Part 2. Sums of SquaresI. Sums of two squares
- x
^{2}+y^{2}= z^{k}a. (a+1/a) ^{2}+ (b+1/b)^{2}= c^{2}b. a ^{2}(b^{2}-1)^{2 }+ b^{2}(a^{2}-1)^{2}= c^{2}c. (a ^{2}+b^{2})^{2}+ (c^{2}+d^{2})^{2}= e^{2}d. a ^{2}+ (a+n)^{2}= b^{2}e. a ^{2}+ b^{2}= (b+n)^{2}f. (2ab) ^{2}+ (2cd)^{2}= (a^{2}+b^{2}-c^{2}-d^{2})^{2}g. (ac+bd) ^{2}+ (ad-bc)^{2}= (a^{2}+b^{2})^{2} - x
^{2}+ny^{2}= z^{k}; 2b. ax^{2}+by^{2}= cz^{2} - ad-bc = ±1
- x
^{2}+y^{2}= z^{2}+1 - x
^{2}+y^{2}= z^{2}-1 - x
^{2}+y^{2}= z^{2}+nt^{2} - x
^{2}+y^{2}= z^{2}+nt^{k} - x
^{2}+y^{2}= mz^{2}+nt^{2} - c
_{1}(x^{2}+ny^{2}) = c_{2}(z^{2}+nt^{2}) - mx
^{2}+ny^{2}= mz^{2}+nt^{2} - x
^{2}+y^{2}= z^{k}+t^{k}
A. IntroductionWhile integers
a,b,c that satisfy a are called ^{2}+b^{2} = c^{2}Pythagorean triples, the ancient Babylonians already knew there were triangles whose sides satisfy that relationship more than a thousand years earlier. The famous tablet Plimpton 322 (pre-1500 BC, now kept in Columbia University) contains pairs of numbers in sexigesimal which can be seen as part of a Pythagorean triple. The largest pair is (18541, 12709) and a quick calculation shows that the difference of their squares is also a square, 18541^{2}-12709^{2} = 13500^{2}. Quick for us using a calculator but the size of this example shows the Babylonians must have known of a method to generate solutions other than randomly scribbling figures in the sand. The smallest primitive solns (where a,b,c have no common factor) are: {3, 4, 5}, {5, 12, 13}, {7, 24, 25}, {8, 15, 17}, etc.
1) one of 2) one of 3) one of
which shows the importance of the triple {3,4,5}.
Note: It is possible the divisibility by 3,4,5 is contained in just one term, such as in the first term of,(60v)
^{2} + (900v^{2}-1)^{2} = (900v^{2}+1)^{2}Theorem: “One soln can lead to another, or if a^{2}+b^{2} = c^{2}, then (a+2b+2c)^{2} + (2a+b+2c)^{2} = (2a+2b+3c)^{2} is another triple.”
Starting with {a, b, c} = {±3, ±4, 5}, one can iteratively generate
(2m+1) (4m)
((a
Proof: For any where
x_{1}x_{2 }y_{1}_{ }≠ 0, one can always find rational {a,b,t} using the formulas: {a,b,t} = {x_{1}+y_{1}, x_{2}, 1/(2(x_{1}+y_{1}))}. In general, one can completely solve x_{1}^{2}+x_{2}^{2}+...+ x_{n}^{2} = y_{1}^{2}. See Sums of Three Squares. Update (7/2/09): Adam Bailey gave a variant of the complete soln to a
^{2}+b^{2} = c^{2} as, (p+2) ^{2}r^{2} + (2/p+2)^{2}r^{2} = (p+2/p+2)^{2}r^{2}where r is a scaling factor, since for any {a,b,c} > 0, one can always find rational {p,r} using the formulas {p,r} = {-(a+b-c)/(a-c), 2/(a+b-c). Thus, this shows that the complete soln to a^{2}+b^{2} = c^{2}, as well as Binet's complete one for a^{3}+b^{3}+c^{3 }= d^{3}, is expressible (with a scaling factor) in just n-1 parameters, where n is the degree of the equation. (End note).Pythagorean triples were already known by the classical Greek mathematicians. But the Babylonians and Greeks didn’t know about
complex numbers which makes it easy for us to identically solve the more general equation x^{2}+y^{2} = z^{k} for any positive integer k, though for k > 2, it is not necessarily the complete soln. For example, when k = 3, the method yields, with a rational scaling factor u,
(a
^{3}-3ab^{2})^{2}u^{6} + (3a^{2}b-b^{3})^{2}u^{6} = (a^{2}+b^{2})^{3}u^{6}but does
not always cover the alternative soln with scaling factor v,(a ^{3}+ab^{2})^{2}v^{6} + (a^{2}b+b^{3})^{2}v^{6} = (a^{2}+b^{2})^{3}v^{6}This will be discussed more in k, given the expansion of the complex number (a+bi)^{k} = A+Bi where {A,B} are polynomials in terms of {a,b}, and i = Ö-1, then,
A
Similarly, for the case of
A
though there is an easier method to be discussed later. One can also solve the equation A
(A+Bi)(A-Bi) = (a+bi)
Equating factors yields a system of two equations in two unknowns A,B which, being only linear, can then easily be solved for. This yields,
A = (p
where {p,q} = {a+bi, a-bi} and B, after simplification, is a real value. This technique of equating factors will prove useful for similar equations.
Update (6/17/09): Pythagorean triples {a,b,c} = {x
^{2}-1, 2x, x^{2}+1} can be used in the form,a
^{k}+b^{k}+c^{k} = 2(x)^{8}+14x^{4}+1^{k/4}, for k = 4,8The expression x
^{8}+14x^{4}+1^{ }is no ordinary polynomial. Equated to zero, it is one of the octahedral equations, involved in the projective geometry of the octahedron. It also appears in a formula for the beautiful j-function j(τ) in terms of the Dedekind eta function η(τ) as, j(τ) = 16(x)^{8}+14x^{4}+1^{3}/(x^{5}-x)^{4}where x = 2
η^{2}(τ) η^{4}(4τ) / η^{6}(2τ)If you have
Mathematica, you can easily verify this. For ex, choosing the particular value d = (1+Sqrt[-163])/2 which conveniently yields an integer, j(d) = N[16(x^{8}+14x^{4}+1)^{3}/(x^{5}-x)^{4 }/. x → 2*DedekindEta[d]^{2} DedekindEta[4d]^{4} / DedekindEta[2d]^{6} /. d → (1+Sqrt[-163])/2, 100] = -640320^{3 }Furthermore, using
-Numerator+1728Denominator of the j-function formula above (note that 1728 = 12^{3}), one gets another octahedral equation. Expressed in terms of the same Pythagorean triples {a,b,c}, this is,(a ^{4}+b^{4}+c^{4})^{3} - 54(abc)^{4} = 8(x)^{12}-33x^{8}-33x^{4}+1^{2}Why Pythagorean triples, when expressed in a constrained manner, spell out the two octahedral equations, I do not know. (
End note.) Update (10/11/09): Pythagorean triples also appear in the equation 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
In a
Quaternion[a,b,c,d] ** Quaternion[a,b,c,d] = Quaternion[a
One can see two immediate consequences. First, as expected, (a
Quaternion[a
Since {a,b,c,d} are to be real, then
b
which have an infinite number of rational solns, a subset of which is d = 0 and {b,c} as rationalized Pythagorean triples. (If c = d = 0, then the only soln is b = ±1 in which case the quaternion reduces to the imaginary unit.) This will also be discussed in the section on
y
( B. Special Forms1. (a+1/a)
^{2} + (b+1/b)^{2} = c^{2}2. a
^{2}(b^{2}-1)^{2 }+ b^{2}(a^{2}-1)^{2} = c^{2}3. (a
^{2}+b^{2})^{2} + (c^{2}+d^{2})^{2} = e^{2}4. a
^{2} + (a+n)^{2} = b^{2}5. a
^{2} + b^{2} = (b+n)^{2}6. (2ab)
^{2} + (2cd)^{2} = (a^{2}+b^{2}-c^{2}-d^{2})^{2}7. (ac+bd)
^{2} + (ad-bc)^{2} = (a^{2}+b^{2})^{2}If you know of any other form and its soln, pls submit them. Form 1: (a+1/a)^{2} + (b+1/b)^{2} = c^{2}Euler
(x+1/x)
{x,y} = {4p/(p Form 2: a^{2}(b^{2}-1)^{2 }+ b^{2}(a^{2}-1)^{2} = c^{2}Euler
x
{x,y} = {4p/(p
with a very similar soln to the previous and which is useful in making {x
^{2}+y^{2}, x^{2}+z^{2}, y^{2}+z^{2}} all squares, or the Euler Brick Problem, to be discussed more in the section on Simultaneous Polynomials. Form 3: (a ^{2}+b^{2})^{2} + (c^{2}+d^{2})^{2} = e^{2}Euler
(a
This can be generalized as, let {p,q} = {2n
(a
for any
(a
where {a,b,p,q}= {u
One soln to the conditional eqn is to let {v,w} = {e/(2u), u}, and it becomes c
Update (6/20/09): After this author posted the question in
sci.math, Dave Rusin gave a very elegant five-parameter soln. He notes that one way to approach the problem is to treat the legs (each of which are sums of two squares) as the product of two sums of two squares since we know that,(x
_{1}x_{3}+x_{2}x_{4})^{2} + (x_{1}x_{4}-x_{2}x_{3})^{2} = (x_{1}^{2}+x_{2}^{2})(x_{3}^{2}+x_{4}^{2})Thus we get, ((x and, after some clever manipulation of some eqns describing a 5-dimensional projective variety over {x {y where z = (a ((u has, {u {v where z
_{2} = (a^{2}-b^{2})^{2}+6(a^{2}-b^{2})c^{2}+c^{4}-d^{2} with parameters {a,b,c,d}. (End update.) Form 4: a^{2} + (a+n)^{2} = b^{2}Given a triangle with legs as consecutive integers {p, p+1}, then a second can be found,
Theorem: “If p^{2} + (p+1)^{2} = r^{2}, then q^{2} + (q+1)^{2} = (p+q+r+1)^{2}, where q = 3p+2r+1.” (Fermat)(Update, 6/24/09): Maurice Mischler pointed out that all integer solns to the triple (p, p+1, r) is given by,
((x-1)/2)
^{2} + ((x+1)/2)^{2} = y^{2}where {x,y} satisfy the
Pell equation x^{2}-2y^{2} = -1. Since x is always odd, then terms are integers. (End note).(Update, 12/25/09): Given the
co-prime triple {a,b,c}, the difference between the hypotenuse and a leg, or c-a = n, can assume any non-zero value n. In contrast, the difference between the two legs, or b-a = n, is only IF the prime factors of n is of form 8m ±1 which is sequence n = {1, 7, 17, 23, 31,...} in the OEIS. These triples can be completely parameterized by the form,((x-n)/2)
^{2} + ((x+n)/2)^{2} = y^{2}where {x,y} are
co-prime solns of the Pell-like equation x^{2}-2y^{2} = -n^{2}. Thus, x^{2}-2y^{2} = -7^{2} has {x,y} = {1,5}, etc, and it is easily shown that if there is one co-prime soln, then there is an infinite number of them by generalizing Fermat's result as,Theorem: “If p^{2} + (p+n)^{2} = r^{2}, then q^{2} + (q+n)^{2} = (p+q+r+n)^{2}, where q = 3p+2r+n.” (End update.)(Update, 12/30/09): Given the first four primitive Pythagorean triples, {3,4,5}, {5,12,13}, {7,24,25}, {8,15,17}, it can be noticed that the sum of their legs, {3+4, 5+12, 7+24, 8+15} = {7, 17, 31, 23} are primes of the form 8n±1. The relevant theorem is this:
Thus, no matter how many two-part positive partitions {a,b} of p there may be, there is one and only one such that a
(m
If solvable, eq.1 has an infinite number of integer solns, but it can be shown that,
a) For p a prime, eq.1 has only one fundamental soln {x b) The recursion can show that {x
If p is not a prime, then (a) will not be true. For example p = 7(23) = 161 has two, leading to two primitive triples {17, 144, 145} and {44, 117, 125} with the sum of their legs as 17+144 = 44+117 = 7(23) = 161. ( Form 5: a^{2} + b^{2} = (b+n)^{2}When a leg and the hypotenuse differ by a constant
n, a general formula is,(nx)
^{2} + y^{2} = (y+n)^{2} where 2y = n(x
^{2}-1), for arbitrary {n,x}. (Of course, for y to be integral, then either n should be even, or x is odd.)(Update, 12/24/09)
Form 6: (2ab)^{2} + (2cd)^{2} = (a^{2}+b^{2}-c^{2}-d^{2})^{2}Euler observed that the form,
2(p
^{2}v^{4}+x^{4}+y^{4}+z^{4}) = (pv^{2}+x^{2}+y^{2}+z^{2})^{2} (eq.1)
(with
p(2vx) p(2vy) p(2vz)
When
Piezas
Let
{v,x,y,z} = {2(ab-cd)(ab+cd), (a
This is discussed more in Form 18 here. ( Form 7: (ac+bd)^{2} + (ad-bc)^{2} = (a^{2}+b^{2})^{2}One can also find solutions to x
^{2}+y^{2} = z^{2} using the equation p^{2}+q^{2} = r^{2}+s^{2} and vice-versa,
Fibonacci
(ad+be)
^{2} + (ae-bd)^{2} = (dc)^{2} + (ec)^{2}, if a^{2}+b^{2} = c^{2}Conversely, if a
^{2}+b^{2} = c^{2}+d^{2}, then,P. Volpicelli
(ac+bd)
^{2} + (ad-bc)^{2} = (a^{2}+b^{2})^{2} A. Fleck
(a
^{2}c-b^{2}c+2abd)^{2} + (a^{2}d-b^{2}d-2abc)^{2} = (a^{2}+b^{2})^{3} The solns by Volpicelli and Fleck suggest a generalization can be found for all positive integer
k. Using a variation of the technique introduced earlier, assume,
A
and equating factors,
A+Bi = (a+bi)
call this as
A
where A,B are polynomials in {a,b,c,d}
(ac
which continues Volpicelli's and Fleck's identities, and so on for all
(pr+qs)
plus the fact that for non-negative integers {g,h} there are k+1 ways to get the sum g+h = k, this gives the more general result:
Proof: It is well known
(pr+qs)
thus easily proving that for k = 1, there are indeed two ways to express
A
where A,B are polynomials in {a,b,c,d}, the solns are dependent on
A
By Euler’s soln {a,b} = {pr+qs, ps-qr}, so
A
in k+1 ways. (
End proof.) For example, for F^{3}, let exponents {g,h} of the factors be in turn {0,3}, {1,2}, {2,1}, {3,0}. Solving for A,B in System 1, these yield the four identities,
{A {A {A {A
where {a,b,c,d} = {pr+qs, ps-qr, pr-qs, ps+qr}. To give a random numerical example, let {p,q,r,s} = {1, 2, -2, 3} so (p
^{2}+q^{2})(r^{2}+s^{2}) = 65. These then yield {A_{i}, B_{i}} as {488, 191}, {140, 505}, {320, 415}, {524, 7} such that A^{2}+B^{2} = 65^{3}. And so on for other k, for arbitrary {p,q,r,s}. An interesting variant to Pythagorean triples is x
^{2}+y^{3} = z^{4}:
H. Mathieu
(q
((p
Solns can be given by the nth
Piezas
(4q
(4p
thus it is
K. Brown
p
(See Brown’s
More generally,
p
^{4} + (dq^{2}-1)^{3} = d(dq^{3}+3q)^{2}, if p^{2}-3dq^{2} = 1Note: As pointed out by Michael Somos, the above equation is also true if p^{2}+3dq^{2} = -1. (Even for negative integer d, this is not solvable for integer {p,q}, but is easily solvable in the rationals for any d.) |