IV. Some Identities of Squares
(Update, 10/26/09): The Lebesgue Polynomial Identity is given by,
(a^{2}+b^{2}c^{2}d^{2})^{2} + (2ac+2bd)^{2} + (2ad2bc)^{2} = (a^{2}+b^{2}+c^{2}+d^{2})^{2}
We can generalize this to the form,
x_{1}^{2}+x_{2}^{2}+… +x_{m}^{2} = (y_{1}^{2}+y_{2}^{2}+… +y_{n}^{2})^{2}
It can be proven there are polynomial identities with integer coefficients for all m = n (discussed in Sums of Three Squares). But what about when m < n? It turns out we can generalize the one by Lebesgue by looking at its underlying structure. Note that this can be expressed as,
(p_{1}^{2}+p_{2}^{2}+q_{1}^{2}+q_{2}^{2})^{2}  (p_{1}^{2}+p_{2}^{2}q_{1}^{2}q_{2}^{2})^{2} = 4(p_{1}^{2}+p_{2}^{2})(q_{1}^{2}+q_{2}^{2}) (eq.1)
Let {a,b} = {p_{1}^{2}+p_{2}^{2}, q_{1}^{2}+q_{2}^{2}} and this reduces to the basic Difference of Two Squares Identity,
(a+b)^{2}  (ab)^{2} = 4ab
a special case of Boutin's Theorem also given at the end of this section which generalizes it to a sum and difference of kth powers. Since (p_{1}^{2}+p_{2}^{2})(q_{1}^{2}+q_{2}^{2}) = r_{1}^{2}+r_{2}^{2} by the BramaguptaFibonacci TwoSquare Identity, then the RHS of eq.1 can be expressed as the sum of two squares, which explains the Lebesgue Identity. But there are the higher Euler FourSquare Identity and Degen EightSquare Identity. (See also the article, Pfister's 16Square Identity.) Thus, let {a,b} = {p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}, q_{1}^{2}+q_{2}^{2}+q_{3}^{2}+q_{4}^{2}} and we get,
(p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}+q_{1}^{2}+q_{2}^{2}+q_{3}^{2}+q_{4}^{2})^{2}  (p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}q_{1}^{2}q_{2}^{2}q_{3}^{2}q_{4}^{2})^{2} = 4(p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2})(q_{1}^{2}+q_{2}^{2}+q_{3}^{2}+q_{4}^{2})
Since the RHS can be expressed as four squares, this gives an identity for the square of 8 squares as the sum of five squares. Using the Degen EightSquare, this gives one for the square of 16 squares as nine squares. In general, we can have a Theorem 3 to complement the first two in Sums of Three Squares.
Theorem 3: Given x_{1}^{2}+x_{2}^{2}+… +x_{m}^{2} = (y_{1}^{2}+y_{2}^{2}+… +y_{n}^{2})^{2}, where m ≤ n, one can identically express the square of n squares as m squares for:
I. All m = n.
II. For m < n, with n even (v > 1):
a) {m,n} = {2v±1, 2v}
b) {m,n} = {4v3, 4v}
c) {m,n} = {8v7, 8v}
III. For m < n, with n odd (v > 1):
a) For n = 4v1, then m = 4v3.
b) For n = {8v1, 8v3, 8v5}, then m = 8v7.
Thus, for example, (y_{1}^{2}+y_{2}^{2}+… +y_{16}^{2})^{2} can be identically expressed as m squares for m = {1, 9, 13, 15, 16}. (Q. Is this the most number of m when m ≤ n?) However, when m < n, and after eliminating the possibility of (x_{1}^{2}+x_{2}^{2}+x_{3}^{2})^{2} expressed as two nonzero squares, two cases are still unresolved: n = 5 and n = 8v7. Whether or not there are identities for these remains to be seen.
Proof: One simply uses the difference of two squares identity:
(a+b)^{2} + (ab)^{2} = 4ab (eq.1)
Let a = (p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}+ ... +p_{2u}^{2}). Disregarding the square numerical factor, the RHS of eq.1 becomes (by distributing in pairs) as,
b(p_{1}^{2}+p_{2}^{2}) + b(p_{3}^{2}+p_{4}^{2}) + ...
Let b = (q_{1}^{2}+q_{2}^{2}). Since by the BramaguptaFibonacci Identity (x_{1}^{2}+x_{2}^{2})(y_{1}^{2}+y_{2}^{2}) = z_{1}^{2}+z_{2}^{2}, then the RHS can be expressed as 2u squares. Thus, {m,n} = {2u+1, 2u+2} or, equivalently, {2v1, 2v}. However, if the identity is not used on one pair which is instead expressed as four squares, then the RHS has 2u+2 squares. Thus {m,n} = {2u+3, 2u+2} or, {2v+1, 2v}, proving part (Ia) of Theorem 3.
Similarly, let a = (p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}+ ... +p_{4u}^{2}). Disregarding again the square numerical factor, the RHS (by distributing fourwise) is,
b(p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+p_{4}^{2}) + b(p_{5}^{2}+p_{6}^{2}+p_{7}^{2}+p_{8}^{2}) + ...
Let b = (q_{1}^{2}+q_{2}^{2}+q_{3}^{2}+q_{4}^{2}). Since by the Euler FourSquare Identity (x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2})(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+y_{4}^{2}) = z_{1}^{2}+z_{2}^{2}+z_{3}^{2}+z_{4}^{2}, then the RHS can be expressed as 4u squares. Thus, {m,n} = {4u+1, 4u+4}, or {4v3, 4v}, proving part (Ib). The last part is proven using the last such identity (by the Hurwitz Theorem) namely, the Degen EightSquare Identity. Finally, part II of the Theorem can be proven by simply setting the appropriate number of variables y_{i} as equal to zero. For example, we can reduce the {5,8} Identity to a {5,7} by letting one of the eight squares as zero to get,
(a^{2}+b^{2}+c^{2}+d^{2}e^{2}f^{2}g^{2})^{2} + 4(aebfcg)^{2} + 4(af+bedg)^{2} + 4(ag+ce+df)^{2} + 4(bgcf+de)^{2} = (a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2})^{2} and so on. (End update.)
1. EulerAida Ammei Identity
Theorem: “The square of the sum of n squares is itself a sum of n squares.”
(x_{1}^{2}x_{2}^{2}…x_{n})^{2} + å(2x_{1}x_{n})^{2} = (x_{1}^{2}+x_{2}^{2}+…+x_{n}^{2})^{2}
Examples:
(a^{2}b^{2})^{2} + (2ab)^{2} = (a^{2}+b^{2})^{2}
(a^{2}b^{2}c^{2})^{2} + (2ab)^{2} + (2ac)^{2} = (a^{2}+b^{2}+c^{2})^{2}
(a^{2}b^{2}c^{2}d^{2})^{2} + (2ab)^{2} + (2ac)^{2} + (2ad)^{2} = (a^{2}+b^{2}+c^{2}+d^{2})^{2}
and so on. Note that these can be alternatively expressed as the basic identity,
(x_{1}^{2}x_{0})^{2} + (2x_{1})^{2}x_{0}^{ } = (x_{1}^{2}+x_{0})^{2}
for arbitrary x_{1} and x_{0}, where for the theorem it was set x_{0} = x_{2}^{2}+x_{3}^{2}+…+x_{n}^{2}. By letting x_{0} = x_{2}^{2}, one can see this basic identity is essentially the formula for Pythagorean triples. A stronger result by M. Moureaux is that, “The kth power of the sum of n squares, for k a power of 2, is itself a sum of n squares.” After some time with Mathematica, I observed that there seems to be this beautifully consistent pattern in the identities based on a certain algebraic form. Let the summation å be from m = {2 to n}, then,
(a)^{2} + å(2x_{1}x_{n})^{2} = (x_{1}^{2}+x_{2}^{2}+…+x_{n}^{2})^{2}
(b)^{2} + å(4ax_{1}x_{n})^{2} = (x_{1}^{2}+x_{2}^{2}+…x_{n}^{2})^{4},
(c)^{2} + å(8abx_{1}x_{n})^{2} = (x_{1}^{2}+x_{2}^{2}+…x_{n}^{2})^{8},
(d)^{2} + å(16abcx_{1}x_{n})^{2} = (x_{1}^{2}+x_{2}^{2}+…x_{n}^{2})^{16},
(e)^{2} + å(32abcdx_{1}x_{n})^{2} = (x_{1}^{2}+x_{2}^{2}+…x_{n}^{2})^{32},
and so on, where a = (x_{1}^{2}x_{2}^{2}…x_{n}) and b = (4x_{1}^{4}+4ax_{1}^{2}a^{2}), c = (4a^{4}+4a^{2}bb^{2}), d = (4b^{4}+4b^{2}cc^{2}), e = (4c^{4}+4c^{2}dd^{2}), etc. It seems there is this “recurrence relation” involving the algebraic form 4u^{4}+4u^{2}vv^{2}. I have no proof this in fact goes on, though it would be odd if the pattern stops. (And the algebraic form factors over √2, which is only appropriate since we are dealing with powers of two.)
To generalize the EulerAida Ammei identity: For what k is the kth power of n diagonal quadratic forms identically a sum of like form? Or,
c_{1}w_{1}^{2} + c_{2}w_{2}^{2} + …+ c_{n}w_{n}^{2} = (c_{1}x_{1}^{2} + c_{2}x_{2}^{2} + …+ c_{n}x_{n}^{2})^{k}
It turns out for the monic case, or when c_{1} = 1, the answer is for all positive integer k. In a previous section, it was proven that, “the kth power of the sum of n squares (x_{1}^{2}+x_{2}^{2}+… x_{n}^{2})^{k} is itself the sum of n squares”. It takes only a very small modification of the proof to generalize this.
Theorem 1. “The expression (c_{1}x_{1}^{2} + c_{2}x_{2}^{2} + …+ c_{n}x_{n}^{2})^{k}, for c_{1 }= 1, is identically a sum of n squares of like form for all positive integer n and k.”
Proof: (Piezas) To recall, 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. Their product, or norm, is,
U^{2}+V^{2} = (a^{2}+b^{2})^{k}
Since b can be factored out in V, or V = V_{1}b, if we let {a,b} = {p_{1},√p_{0}}, then,
U^{2 }+ p_{0}V_{1}^{2} = (p_{1}^{2}+p_{0})^{k}
All these are familiar. The slightly different step is that since p_{0} is arbitrary, one can choose it to be the sum of squares of form p_{0} = c_{2}p_{2}^{2}+c_{3}p_{3}^{2}+…+c_{n}p_{n}^{2} and distributing terms, we get,
U^{2 }+ c_{2}p_{2}^{2}V_{1}^{2} + … + c_{n}p_{n}^{2}V_{1}^{2} = (p_{1}^{2} + c_{2}p_{2}^{2} +…+ c_{n}p_{n}^{2})^{k}
thus proving the kth power of the right hand side of the eqn is a sum of squares of like form. (End proof.) For k = 2 and all c_{i} = 1, this is of course the EulerAida Ammei identity and the inspiration for the proof. For k = 3, this is,
(p_{1}^{3}3p_{0}p_{1})^{2} + p_{0}(3p_{1}^{2}p_{0})^{2} = (p_{1}^{2}+p_{0})^{3}
for p_{0} = c_{2}p_{2}^{2 }+…+ c_{n}p_{n}^{2} and so on. For the more general nonmonic case (c_{1}y_{1}^{2} + …+ c_{n}y_{n}^{2})^{k}, some particular identities are also known for the case k=3,
J. Neuberg
ax_{1}^{2}+bx_{2}^{2}+cx_{3}^{2} = (ap^{2}+bq^{2}+cr^{2})^{3}
{x_{1},_{ }x_{2}, x_{3}} = {p(y2z), q(y2z), ry}, if y = 4(ap^{2}+bq^{2})z, z = ap^{2}+bq^{2}+cr^{2}
G. de Longchamps
ax_{1}^{2}+bx_{2}^{2}+cx_{3}^{2}+dx_{4}^{2} = (ap^{2}+bq^{2}+cr^{2}+ds^{2})^{3}
{x_{1},_{ }x_{2}, x_{3}, x_{4}} = {p(y2z), q(y2z), ry, sy}, if y = 4(ap^{2}+bq^{2})z, z = ap^{2}+bq^{2}+cr^{2}+ds^{2}
This author observed that this can be generalized as,
ax_{1}^{2}+bx_{2}^{2}+cx_{3}^{2}+dx_{4}^{2}+ex_{5}^{2} = (ap^{2}+bq^{2}+cr^{2}+ds^{2}+et^{2})^{3}
{x_{1},_{ }x_{2}, x_{3}, x_{4}, x_{5}} = {p(y2z), q(y2z), ry, sy, ty}, if y = 4(ap^{2}+bq^{2})z, z = ap^{2}+bq^{2}+cr^{2}+ds^{2}+et^{2}
and so on for n variables by simply modifying z. A more systematic approach is given below.
Theorem 2. “The nonmonic expression (c_{1} x_{1}^{2} + c_{2}x_{2}^{2} + …+ c_{n}x_{n}^{2})^{k} is identically a sum of n squares of like form for all positive integer n and odd k.”
Proof: The proof is a variation of the one above. One simply solves the equation,
U^{2}+c_{1}V^{2} = (x^{2}+c_{1}y^{2})^{k},
by equating its linear factors,
U+V√c_{1} = (x+y√c_{1})^{k}, UV√c_{1} = (xy√c_{1})^{k},
and easily solving for U,V as expressions in x,y. Since, for odd k, x can be factored out in U, or U = U_{1}x, if we let {x,y} = {√p_{0}, p_{1}} then,
p_{0}U_{1}^{2 }+ c_{1}V^{2} = (p_{0}+c_{1}p_{1}^{2})^{k},
where p_{0} can then be set as the sum of squares p_{0} = c_{2}p_{2}^{2 }+…+ c_{n}p_{n}^{2}, giving,
c_{1}V^{2 }+ (c_{2}p_{2}^{2} +… + c_{n}p_{n}^{2})U_{1}^{2} = (c_{1}p_{1}^{2 }+ c_{2}p_{2}^{2 }+…+ c_{n}p_{n}^{2})^{k}
proving that, for odd k, the right hand side is identically the sum of squares of like form. (End of proof.) For k = 3, this is,
c_{1}(c_{1}p_{1}^{3}3p_{0}p_{1})^{2} + p_{0}(3c_{1}p_{1}^{2}p_{0})^{2} = (c_{1}p_{1}^{2}+p_{0})^{3}
for p_{0} = c_{2}p_{2}^{2 }+…+ c_{n}p_{n}^{2}, and so on for all odd k.
2. BrahmaguptaFibonacci TwoSquare Identity
(ac+bd)^{2} + (adbc)^{2} = (a^{2}+b^{2})(c^{2}+d^{2})
This can be generalized as,
(ac+nbd)^{2} + n(adbc)^{2} = (a^{2}+nb^{2})(c^{2}+nd^{2})
From the TwoSquare we can derive the EulerLebesgue ThreeSquare,
(a^{2}+b^{2}c^{2}d^{2})^{2} + (2ac+2bd)^{2} + (2ad2bc)^{2} = (a^{2}+b^{2}+c^{2}+d^{2})^{2}
This can be generalized by the Fauquembergue nSquares Identity. It is a bit difficult to convey with limited notation but in one form can be seen as,
(a^{2}+b^{2}c^{2}d^{2}+x)^{2} + (2ac+2bd)^{2} + (2ad2bc)^{2} + 4x(c^{2}+d^{2}) = (a^{2}+b^{2}+c^{2}+d^{2}+x)^{2}
where x is arbitrary and can be chosen as any sum of n squares. Note that for x = 0 this reduces to the EulerLebesgue. For the case x as a single square this gives, after minor changes in variables,
(a^{2}+b^{2}+c^{2}d^{2}e^{2})^{2} + (2ad+2ce)^{2} + (2ae2cd)^{2} + (2bd)^{2} + (2be)^{2} = (a^{2}+b^{2}+c^{2}+d^{2}+e^{2})^{2}
distinct from the EulerAida Ammei identity for n = 5 which is given by,
(a^{2}b^{2}c^{2}d^{2}e^{2})^{2} + (2ab)^{2} + (2ac)^{2} + (2ad)^{2} + (2ae)^{2} = (a^{2}+b^{2}+c^{2}+d^{2}+e^{2})^{2}
For x = e^{2}+f^{2}, it results in seven squares whose sum is the square of six squares:
(a^{2}+b^{2}+c^{2}+d^{2}e^{2}f^{2})^{2} + (2ae+2df)^{2} + (2af2de)^{2} + (2be)^{2} + (2bf)^{2} + (2ce)^{2} + (2cf)^{2} = (a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})^{2}
and so on for other x.
3. Euler FourSquare Identity
(a^{2}+b^{2}+c^{2}+d^{2}) (e^{2}+f^{2}+g^{2}+h^{2}) = u_{1}^{2} + u_{2}^{2} + u_{3}^{2} + u_{4}^{2}
u_{1} = aebfcgdh u_{2} = af+be+chdg u_{3} = agbh+ce+df u_{4} = ah+bgcf+de
Note that a cubic version, in fact, is possible, (x_{1}^{3}+x_{2}^{3}+x_{3}^{3}+x_{4}^{3}) (y_{1}^{3}+y_{2}^{3}+y_{3}^{3}+y_{4}^{3}) = z_{1}^{3}+ z_{2}^{3}+z_{3}^{3}+ z_{4}^{3}, to be discussed below. Also, by Lagrange's Identity, the product can be expressed as the sum of seven squares,
(a^{2}+b^{2}+c^{2}+d^{2}) (e^{2}+f^{2}+g^{2}+h^{2}) = (ae+bf+cg+dh)^{2} + (afbe)^{2} + (agce)^{2} + (ahde)^{2} + (bgcf)^{2} + (bhdf)^{2} + (chdg)^{2}
A more general version for squares was also given by Lagrange as,
(a^{2}+mb^{2}+nc^{2}+mnd^{2}) (p^{2}+mq^{2}+nr^{2}+mns^{2}) = x_{1}^{2}+mx_{2}^{2}+nx_{3}^{2}+mnx_{4}^{2}
x_{1} = apmbqncr+mnds, x_{2} = aq+bpncsndr, x_{3} = ar+mbs+cp+mdq, x_{4} = asbr+cqdp
In analogy to the ThreeSquare, we can also find a FiveSquare (by yours truly),
(a^{2}+b^{2}+c^{2}+d^{2}e^{2}f^{2}g^{2}h^{2})^{2} + (2u_{1})^{2} + (2u_{2})^{2} + (2u_{3})^{2} + (2u_{4})^{2} = (a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2}+h^{2})^{2}
with the u_{i} as defined above. Hence this is another case of a square of n squares expressed in less than n squares. And, in analogy to Fauquembergue’s n squares, another kind of n squares identity can be derived from the FiveSquare as,
(a^{2}+b^{2}+c^{2}+d^{2}e^{2}f^{2}g^{2}h^{2}+x)^{2} + (2u_{1})^{2} + (2u_{2})^{2} + (2u_{3})^{2} + (2u_{4})^{2} + 4x(e^{2}+f^{2}+g^{2}+h^{2}) = (a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2}+h^{2}+x)^{2}
where x again can be any number of squares. For x a square, this can give a 9square identity.
4. DegenGravesCayley EightSquares Identity (DGC)
(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2}+h^{2}) (m^{2}+n^{2}+o^{2}+p^{2}+q^{2}+r^{2}+s^{2}+t^{2}) = v_{1}^{2}+v_{2}^{2}+v_{3}^{2}+v_{4}^{2}+v_{5}^{2}+v_{6}^{2}+v_{7}^{2}+v_{8}^{2}
v_{1} = ambncodpeqfrgsht v_{2} = bm+an+docp+fqerhs+gt v_{3} = cmdn+ao+bp+gq+hresft v_{4} = dm+cnbo+ap+hqgr+fset v_{5} = emfngohp+aq+br+cs+dt v_{6} = fm+enho+gpbq+ards+ct v_{7} = gm+hn+eofpcq+dr+asbt v_{8} = hmgn+fo+epdqcr+bs+at
For convenience, let {a, b,…t} = {a_{1}, a_{2},…a_{16}}. This can also give a NineSquare Identity (distinct from the one in the previous section) as,
(a_{1}^{2}+…+ a_{8}^{2} a_{9}^{2}… a_{16}^{2})^{2} + (2v_{1})^{2} + …+ (2v_{8})^{2} = (a_{1}^{2}+ a_{2}^{2 }+… + a_{16}^{2})^{2}
and the DGC nSquares identity,
(a_{1}^{2}+…+ a_{8}^{2} a_{9}^{2}… a_{16}^{2 }+ x)^{2} + (2v_{1})^{2} + …+ (2v_{8})^{2} + 4x(a_{9}^{2 }+… + a_{16}^{2}) = (a_{1}^{2}+ a_{2}^{2 }+… + a_{16}^{2 }+ x)^{2}
Since the DGC is the last bilinear nsquares identity, these two should also be the last of their kind.
(Update, 10/26/09): Just like the TwoSquare and FourSquare, the EightSquare Identity can be generalized. For arbitrary {u, v},
(a^{2}+ ub^{2}+c^{2}+ud^{2}+ve^{2}+uvf^{2}+vg^{2}+uvh^{2}) (m^{2}+un^{2}+o^{2}+up^{2}+vq^{2}+uvr^{2}+vs^{2}+uvt^{2}) = x_{1}^{2}+ux_{2}^{2}+x_{3}^{2}+ux_{4}^{2}+vx_{5}^{2}+uvx_{6}^{2}+vx_{7}^{2}+uvx_{8}^{2}
x_{1} = ambnucodpueqvfruvgsvhtuv x_{2} = bm+an+docp+fqvervhsv+gtv x_{3} = cmdnu+ao+bpu+gqv+hruvesvftuv x_{4} = dm+cnbo+ap+hqvgrv+fsvetv x_{5} = emfnugohpu+aq+bru+cs+dtu x_{6} = fm+enho+gpbq+ards+ct x_{7} = gm+hnu+eofpucq+dru+asbtu x_{8} = hmgn+fo+epdqcr+bs+at
(End update)
5. V. Arnold’s Perfect Forms
Let {a,b,c} be in the integers. Given the equation (au^{2}+buv+cv^{2})(ax^{2}+bxy+cy^{2}) = az_{1}^{2}+bz_{1}z_{2}+cz_{2}^{2}. If for any integral integral {u,v,x,y} one can always find integral {z_{1}, z_{2}}, then the binary quadratic form F(a,b,c) is defined as a perfect form.
Theorem 1: “The product of three binary quadratic forms F(a,b,c) is of like form.”
Proof: (an_{1}^{2}+bn_{1}n_{2}+cn_{2}^{2})(au^{2}+buv+cv^{2})(ax^{2}+bxy+cy^{2}) = az_{1}^{2}+bz_{1}z_{2}+cz_{2}^{2}
where,
z_{1} = u(n_{3}x+cn_{2}y)+cv(n_{2}xn_{1}y)
z_{2} = v(an_{1}x+n_{4}y)au(n_{2}xn_{1}y)
and {n_{3}, n_{4 }} = {an_{1}+bn_{2}, bn_{1}+cn_{2}} from which immediately follows,
Corollary:“If there are integers {n_{1}, n_{2}} such that F(a,b,c) = 1, then F(a,b,c) is a perfect form.”
Note that if F(a,b,c) is monic, the soln {x,y} = {1,0} immediately implies this form is perfect. But by dividing z_{1}, z_{2} with c, a, respectively, and modifying the expressions for {n_{3}, n_{4}} will result in a second theorem,
Theorem 2: “If there are integers {n_{1}, n_{2}, n_{3}, n_{4}} such that an_{1}^{2}+bn_{1}n_{2}+cn_{2}^{2} = ac, n_{3} = (an_{1}+bn_{2})/c, n_{4} = (bn_{1}+cn_{2})/a, then F(a,b,c) is a perfect form.”
Proof: (an_{1}^{2}+bn_{1}n_{2}+cn_{2}^{2})(au^{2}+buv+cv^{2})(ax^{2}+bxy+cy^{2}) = (az_{1}^{2}+bz_{1}z_{2}+cz_{2}^{2})(ac)
where,
z_{1} = v(n_{1}x+n_{4}y)u(n_{2}xn_{1}y)
z_{2} = u(n_{3}x+n_{2}y)+v(n_{2}xn_{1}y)
and {n_{3}, n_{4 }} = {(an_{1}+bn_{2})/c, (bn_{1}+cn_{2})/a}.
The expressions are essentially the same as in Theorem 1 but have been divided by c,a. This second class is relevant to quadratic discriminants d with class number h(d) = 3m. For imaginary fields with h(d) = 3, there are sixteen fundamental d, all of which have its associated F(a,b,c) as perfect forms. For brevity, only the first three will be given and in the format {a,b,c}, {n_{1}, n_{2}, n_{3}, n_{4}}:
d = 23; {2,1,3}, {1,1,1,2} d = 31; {2,1,4}, {2,0,1,1} d = 59; {3,1,5}, {2,1,1,1}
For real fields with h(d) = 3, there are fortytwo d in the Online Encyclopedia of Integer Sequences, all of which also have F(a,b,c) as perfect. However, while most of the n_{i} for negative d with h(d) = 3,6 were only single digits, for positive d these can get quite large. For ex,
d = 2857; {2,51,32}, {3326866, 127404, 4879, 86873547}
Q: Any other theorems regarding the product of two or three binary quadratic forms? We can generalize this somewhat and go to diagonal nnary quadratic forms (one without cross terms),
F(a_{1},a_{2},…a_{n}):= a_{1}x_{1}^{2 }+ a_{2}x_{2}^{2 }+…+ a_{n}x_{n}^{2}
If we consider the equation,
(a_{1}x_{1}^{2 }+ a_{2}x_{2}^{2 }+…+ a_{n}x_{n}^{2})(a_{1}y_{1}^{2 }+ a_{2}y_{2}^{2 }+…+ a_{n}y_{n}^{2}) = a_{1}z_{1}^{2 }+ a_{2}z_{2}^{2 }+…+ a_{n}z_{n}^{2}
then for what constants {a_{1}, a_{2},…a_{n}} is there such that the product of two diagonal nnary quadratic forms is of like form? Most of the results have been limited to the special case of all a_{i }= 1 and n = 2,4,8, namely the BrahmaguptaFibonacci, Euler, and DegenGraves identities discussed above. The first can be generalized to the form {1,p}, the second to {1, p, q, pq} by Lagrange, and the third to {1,1, p,p, q,q, pq, pq}. Ramanujan in turn generalized Lagrange’s FourSquare Theorem and found 54 {a,b,c,d} such that ax_{1}^{2}+bx_{2}^{2}+cx_{3}^{2}+dx_{4}^{2} can represent all positive integers, namely,
{1,1,1,v}; v = 17 {1,1,2,v}; v = 214 {1,1,3,v}; v = 36 {1,2,2,v}; v = 27 {1,2,3,v}; v = 310 {1,2,4,v}; v = 414 {1,2,5,v}; v = 610
which is the complete list. (Note: Incidentally, it would have been expected that the last would be for v = 510. What is the smallest positive integer not expressible by the form {1,2,5,5}?) All 54 are then perfect quaternary forms since, needless to say, the product of two positive integers is always a positive integer. For the first case {1,1,1,1} this is just Euler’s foursquare identity and the z_{i} have a bilinear expression in terms of the x_{i} and y_{i}. It might be interesting to know if the other 53 {a,b,c,d} have similar formulas for their z_{i}. In general, for a_{i} not all equal to unity, what other results are there for the product of two nnary quadratic forms, especially for n not a power of two?
Update, 9/21/09: Turns out the form {1,2,5,5} cannot express the number 15. See "The 15 and 290 Theorems" by Conway, Schneeberger, and Bhargava.
6. Lagrange’s Identity
A faintly similar identity to the sumproduct of n squares given previously is,
(x_{1}y_{1 }+ … + x_{n}y_{n })^{2} + S (x_{k}y_{j }– x_{j}y_{k})^{2} = (x_{1}^{2}_{ }+ … + x_{n}^{2}) (y_{1}^{2}_{ }+ … + y_{n}^{2}_{ }) for 1£ k<j£n.
For n=3, this has 4 addends,
(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3 })^{2} + (x_{1}y_{2 }– x_{2}y_{1 })^{2} +(x_{1}y_{3 }– x_{3}y_{1 })^{2} + (x_{2}y_{3 }– x_{3}y_{2 })^{2} = (x_{1}^{2}+x_{2}^{2}+x_{3}^{2}) (y_{1}^{2}+y_{2}^{2}+y_{3}^{2})
while n=4 already involves 7 addends, and is an alternative way to express Euler's FourSquare Identity. A general identity was found by A. Cauchy and J. Young. The case n=3 was also rediscovered by T.Weddle while studying the semiaxes of an ellipsoid.
7. Difference of Two Squares Identity
A differenceproduct identity on the other hand is given by,
(x_{1}^{2}+ … +x_{n}^{2} + (y_{1}^{2}+…+y_{n}))^{2} – (x_{1}^{2}+ … +x_{n}^{2} – (y_{1}^{2}+…+y_{n}))^{2} = 4(x_{1}^{2}+ … +x_{n}^{2})(y_{1}^{2}+ … +y_{n}^{2}_{ })
Proof: Let a = x_{1}^{2}+ … +x_{n}^{2}, b = y_{1}^{2}+ … +y_{n}^{2}, then this is just the basic identity, (a+b)^{2}(ab)^{2} = 4ab. (End proof.) If the right hand side of the identity is nontrivially expressible as the sum of n squares, as is the case for n = 2,4,8, this automatically implies a square of 2n squares expressible as the sum of n+1 squares, thus explaining the Three, Five, NineSquare Identities above. (The case n = 1 just gives the formula for Pythagorean triples.) Let,
(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})^{2}  (a^{2}+b^{2}+c^{2}d^{2}e^{2}f^{2})^{2 } = 4(a^{2}+b^{2}+c^{2})(d^{2}+e^{2}+f^{2})
(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2}+h^{2})^{2}  (a^{2}+b^{2}+c^{2}+d^{2}e^{2}f^{2}g^{2}h^{2})^{2 } = 4(a^{2}+b^{2}+c^{2}+d^{2})(e^{2}+f^{2}+g^{2}+h^{2}) and together with Lagrange’s Identity for n = 3,4 applied on the RHS respectively, this can be used to prove that the square of 6 squares can be expressed as the sum of five squares,
(a^{2}+b^{2}+c^{2}d^{2}e^{2}f^{2})^{2 }+ 4(ad+be+cf)^{2} + 4(aebd)^{2} + 4(afcd)^{2} + 4(bfce)^{2} = (a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})^{2} and provide an alternative soln to expressing the square of 8 squares as a sum of 8 squares,
(a^{2}+b^{2}+c^{2}+d^{2}e^{2}f^{2}g^{2}h^{2})^{2 }+ 4((ae+bf+cg+dh)^{2} + (afbe)^{2} + (agce)^{2} + (ahde)^{2} + (bgcf)^{2} + (bhdf)^{2} + (chdg)^{2}) = (a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2}+h^{2})^{2} since the EulerAida Ammei gives it as,
(a^{2}b^{2}c^{2}d^{2}e^{2}f^{2}g^{2}h^{2})^{2 }+ 4a^{2}(b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2}+h^{2}) = (a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2}+g^{2}+h^{2})^{2}
Note 1: By not using the EulerAida Ammei identity, are there always alternative solns to the square of n squares as the sum of n squares? Note 2: For what n is the square of n squares identically the sum of less than n squares? (It can be for n = 4, 8, 16 (as 3, 5, 9 squares, respectively). In fact, by setting appropriate variables equal to zero, it is the case for all n ≤ 16 other than n = 3, 5, 9.) (See update at start of this section.)
A. Boutin
Boutin’s Identity: S ± (x_{1 }± x_{2 }±…± x_{k})^{k} = k! 2^{k1}x_{1}x_{2}…x_{k}
where the exterior sign is the product of the interior signs. (Or, the term is negative if there is an odd number of negative interior signs.) The case k=2 gives the wellknown,
(a+b)^{2}  (ab)^{2} = 4ab
while for k = 3,4,
(a+b+c)^{3}  (ab+c)^{3}  (a+bc)^{3} + (abc)^{3} = 24abc
(a+b+c+d)^{4}  (ab+c+d)^{4}  (a+bc+d)^{4}  (a+b+cd)^{4} + (abc+d)^{4} + (ab+cd)^{4} + (a+bcd)^{4}  (abcd)^{4} = 192abcd
and so on for other kth powers. The case k=3 then implies that,
Piezas
(x_{1}^{3}+x_{2}^{3}+…+x_{n}^{3}) (y_{1}^{3}+y_{2}^{3}+…+y_{n}^{3}) = z_{1}^{3}+ z_{2}^{3}+z_{3}^{3}+ z_{4}^{3}
or, “The product of two sums of n cubes is the sum of four cubes.”
Proof: Simply let a = x_{1}^{3}+x_{2}^{3}+…+x_{n}^{3}, b = y_{1}^{3}+y_{2}^{3}+…+y_{n}^{3}, and c = 9 in Boutin's Identity for k = 3.
