II. Sum / Sums of three squares
1. Form: x^{2}+y^{2}+z^{2} = t^{k}
Geometrically, the case k = 2 of Form 1 can be seen as a cuboid with sides {x,y,z} and space diagonal t,
or, for t = r = 1, a sphere of unit radius,
By analogy to Pythagorean triples, the numbers {x,y,z,t} is also known as a Pythorean quadruple. The simplest integer soln is,
n^{2} + (n+1)^{2} + (n^{2}+n)^{2} = (n^{2}+n+1)^{2}
which easily proves that, unlike in Pythagorean triples, every integer appears at least once in an integer Pythagorean quadruple. Similarly for quintuples,
n^{2} + (n2)^{2} + (2n+1)^{2} + (3n^{2}+2)^{2} = (3n^{2}+3)^{2}
and so on.
Theorem: The complete soln to x_{1}^{2}+x_{2}^{2}+x_{3}^{2} = y_{1}^{2} is given by,
((a^{2}b^{2}c^{2})t)^{2} + (2abt)^{2} + (2act)^{2} = ((a^{2}+b^{2}+c^{2})t)^{2}
where t is just a scaling factor. Proof: For any soln x_{i} one can always find rational {a,b,c,t} using the formulas: {a,b,c,t} = {x_{1}+y_{1}, x_{2}, x_{3}, 1/(2(x_{1}+y_{1}))}. Similarly, Theorem: The complete nontrivial soln to x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2} = y_{1}^{2} is given by the scaled EulerAida Ammei identity, ((a^{2}b^{2}c^{2}d^{2})t)^{2} + (2abt)^{2} + (2act)^{2} + (2adt)^{2}= ((a^{2}+b^{2}+c^{2}+d^{2})t)^{2}
Proof: For any soln x_{i} one can always find rational {a,b,c,d,t} using the formulas: {a,b,c,d,t} = {x_{1}+y_{1}, x_{2}, x_{3}, x_{4}, 1/(2(x_{1}+y_{1}))}.
and so on. One can easily see the pattern for other n sums of squares.
Update (10/11/09): The eqn x^{2}+y^{2}+z^{2} = 1, which also defines a sphere of unit radius centered at the origin, also appears in the eqn x^{2} = 1 where x, instead of the imaginary unit, is to be a quaternion. To recall, given the expansion of (a+bi+cj+dk)^{n} = A+Bi+Cj+Dk, then,
A^{2}+B^{2}+C^{2}+D^{2} = (a^{2}+b^{2}+c^{2}+d^{2})^{n }
In a Mathematica addon package, this object is given as Quaternion[a,b,c,d] and one can use NonCommutativeMultiply to multiply them, and for the case n = 2, we get,
Quaternion[a,b,c,d] ** Quaternion[a,b,c,d] = Quaternion[a^{2}b^{2}c^{2}d^{2}, 2ab, 2ac, 2ad] (eq.1)
One can note two immediate consequences. First, as expected, (a^{2}b^{2}c^{2}d^{2})^{2} + (2ab)^{2} + (2ac)^{2} + (2ad)^{2} = (a^{2}+b^{2}+c^{2}+d^{2})^{2}. Second, while the Fundamental Theorem of Algebra states that an nth degree univariate equation with complex coefficients has n complex roots, this does not apply when the roots are quaternions. For example, the nth roots of plus/minus unity or, x^{n} = ±1 have n complex solns. But if x is to be a quaternion, then there can be an infinite number of them. Let n = 2 and take the negative case: x^{2} = 1. Thus, we wish (eq.1) to have the product: Quaternion[1,0,0,0]. Since the {a,b,c,d} are to be real, then a = 0, and {b,c,d} should satisfy,
b^{2}+c^{2}+d^{2} = 1
which have an infinite number of real and rational solns, a subset of which are rationalized Pythagorean triples. The rational soln with the smallest denominator is {b,c,d} = ±{1,0,0}, in which case the quaternion reduces to the imaginary unit. The next smallest is ±{1,2,2}/3. Of course, the most general setting for solving x^{2} = 1 would be x as the octonions, where the coefficients of the nonreal part would have to satisfy the sum of seven squares,
y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+y_{4}^{2}+y_{5}^{2}+y_{6}^{2}+y_{7}^{2} = 1
and, if all y_{i} are to be nonzero, the soln with the smallest denominator is ±{1,1,1,1,2,2,2}/4. (End update.) Theorem (H. Monck): If a^{2}+b^{2}+c^{2} = d^{2}, then (a+b+d)^{2} + (a+c+d)^{2} + (b+c+d)^{2} = (a+b+c+2d)^{2}. Thus, one soln can generate others (with sign changes in a,b,c,d yielding several distinct ones), just like for Pythagorean triples. Q: Any similar identity for four squares or more?
Other versions for sums of three squares are,
A. Desboves (complete soln)
a^{2}+b^{2}+c^{2} = d^{2}
{at, bt, ct, dt} = {2(p^{2}q^{2}+r^{2}), 2(pq)^{2}2r^{2}2p(qr), p^{2}(qr)^{2}4r(pq), 3(p^{2}+q^{2}+r^{2})2q(2p+r)}
for arbitrary p,q,r and where t is just a scaling factor. (V. Lebesgue earlier gave a more complicated 6parameter soln.) Proof (Piezas): For any soln a,b,c,d, one can always find rational p,q,r,t using the formulas,
{p,q,r,t} = {2a+2b, 3a2d, a2c, 8(2a2b+c3d)}
Piezas (complete)
t^{2}(2uv+2w)^{2} + t^{2}(u+2v2w)^{2} + t^{2}(1+w)^{2} = t^{2}(2u2v+3w)^{2}, where w = (u^{2}+v^{2}+1)/2
and t is a scaling factor. Since the eqn is homogenous, after some changes one can also use w = u^{2}+v^{2}+x^{2} though for simplicity it can be set x=1 without loss of generality. Proof: For any particular soln a,b,c,d, one can always find rational u,v,t given by,
{u,v} = {(2ab2d)/t, (a+2b+2d)/t}, where t = (2a2b+c3d)
Note that the scaling factor t is essentially the same as the one for Desboves. Partial solns in terms of binary quadratic forms can also be given,
A. Gerardin
(6x^{2}14xy+6y^{2})^{2} + (3x^{2}3y^{2})^{2} + (2x^{2}2y^{2})^{2} = (7x^{2}12xy+7y^{2})^{2}
Notice that 6^{2}+3^{2}+2^{2} = 7^{2}. This can be generalized as,
Piezas
(ax^{2}2dxy+ay^{2})^{2} + (bx^{2}by^{2})^{2} + (cx^{2}cy^{2})^{2} = (dx^{2}2axy+dy^{2})^{2}, if a^{2}+b^{2}+c^{2} = d^{2}
Euler, Lebesgue
(a^{2}b^{2}c^{2}+d^{2})^{2} + (2ab2cd)^{2 }+ (2ac+2bd)^{2} = (a^{2}+b^{2}+c^{2}+d^{2})^{2}
which is also known as the Lebesgue ThreeSquare Identity.
The square (a^{2}+b^{2}+c^{2}+ab+ac+bc)^{2} can be identically expressed as the sum of 3 or 4 squares,
Catalan
((a+b)(a+c))^{2} + ((a+b)(b+c))^{2} + (abacbcc^{2})^{2} = (a^{2}+b^{2}+c^{2}+ab+ac+bc)^{2}
J. Neuberg:
a^{2}(a+b+c)^{2} + b^{2}(a+b+c)^{2} + c^{2}(a+b+c)^{2} + (ab+ac+bc)^{2} = (a^{2}+b^{2}+c^{2}+ab+ac+bc)^{2}
Likewise, (a^{2}+b^{2}+2c^{2})^{2} as 3 squares (in two ways), or 4 squares,
E. Catalan:
(2(a+b)c)^{2} + (a^{2}b^{2})^{2} + (2ab2c^{2})^{2} = (a^{2}+b^{2}+2c^{2})^{2}
(2(a+b)c)^{2} + (2(ab)c)^{2} + (a^{2}+b^{2}2c^{2})^{2 }= (a^{2}+b^{2}+2c^{2})^{2}
(a^{2}c^{2})^{2} + ((a+c)(b+c))^{2} + ((a+c)(bc))^{2} + (2acb^{2}c^{2})^{2} = (a^{2}+b^{2}+2c^{2})^{2} (Update, 6/25/10): Alain Verghote gave a similar identity,
(2(a+b)c)^{2} + (2(ab)c)^{2} + (2a^{2}+2b^{2}c^{2})^{2 }= (2a^{2}+2b^{2}+c^{2})^{2}
though this can be derived from Catalan's by letting c → c'/2.
J. Euler
((a^{2}+c^{2})(b^{2}c^{2}))^{2} + (2bc(a^{2}c^{2}))^{2} + (4abc^{2})^{2} = ((a^{2}+c^{2})(b^{2}+c^{2}))^{2}
Just like one can solve the sum of two squares equal to a kth power in the form,
A^{2}+B^{2} = (a^{2}+b^{2})^{k}
it can also be done for three squares,
A^{2}+B^{2}+C^{2} = (a^{2}+b^{2}+c^{2})^{k}
for any positive integer k. In fact, it can be done for n squares and the general result is that:
Theorem 1: “For all n and k, then x_{1}^{2}+x_{2}^{2}+… +x_{n}^{2} = (y_{1}^{2}+y_{2}^{2}+… +y_{n}^{2})^{k}, or the kth power of the sum of n squares is identically the sum of n squares.”
Theorem 2: “For all n > 1 and k > 2, then x_{1}^{2}+x_{2}^{2}+… +x_{n+1}^{2} = (y_{1}^{2}+y_{2}^{2}+… +y_{n}^{2})^{k}, or the kth power of the sum of n squares is identically the sum of n+1 squares.”
Note 1: The case k = 2 is discussed in Sums of n Squares.
Note 2: The cubic version A^{3}+B^{3}+C^{3}3ABC = (a^{3}+b^{3}+c^{3}3xyz)^{k} is discussed as Form 21 in Cubic Polynomial as a kth Power.
Proof for Theorem 1: (Piezas) The proof is simple. Let the expansion of the complex number (a±bi)^{k} be,
U+Vi = (a+bi)^{k}; UVi = (abi)^{k}
where U,V are expressions in the arbitrary a,b. It is easy to see their product, or norm, is,
U^{2}+V^{2} = (a^{2}+b^{2})^{k}
Note that b can be factored out in V, or V = V_{1}b, so if we let {a,b} = {p_{1},√p_{0}}, then,
U^{2 }+ p_{0}V_{1}^{2} = (p_{1}^{2}+p_{0})^{k}
And since p_{0} is arbitrary, one can choose it to be the sum of squares p_{0} = p_{2}^{2}+p_{3}^{2}+…+p_{n}^{2} giving,
U^{2 }+ V_{1}^{2}p_{2}^{2 }+ … + V_{1}^{2}p_{n}^{2} = (p_{1}^{2}+ p_{2}^{2}+…+p_{n}^{2})^{k}
Thus proving the kth power of n squares is itself the sum of n squares. (End proof.) For k = 2, this is,
(p_{1}^{2}p_{0})^{2} + p_{0}(2p_{1})^{2} = (p_{1}^{2}+p_{0})^{2}
so if p_{0 }= p_{2}^{2}, this gives the classic formula for Pythagorean triples. But if p_{0 }= p_{2}^{2}+p_{3}^{2}, and changing variables, yields an analogue for three squares,
(a^{2}b^{2}c^{2})^{2} + (2ab)^{2} + (2ac)^{2} = (a^{2}+b^{2}+c^{2})^{2}
which, after scaling, is the complete soln. For k = 3,
(a^{3}3ab^{2}3ac^{2})^{2} + b^{2}(3a^{2}+b^{2}+c^{2})^{2} + c^{2}(3a^{2}+b^{2}+c^{2})^{2} = (a^{2}+b^{2}+c^{2})^{3}
and so on for all positive integer k and n. (Using identities by Mehmed Nadir and Matsunago, we will see that the kth power of the sum of n squares is also identically the sum of n+1 squares for k = 3m or 4m. Update: It can be done for all k > 2. See further below.)
Mehmed Nadir (also known as Mehmet Nadir)
a^{2}(a^{2}+b^{2})^{2} + (2ab^{2})^{2} + (a^{2}b^{2})^{2}b^{2} = (a^{2}+b^{2})^{3}
Matsunago:
(a^{4}b^{4})^{2} + (4a^{2}b^{2})^{2} + (2ab(a^{2}b^{2}))^{2} = (a^{2}+b^{2})^{4}
Q: Is there one for k = 5? (See update below.) While we already know that (a^{2}+b^{2})^{k} is expressible as the sum of two squares, for what k is it identically the sum of three nonzero squares? By solving a^{2}+b^{2} = (p^{2}+q^{2})^{m}, one can see there are solns when k = 3m, 4m for all integer m > 0. Letting d = b^{2}, we can also express these as,
a^{2}(a^{2}+d)^{2} + (2ad)^{2} + (a^{2}d)^{2}d = (a^{2}+d)^{3} (a^{4}d^{2})^{2} + (4a^{2}d)^{2} + (2a(a^{2}d))^{2}d = (a^{2}+d)^{4}
where d can then be set as the sum of n squares d = x_{2}^{2}+x_{3}^{2}+…+x_{n}^{2} thus proving, “ ...The kth power of the sum of n squares is the sum of n+1 squares for k = 3m or 4m.” But for what other k is there? One limited form with b = 1, is
A. Gerardin
(y1)^{2} + y^{2} + (y+1)^{2} = x^{2} + 1, if x^{2}3y^{2} = 1
Thus the kth power of (x^{2}+1)^{k}, for all solns x to x^{2}3y^{2} = 1, should be expressible as the sum of three squares for any positive integer k. For ex, substituting {p,q,r} = {y1, y, y+1} into the general identity, then for k = 2 this is,
(y^{2}+4y)^{2} + (2y^{2}2y)^{2} +(2y^{2}2)^{2} = (x^{2}+1)^{2}, if x^{2}3y^{2} = 1
and so on for all k. A closer one is,
(a2b)^{2} + (2ab)^{2 }+ (2a+2b)^{2} = 9(a^{2}+b^{2})
though this does not give MehmedNadir’s or Matsunago’s elegant forms.
Update (6/18/09, 10/12/09) Theorem: "For all k > 2, the expression (a^{2}+b^{2})^{k} is the sum of three squares as polynomials in {a,b} with integral coefficients." Proof: Given the tautology,
a^{2}+b^{2} = a^{2}+b^{2}
Multiply both sides by (a^{2}+b^{2})^{2n},
(a^{2}+b^{2}) (a^{2}+b^{2})^{2n} = (a^{2}+b^{2})^{2n+1}
Distribute,
a^{2}(a^{2}+b^{2})^{2n} + b^{2}(a^{2}+b^{2})^{2n} = (a^{2}+b^{2})^{2n+1}
and it is easy to express either of the squares on the LHS as the sum of two squares. And since the eqn x_{1}^{2}+x_{2}^{2}+… +x_{n}^{2} = (y_{1}^{2}+y_{2}^{2}+… +y_{n}^{2})^{m} can be solved for any positive integer m (Theorem 1 discussed at the start of this Section), then one can also solve,
a^{2}(a^{2}+b^{2})^{2n} + b^{2}(a^{2}+b^{2})^{2n} = (p^{2}+q^{2})^{(2n+1)m}
where {a,b} are polynomials in {p,q}. Together with the identity for k = 4, which is effectively for all k = 4m, then it is possible to express (a^{2}+b^{2})^{k} as the sum of three squares for all k > 2. (End proof.) For k = 5, this gives,
a^{2}(a^{2}+b^{2})^{4} + (4a^{3}b^{2}4ab^{4})^{2} + (a^{4}b6a^{2}b^{3}+b^{5})^{2} = (a^{2}+b^{2})^{5}
One can then always find expressions {a,b} such that the sum is (p^{2}+q^{2})^{5m}. And so on for other prime k > 2. However, more generally,
Theorem 2: “For all n > 1 and k > 2, then x_{1}^{2}+x_{2}^{2}+… +x_{n+1}^{2} = (y_{1}^{2}+y_{2}^{2}+… +y_{n}^{2})^{k}, or the kth power of the sum of n squares is identically the sum of n+1 squares.”
Proof: As before, one simply expresses b^{2 }= d, set d as any nonzero number of squares, distribute the third term appropriately, then use Theorem 1. For example, for k = 5, this is,
a^{2}(a^{2}+d)^{4} + (4a^{3}d4ad^{2})^{2} + (a^{4}6a^{2}d+d^{2})^{2}d = (a^{2}+d)^{5}
a^{2}(a^{2}+b^{2}+c^{2})^{4} + (4a^{3}(b^{2}+c^{2})4a(b^{2}+c^{2})^{2})^{2} + b^{2}(a^{4}6a^{2}(b^{2}+c^{2})+(b^{2}+c^{2})^{2})^{2} + c^{2}(a^{4}6a^{2}(b^{2}+c^{2})+(b^{2}+c^{2})^{2})^{2} = (a^{2}+b^{2}+c^{2})^{5}
or 4 squares equal to the 5th power of 3 squares, one can then use Theorem 1 to find {a,b,c} such a^{2}+b^{2}+c^{2} = (p^{2}+q^{2}+r^{2})^{m}, for any m. For m = 2, this would be {a,b,c} = {p^{2}q^{2}r^{2}, 2pq, 2pr}. And so on for any n > 1 and k > 2. (End update.)
For (a^{2}+b^{2})^{k} as the sum of four squares, Barisien gave several examples for small k. These are easy to find since solns to u^{2}+v^{2 }= z^{m} and x^{2}+y^{2} = z^{n} immediately imply,
(u^{2}+v^{2})(x^{2}+y^{2}) = (ux)^{2}+(vx)^{2}+(uy)^{2}+(vy)^{2} = z^{m+n}.
E. Barisien:
a^{2}(a^{2}b^{2}c^{2})^{2} + b^{2}(a^{2}+b^{2}3c^{2})^{2} + c^{2}(a^{2}3b^{2}+c^{2})^{2} + (2a^{2}b)^{2} + (2a^{2}c)^{2} + (4abc)^{2} = (a^{2}+b^{2}+c^{2})^{3}
Q: How about the kth power of three squares (a^{2}+b^{2}+c^{2})^{k} identically the sum of n squares with n > 4? Any other prime k?
F. Hrodmadko
n^{2} + (n+1)^{2} + (nx)^{2} = (nx+1)^{2}, if x = n+1.
There is a similar identity for third powers, by Jandacek,
n^{3} + (xn1)^{3} + (nx)^{3} = (nx+1)^{3}, if x = 3n^{2}+3n+2.
Proving that any integer n appears at least once as a soln to a^{k}+b^{k}+c^{k} = (c+1)^{k} for k = 2 or 3. Finally, the system of eqns,
x_{1}^{k}+x_{2}^{k}+…+ x_{n}^{k} = y_{1}^{k}+y_{2}^{k}+…+ y_{n}^{k}, for k = 1,2
by Bastien’s theorem (to be discussed more in the section on Equal Sums of Like Powers) has nontrivial solns only for n > 2. It is possible two terms on one side are equal to zero and hence reduces to the minimal form,
a^{k}+b^{k}+c^{k} = d^{k},
the complete soln of which obeys the relationship ab+ac+bc = 0 and is given by,
(p^{2}+pq)^{k} + (pq+q^{2})^{k} + (pq)^{k} = (p^{2}+pq+q^{2})^{k}, for k = 1,2
implying at least one term is always negative. There are analogous cases for multigrade third, fourth, fifth powers to be given later.
2. Form: x^{2}+y^{2}+z^{2 }= u^{2}+v^{2}
One can set any of its terms equal to 1, but depending on which side of the equation it is chosen, it may need slightly different Pell equations. To set this form either as p^{2}+q^{2}+1^{ }= s^{2}+t^{2}, or p^{2}+q^{2}+r^{2 }= s^{2}+1, can involve,
x^{2}(d+1)y^{2} = ±1; or x^{2}(d^{2}+1)y^{2} = ±1
respectively. To find the complete soln of this form, use the more general one for x_{1}^{2}+x_{2}^{2}+x_{3}^{2} = y_{1}^{2}+y_{2}^{2}+y_{3}^{2},
(ab)^{2} + (c+d)^{2} + (ef)^{2} = (a+b)^{2} + (cd)^{2} + (e+f)^{2}
where abcd+ef = 0. It is easy to make one of the terms vanish.
(ab)^{2} + (c+d)^{2} + 4n = (a+b)^{2} + (cd)^{2}
where n = abcd. Let {a,b,c} = {x+y, xy, y^{2}} and this condition becomes,
x^{2}(d+1)y^{2} = n
where one can then set n = ±1. If d,x,y are odd, then all the terms above are even and can be reduced. Alternative versions are,
Catalan (complete)
(ad+qr)^{2} + (bdpr)^{2} + w^{2} = (a+bdpr)^{2} + (adb+qr)^{2}, where r = (a^{2}+b^{2}w^{2})/2, ap+bq = 1.
If we set p = q = 0, this reduces to,
(ad)^{2} + (bd)^{2} + w^{2} = (a+bd)^{2} + (adb)^{2}, if a^{2}+b^{2} = w^{2}
or, explicitly,
(u^{2}+v^{2})^{2} + (2duv)^{2} + (du^{2}dv^{2})^{2} = (du^{2}2uvdv^{2})^{2} + (u^{2}+2duvv^{2})^{2}
Integral solns to p^{2}+q^{2}+r^{2 }= s^{2}+1 can then be obtained by setting the last term u^{2}+2duvv^{2} = ±1. Let {u,v} = {xdy, y} and this transforms into the Pell equation x^{2}(d^{2}+1)y^{2} = ±1. Other solns 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}
(a^{2}+c^{2}+a1)^{2} + (2c)^{2} + (2a+1)^{2} = (a^{2}+c^{2}+a+1)^{2} + 1
which were derived using the complete soln of x_{1}^{2}+x_{2}^{2}+x_{3}^{2} = y_{1}^{2}+y_{2}^{2}+y_{3}^{2}. For the special case c = 0, these two become the same nontrivial identity,
(a^{2}+a1)^{2} + (2a+1)^{2} = (a^{2}+a+1)^{2} + 1
while Fauquebergue’s for either {a,b} = 0 reduces to a tautology. The ff has been discussed previously,
A. Gerardin
(y1)^{2} + y^{2} + (y+1)^{2} = x^{2} + 1, if x^{2}3y^{2} = 1.
Q: Does this generalize to an identity needing Pell equations distinct from the one derived from Catalan’s?
E. Lucas
(4m)^{2} + (8m+1)^{2} + (8m+2)^{2} = (12m+2)^{2} + 1
3. Form: (x^{2}1)(y^{2}1) = (z^{2}1)^{2}
This eqn was studied by Schinzel and Sierpinski. Expanded out, this is just a special case of the previous form being,
x^{2 }+ y^{2 }+ (z^{2}1)^{2} = x^{2}y^{2} +1
This has an infinite number of integral solns. A pair is given by,
{x,y} = {z+t, zt}, if t^{2}2z^{2} = 2. (Smallest soln is {t,z} = {10,7}.) {x,y} = {z+2t, z2t}, if z^{2}2t^{2} = 1
Q: Any others? As with Gerardin’s and others, this can also be solved as x^{2 }+ y^{2 }+ (z^{2}1)^{2} = (x^{2}y^{2} +1)^{k }for any k > 0.
4. Form: x^{2}+y^{2}+z^{2} = u^{2}+v^{2}+w^{2}
This form is important since certain equal sums of like powers for k = 4,5,6 may have this as a “side” condition. The complete soln already has been given earlier,
(a+b)^{2} + (c+d)^{2} + (e+f)^{2} = (ab)^{2} + (cd)^{2} + (ef)^{2}
where ab+cd+ef = 0. It is easy to give other conditions such that one of the terms vanish or has special forms like equal to unity, etc. The complete soln for k = 1,2 is,
L. Dickson
(ad+e)^{k} + (bc+e)^{k} + (ac+bd+e)^{k} = (ac+e)^{k} + (bd+e)^{k} + (ad+bc+e)^{k}, for k = 1,2
Note: This can be true for k = 4 as well if (a+b)(c+d) = 3e.
(Update, 12/29/09): Alain Verghote
Define {p,q} = {x^{2}+2y^{2}, 2x+xy^{2}}, then,
p^{2} = (x^{2})^{2} + (2xy)^{2} + (2y^{2})^{2} q^{2} = (2x)^{2} + (2xy)^{2} + (xy^{2})^{2}
Since there is a common middle term, then,
(2x)^{2} + (xy^{2})^{2} + (x^{2}+2y^{2})^{2} = (2x+xy^{2})^{2} + (x^{2})^{2} + (2y^{2})^{2}
This belongs to the general form,
a^{k} + b^{k} + (c+d)^{k} = (a+b)^{k} + c^{k} + d^{k},
which is already for k = 1, and will also be true for k = 2 if abcd. (End update.)
C. Goldbach
p^{2} + q^{2} + (p+q+3r)^{2} = (p+2r)^{2} + (q+2r)^{2} + (p+q+r)^{2}
Note: Can be simultaneous for k = 2,3 if 6(pq+pr+qr)+5r^{2} = 0. Alternatively, this can be more simply expressed as, (pr)^{k} + (qr)^{k} + (p+q+r)^{k} = (p+r)^{k} + (q+r)^{k} + (p+qr)^{k} which is already for k = 2, but is also valid for k = 3 if 6pq = r^{2}.
Euler:
a^{2} + b^{2} + c^{2} = (a+2d)^{2} + (b+2d)^{2} + (c+2d)^{2}, if a+b+c+3d = 0.
In general,
a^{2} + b^{2} + c^{2} = (a+pm)^{2} + (b+qm)^{2} + (c+rm)^{2}, if m = 2(ap+bq+cr)/(p^{2}+q^{2}+r^{2})
R. Davis
a^{2}+b^{2}+c^{2} = (ap+bq+cr)^{2} + (aq+br+cp)^{2} + (ar+bp+cq)^{2}, if p+q+r =1 and 1/p+1/q+1/r = 0.
Q: This seems a particularly elegant soln. In what context does it appear?
L. Lander
x_{1}^{2}+x_{2}^{2}+x_{3}^{2} = y_{1}^{2}+y_{2}^{2}+y_{3}^{2}
{x_{1}, x_{2}, x_{3}} = {b(ac+d), (ac)(acd), ab(a+c)} {y_{1}, y_{2}, y_{3}} = {b(acd), (a+c)(ac+d), ab(ac)}
where d = b^{2}c^{2}. Any soln that also satisfies x_{1}x_{2}x_{3} = y_{1}y_{2}y_{3} (which is the case for this example) can then generate solns to,
z_{1}^{k}+z_{2}^{k}+z_{3}^{k}+z_{4}^{k} = z_{5}^{k}+z_{6}^{k}+z_{7}^{k}+z_{8}^{k}, for k = 1,3,5
A. Choudhry
(ay+b)^{2} + (cy+ad)^{2} + (ay+b+cd)^{2} = (ayb)^{2} + (cyad)^{2} + (aybcd)^{2}
for any variable. For certain values a,b,c,d,y, which require solving a particular elliptic curve, this also becomes true for k = 2,3,4.
A. Choudhry
x_{1}^{k}+x_{2}^{k}+x_{3}^{k} = y_{1}^{k}+y_{2}^{k}+y_{3}^{k}
{x_{1}, x_{2}, x_{3}} = {2am+(b+c)n, (a+b)m+(a+c)n, (ab)m(ac)n} {y_{1}, y_{2}, y_{3}} = { 2am+(b+c)n, (a+b)m+(a+c)n, (ab)m(ac)n}
for k = 1,2. This can be extended for k = 6, but a quartic polynomial in c must be made a square as will be discussed in the section on Sixth Powers.
Q: Any other soln for k=2 that can be used for k=6?
5. Form: x^{2}+y^{2}+z^{2} = (u^{2}+v^{2}+w^{2})Poly(t)
H. Burhenne
x_{1}^{2}+x_{2}^{2}+x_{3}^{2} = (p^{2}+q^{2}+r^{2})(x^{2}+y^{2}+z^{2})^{2}
{x_{1},^{ }x_{2}, x_{3}} = {ps2tx, qs2ty, rs2tz}, where {s,t} = {x^{2}+y^{2}+z^{2}, px+qy+rz}
for arbitrary p,q,r and x,y,z. One can also set {p,q,r} = {y,z,x} to solve x_{1}^{2}+x_{2}^{2}+x_{3}^{2} = (x^{2}+y^{2}+z^{2})^{3}. This was generalized as,
S. Realis
ax_{1}^{2}+bx_{2}^{2}+cx_{3}^{2} = (ap^{2}+bq^{2}+cr^{2})(ax^{2}+by^{2}+cz^{2})^{2}
{x_{1},^{ }x_{2}, x_{3}} = {ps2tx, qs2ty, rs2tz} where {s,t} = {ax^{2}+by^{2}+cz^{2}, apx+bqy+crz}
where Burhenne’s is simply the case a=b=c=1. (This can be generalized to n addends as discussed in Section 003.) The next one is also essentially the same.
G. Malfatti
p^{2}+q^{2}+r^{2} = (a^{2}+b^{2}+c^{2})(d^{2}+e^{2}+f^{2})^{2}
{p,q,r} = {(d^{2}+e^{2}f^{2})a+2bdf+2cef, (d^{2}+e^{2}+f^{2})b+2adf2cde, (d^{2}e^{2}+f^{2})c+2aef2bde}
H. Mathieu
x^{2}+y^{2}+z^{2} = (a^{2}+b^{2}+c^{2})(c^{2}m^{2}+c^{2}n^{2}+(am+bn)^{2})
{x,y,z} = {abm+(b^{2}+c^{2})n, abn+(a^{2}+c^{2})m, acnbcm}
6. Form: x^{2}+y^{2}+z^{2 }= 3xyz Known as the Markov equation, Frobenius proved that x^{2}+y^{2}+z^{2 }= nxyz has positive integer solns only for n=1,3 with the former reducible to the latter by scaling variables. Markov gave various parametrizations (?) to this equation though this author hasn’t been able to find them yet. (The similar equation x^{3}+y^{3}+z^{3 }= 3xyz will be discussed later in the section on third powers.) Define {x,y,z} as a Markov triple, with the first few ones as {1,1,1}, {2,1,1}, {5,1,2}, {13,1,5}, etc. It can be shown that one soln leads to another,
Theorem: “If x^{2}+y^{2}+z^{2} = 3xyz, then x^{2}+y^{2}+(3xyz)^{2} = 3xy(3xyz)."
More generally, if we let z = (3xy ± p)/2, it transforms to:
p^{2}(9x^{2}4)y^{2} = 4x^{2} (eq.1)
which is a Pelllike eqn. This approach is essentially solving the Markov equation as a quadratic in z and making its discriminant a square. The problem now is to find an x such that this has a soln {p,y}. If for a particular x there is one, then there is in fact an infinity. For the particular case x = 1, let {x,y,z} = {1, y, (3y ± p)/2}, then eq.1 is,
p^{2 }5y^{2} = 4
which guarantees a bisection (every other term) of the Fibonacci numbers, namely y_{1} = {1, 2, 5, 13, etc.}, will appear in a Markov triple,. Since {p,y} are either both odd or even, then z will always be an integer. For x = 2, let {x,y,z} = {2, y, (3y ± 2r)}, then eq.1 is,
r^{2 }2y^{2} = 1
and a bisection of the Pell numbers as y_{2} = {1, 5, 29, 169, etc} will also be in a triple. So what other x are there? There are x that are neither Fibonacci nor Pell, the smallest of which are x = 194, 433, 1325, etc. Since the Markov eqn is symmetric and the terms {x,y,z} can be interchanged, the elements of the bisections naturally can also be used as an x. For example, if we use x = 5, eq.1 is,
p^{2}221y^{2} = 100
This has three fundamental y, namely y = {1, 2, 13} but a nonfundamental one is y = 194 which explains this smallest nonFibonacci, nonPell, Markov number and others. In turn, we can then use this y as an x. In general, if a number appears in a Markov triple, then it will appear in an infinite number of triples. Q. Any parametric soln?
