}+b^{2} = c^{2}+h
where p = 2(b+c) for arbitrary {n,h}.
Choudhry
(p^{2}q+pq)^{2} + (2pq+1)^{2} = (p^{2}q+p+q)^{2} + 1
(dx)^{2} + (d^{2}y^{2}1)^{2} = (d^{2}y^{2}+d)^{2} + 1, where x^{2}2(d+1)y^{2} = 1
More generally, one can convert this to a Pell equation for a broader class.
Piezas
u^{2} + ((dv^{2})/2)^{2} = ((d+v^{2})/2)^{2} ± 1, where u^{2}dv^{2} = ±1. (With d chosen such that terms are integers.)
Another way would be to use Euler’s complete soln for x^{2}+y^{2} = z^{2}+t^{2},
(ac+bd)^{2} + (adbc)^{2} = (acbd)^{2} + (ad+bc)^{2}
and set one term, say, adbc = ±1, a form discussed previously. Alternatively, using the simple identity,
(2m)^{2} + (2m^{2}n)^{2} = (2m^{2}n+1)^{2} + (2n1)
set n = 1 or 0 to get,
(2m)^{2} + (2m^{2}1)^{2} = (2m^{2})^{2} + 1 (2m)^{2} + (2m^{2})^{2} = (2m^{2}+1)^{2}  1
the second of which is for the next section.
5. Form: x^{2}+y^{2} = z^{2}1
For this form, starting with an initial identity, one can generate an infinite sequence of identities.
Proof (Piezas):
If a^{2}+b^{2} = c^{2 }± d^{2}, then,
(4ac2bc+2t)^{2} + (2ac4bc+2t)^{2} = (4ac4bc+3t)^{2}  d^{4}, where t = a^{2}+b^{2}+c^{2}
which implies one soln to u^{2}+v^{2 }= w^{2}x^{2} leads to another. Furthermore, iterative use leads to u^{2}+v^{2} = w^{2}x^{k} with k a power of two, and x = 1 a special case. For example, the identity given in the previous section, (2m)^{2} + (2m^{2})^{2} = (2m^{2}+1)^{2}  1
yields,
(8m^{4}+16m^{3}+12m^{2}+8m+2)^{2} + (8m^{3}+8m^{2}+4m+2)^{2} = (8m^{4}+16m^{3}+16m^{2}+8m+3)^{2}  1 which, in turn, yields another, ad infinitum. Other solns are,
E. Grigorief
((a^{2}b^{2}+c^{2}d^{2})/2)^{2} + (ab+cd)^{2} = ((a^{2}+b^{2}+c^{2}+d^{2})/2)^{2}  1, where adbc = ±1
One can also use Genocchi’s and Pepin’s method,
(2pq(r^{2}2s^{2}))^{2} + (4pqrs)^{2} = (p^{2}+(r^{4}+4s^{4})q^{2})^{2}  1, if p^{2}(r^{4}+4s^{4})q^{2} = ±1
which for the case r = 1 has a parametric soln to be discussed in Simultaneous Polynomials. (This also discusses the complete soln, by Genocchi and Pepin, such that {x^{2}+y^{2}1, x^{2}y^{2}1} are both squares and also involves solving a Pell eqn.) As an overview, it turns out that,
x^{2}+y^{2} = z^{2}±1 x^{3}+y^{3} = z^{3}±1 x^{4}+y^{4} = z^{2}+1
can be solved using Pell equations, the higher powers to be discussed in their respective sections.
6: Form x^{2}+y^{2} = z^{2}+nt^{2}
This generalizes the two forms above. For the special case when n = 1, this is discussed in mx^{2}+ny^{2} = mz^{2}+nt^{2}, the last section below.
A.Gerardin
(1ma)^{2} + (1mb)^{2} = (mc)^{2} + 2, if m = 2(a+b)/(a^{2}+b^{2}c^{2})
Piezas
(2a+bn)^{2} + (2b+an)^{2} = (2ccn)^{2} + n(2a+2b)^{2}, if a^{2}+b^{2} = c^{2}
for some arbitrary constant n which was derived by modifying Gerardin’s. For the special case t = 1, or Pythagoreanlike triples x^{2}+y^{2} = z^{2}+h, it can be proven that for any h, there is an infinity of solns given by,
(2m+1)^{2} + (2m^{2}+2mn)^{2} = (2m^{2}+2mn+1)^{2} + 2n (2m)^{2} + (2m^{2}n)^{2} = (2m^{2}n+1)^{2} + (2n1)
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 x^{2}+y^{2} = z^{2}, 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 x^{2}+y^{2} = z^{2}+h and define the ratio R(h) = N/B for any h, it seems for h < 0, there might be asymptotics which involve 1/√h. For example, for h = 1, given N with increasing bound B = 10, 10^{2}, 10^{3}, and so on gives the sequence of ratios as,
R(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 squarefree h at bound B = 10^{6} and R(h) rounded to five decimal places, paired with its conjectured asymptotic value:
R(2) = 0.17679 1/(4√2) = 0.17677
R(3) = 0.28868 1/(2√3) = 0.28867
R(5) = 0.22366 1/(2√5) = 0.22360
R(6) = 0.20396 1/(2√6) = 0.20412
R(7) = 0.37799 1/√7 = 0.37796
R(10) = 0.15814 1/(2√10) = 0.15811
is highly suggestive considering the simplicity of the form. Interestingly, for h > 0, the R(h) do not seem to have simple ones involving √h. (Note: Computing R(h) is timeconsuming, 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: x^{2}+y^{2} = z^{2}+nt^{k}
J. Rose
(4n^{2})^{2} + (4n^{3})^{2} = (4n^{2}(n1))^{2} + (2n)^{5}
MehmedNadir
x^{2} + y^{2 }= z^{2} + (a^{2}+b^{2})^{5 }
{x,y,z} = {(a^{2}+b^{2})(a^{2}b^{2})b, ((a^{2}+1)(a^{2}+b^{2})^{2}+4b^{4})/2, ((a^{2}1)(a^{2}+b^{2})^{2}4b^{4})/2}
E. Barisien
((n+2)(n^{2}2n2))^{2} + (4n(n+1))^{2} = (2(n+1)(n+2))^{2} + n^{6}
These can be generalized by the identity given previously by this author for Pythagoreanlike triples where the constant term 2n or 2n1 can be equated to any kth power p^{k}, and n then easily solved for.
Q: Can MehmedNadir’s identity be generalized to x^{2} + y^{2 }= z^{2} + (a^{2}+b^{2})^{k} for other k not using the identity for Pythagoreanlike triples? See also sum of three squares x^{2} + y^{2} + z^{2} = (a^{2}+b^{2})^{k}, one of which is also by MehmedNadir.
8. Form: x^{2}+y^{2} = mz^{2}+nt^{2}
The parallelogram law states that given a parallelogram's four sides {t,t,z,z} (since it necessarily has opposite sides equal) and two diagonals {x,y}, then,
x^{2}+y^{2} = 2z^{2}+2t^{2} (eq.1)
If the diagonals are equal as well, or x = y, then the parallelogram law reduces to the wellknown Pythagorean theorem, a^{2}+b^{2} = c^{2}. (See also the quadrilateral law, on the form a^{2}+b^{2}+c^{2}+d^{2} = x^{2}+y^{2}+z^{2}.) Solns to eq.1 are,
Euler
{x,y,z,t} = {2(prqs), 2(ps+qr), (p+q)r(pq)s, (pq)r+(p+q)s}
M. Gruber
{x,y,z,t} = {4(pq)q2s, 4(pq)r, s, 4p(pq)s}, where s = 2p^{2}q^{2}r^{2}
Paul Cheffers
{x,y} = {z+t, zt}
More completely, by using the transformation {2z, 2t} = {p+q, pq} on (eq.1), it suffices to solve the form x^{2}+y^{2} = p^{2}+q^{2}, which can be done completely. Alternatively, if n is a constant that is the sum of two squares n = a^{2}+b^{2},
x^{2}+y^{2} = n(z^{2}+t^{2})
then this has soln {x,y} = {az+bt, bzat} for constant {a,b} and arbitrary {z,t} since this yields,
x^{2}+y^{2} = (a^{2}+b^{2})(z^{2}+t^{2})
For example, x^{2}+y^{2} = 5(z^{2}+t^{2}) has {x,y} = {2z+t, z2t}, with Cheffer's as the case {a,b} = {1,1}, and so on.
Q: Any soln to the more general form x^{2}+y^{2} = mz^{2}+nt^{2}?
9. Form: c_{1}(x^{2}+ny^{2}) = c_{2}(z^{2}+nt^{2})
In a manner, this generalizes the previous section. For constants m_{1}, m_{2}, given an equation of form,
Case 1: m_{1}(a^{2}+b^{2}) = (r^{2}+s^{2})(c^{2}+d^{2})
Case 2: m_{1}(a^{2}+b^{2}) = m_{2}(c^{2}+d^{2})
these can be solved if m_{1}, m_{2} are sums of two squares using the general identity,
(a^{2}+b^{2})(p^{2}+q^{2}) = (c^{2}+d^{2})(r^{2}+s^{2})
where {a,b,c,d} = {ru+sv, surv, pu+qv, qupv}. For Case 1, it is constant {p,q} with arbitrary {r,s,u,v}; for Case 2, constant {p,q,r,s} with arbitrary {u,v}.
Example 1:
5(a^{2}+b^{2}) = (r^{2}+s^{2})(c^{2}+d^{2}), then {a,b,c,d} = {ru+sv, surv, u+2v, 2uv}.
Example 2:
5(a^{2}+b^{2}) = 13(c^{2}+d^{2}), then {a,b,c,d} = {3u+2v, 2u3v, u+2v, 2uv}.
10. Form: mx^{2}+ny^{2} = mz^{2}+nt^{2}
Theorem: “There is a complete integral solution to ap^{2}+bpq+cq^{2} = ar^{2}+brs+cs^{2}.”
First. we transform this to diagonal form by the transformation {p,q,r,s} = {ubv, 2av, xby, 2ay} to get,
u^{2}Dv^{2} = x^{2}Dy^{2}
where D = b^{2}4ac is the discriminant. This can be solved in an even more general form.
Theorem: “The equation mu^{2}+nv^{2} = mx^{2}+ny^{2} has the complete solution,
m(ac+bdn)^{2} + n(bcadm)^{2} = m(acbdn)^{2} + n(bc+adm)^{2}, for arbitrary {a,b,c,d}.”
The identity is by Euler. Note that either side is equal to (ma^{2}+nb^{2})(mc^{2}+nd^{2}). Proof (Piezas): Given any nontrivial soln {u,v,x,y}, one can always find rational {a,b,c,d} using the formulas,
{a,b,c} = {(yv)/(2dm), (ux)/(2dn), dn(v+y)/(ux)}
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 mu^{2}+nv^{2} = mx^{2}+ny^{2}, 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, bcadm, acbdn}, {a,b,c}]
which finds {a,b,c} and where the resulting radical solns can then be simplified using the relation mu^{2}+nv^{2} = mx^{2}+ny^{2}. Incidentally, if we set the fourth variable y = bc+adm = 0, Euler’s identity reduces to,
m(ma^{2}nb^{2})^{2} + n(2abm)^{2} = m(ma^{2}+nb^{2})^{2}
which for m=n=1, not surprisingly, is the wellknown formula for Pythagorean triples. There are also other versions.
S. Realis (complete)
u^{2}+nv^{2} = x^{2}+ny^{2}
{u,v,x,y} = {a^{2}n(ab)^{2}+n(ac)^{2}, b^{2}(ab)^{2}+n(bc)^{2}, a^{2}+nb^{2}nc^{2}, ^{ }c^{2}(ac)^{2}n(bc)^{2}}
V. Lebesgue
(p+q+rs)^{2} + (p+qr+s)^{2} = (pq+r+s)^{2} + (pqrs)^{2}, if pq = rs
or, alternatively,
(p+q)^{2} + (rs)^{2} = (r+s)^{2} + (pq)^{2}, if pq = rs
A.Desboves
x^{2}+y^{2} = (m^{2}+n^{2})z^{2}
{x,y,z} = {(u^{2}dv^{2})m, nu^{2}2duv+dnv^{2}, u^{2}2nuv+dv^{2}}, where d = m^{2}+n^{2}
11. Form: x^{2}+y^{2} = z^{k}+t^{k}
(Update, 11/9/09): W. Lewis asked if results were known for the eqn x^{2}+y^{2} = z^{5}+t^{5}. 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:
Method 1: Use the BramaguptaFibonacci TwoSquare Identity for the problem in its general case. This is given by,
(pr+qs)^{2} + (psqr)^{2} = (p^{2}+q^{2})(r^{2}+s^{2}) (eq.1) There are 4 free variables {p,q,r,s} to dispose of. Select {p,q} = {x^{k}, y^{k}}. The RHS, after distributing, becomes,
(pr+qs)^{2} + (psqr)^{2} = x^{2k}(r^{2}+s^{2}) + y^{2k}(r^{2}+s^{2})
Then solve for r^{2}+s^{2} = t^{k }which was discussed in the previous page and is easy to do. Eq. 1 then becomes,
(pr+qs)^{2} + (psqr)^{2} = (tx^{2})^{k} + (ty^{2})^{k}
For k = 3, this is satistied by {p,q,r,s,t} = {x^{3}, y^{3}, u^{3}3uv^{2}, 3u^{2}vv^{3}, u^{2}+v^{2}} with 4 free variables. For ex, let {u,v,x,y} = {2,1,3,4}, then,
758^{2} + 169^{2} = (5*9)^{3} + (5*16)^{3}
Method 2: For coprime terms, the small solns given by Mathematica were enough to provide a clue for a parametrization to the eqn,
x^{2}+y^{2} = z^{k}+1
A. For all ODD k in terms of powers and nearpowers of 2. This is given by,
(2^{m+1})^{2} + (2^{2m+1}1)^{2} = (2^{4m+2})^{1} + 1
(2^{3m+2})^{2} + (2^{6m+3}1)^{2} = (2^{4m+2})^{3} + 1
(2^{5m+3})^{2} + (2^{10m+5}1)^{2} = (2^{4m+2})^{5} + 1 etc. The pattern is easily seen and one can notice how z has a constant form regardless of k.
(Update, 2/19/11): Oliver Couto gave an EVEN k counterpart,
(n^{k/2}(n^{k}2))^{2} + (2n^{k}1)^{2} = (n^{3})^{k} + 1
hence,
(n(n^{2}2))^{2} + (2n^{2}1)^{2} = (n^{3})^{2} + 1
(n^{2}(n^{4}2))^{2} + (2n^{4}1)^{2} = (n^{3})^{4} + 1
(n^{3}(n^{6}2))^{2} + (2n^{6}1)^{2} = (n^{3})^{6} + 1
Piezas
For EVEN k = 4m+2:
(n^{3}2n)^{2} + (2n^{2}1)^{2} = n^{6} + 1
(n^{5}2n^{3}+2n)^{2} + (2n^{4}2n^{2}+1)^{2} = n^{10} + 1
(n^{7}2n^{5}+2n^{3}2n)^{2} + (2n^{6}2n^{4}+2n^{2}1)^{2} = n^{14} + 1
and so on.
