Updates Page 07: July

 
 

 

(Update, 7/24/10):  Updates on Mengoli’s Six-Square Problem is condensed in an article here

 

 

(Update, 7/23/10):  Given the Fibonacci numbers Fn = {1, 1, 2, 3, 5, 8, 13, 21, 34,…}.

 

Theorem 1:  Let {a,b,c,d,e} be 5 consecutive Fn  with the first having even index n.  Define f = 4(c3+c).  Then,

 

ac + 1 = b2

ae + 1 = c2

ce + 1 = d2

af + 1 = (ab+c2)2

cf + 1 = (bc-d2)2

ef + 1 = (bc+d2)2

 

For example, starting with F4 = 3, let {a,b,c,d,e} = {3, 5, 8, 13, 21}, and f = 4(83+8) = 2080.  Then,

 

3*8 + 1 = 52

3*21 + 1 = 82

8*21 + 1 = 132

3*2080 + 1 = (3*5+82)2 = 792

8*2080 + 1 = (5*8-132)2 = 1292

21*2080 + 1 = (5*8+132)2 = 2092

 

Theorem 2:  Let {a,b,c,d,e} be 5 consecutive Fn  with the first having odd index n.  Define f differently as f = 4(c3-c).  Then,

 

ac - 1 = b2

ae - 1 = c2

ce - 1 = d2

af + 1 = (ab+c2)2

cf + 1 = (bc-d2)2

ef + 1 = (bc+d2)2

 

For example, starting with F5 = 5, let {a,b,c,d,e} = {5, 8, 13, 21, 34}, and f = 4(133-13) = 8736.  Then,

 

5*13 - 1 = 82

5*34 - 1 = 132

13*34 - 1 = 212

5*8736 + 1 = (5*8+132)2 = 2092

13*8736 + 1 = (8*13-212)2 = 3372

34*8736 + 1 = (8*13+212)2 = 5452

 

Starting with F(n) having even n, the sequences generated by,

 

ab+c2 = {1, 11, 79, 545, …}

bc-d2 = {3, 19, 129, 883,…}

bc+d2 = {5, 31, 209, 1429,…}

 

and starting with F(n) with odd n,

 

bc+d2 = {11, 79, 545, 3739,…}

bc-d2 = {7, 49, 337, 2311,…}

ab+c2 = {5, 31, 209, 1429,…}

 

are bisections (half) of OEIS A110034, A061646, A110035, respectively.  See update below for more details.

 

  

(Update, 7/20/10):  The problem of finding a triple {a,b,c} that solves the system M,

 

ab+1 = x2

ac+1 = y2

bc+1 = z2

 

is an ancient problem, going back to the Greek mathematician Diophantus. 

 

1.  The smallest positive integer soln is {a,b,c} = {1,3,8}, yielding {x,y,z} = {2,3,5}.  The numbers are familiar, and it turns out this has connections to Fibonacci numbers Fn,

 

Fn = 1, 1, 2, 3, 5, 8, 13, 21, 34,…

 

Theorem: A solution to system M can be given in terms of Fibonacci numbers as {a,b,c} = {F2m, F2m+2, F2m+4} where {x,y,z} = {F2m+1, F2m+2, F2m+3}.

 

Proof:  By Catalan’s identity,

 

Fn-rFn+r + (-1)n-r Fr2 = Fn2

 

Let r = 1, n = 2m+1, and this becomes,

 

F2mF2m+2 +1 = F2m+12

 

which is a special case of Cassini’s identity and proves {ab+1, bc+1} are squares. But let r = 2, and n = 2m+2, then,

 

F2mF2m+4 +1 = F2m+22

 

which proves ac+1 is also a square. (End proof.)

 

Question: Can anyone prove this particular soln also obeys (a+b+c)2 - 4(ab+ac+bc) = 4?  This implies the nice identity involving 5 consecutive Fibonacci numbers staring with even index n = 2m,

 

(F2m)2 - 2(F2m+1)2 - (F2m+2)2 - 2(F2m+3)2 + (F2m+4)2  = -2 

 

and its counterpart odd index n = 2m+1,

 

(F2m+1)2 - 2(F2m+2)2 - (F2m+3)2 - 2(F2m+4)2 + (F2m+5)2  = 2 

 

but I don't know how to prove them.

 

Futhermore, following Peter Montgomery (and based on earlier work by Arkin, Hoggatt, and Strauss), if a 4th number is defined as d = a+b+c+2abc+2xyz, then we get further relations,

 

ad+1 = u2

bd+1 = v2

cd+1 = w2

 

Using the values in the theorem, one gets d = 4(F2m+23 + F2m+2).  For m = 1, this yields d = 4(33+3) = 120, hence {a,b,c,d} = {1, 3, 8, 120}, and so on.  All such derived quadruples have the relation,

 

(a+b+c+d)2 - 4(ab+ac+ad+bc+bd+cd) = 4

 

or, equivalently,

 

(a+b-c-d)2 = 4(ab+1)(cd+1)

 

2.  Other parametrizations to system M are,

 

{a,b,c} = {n-1, n+1, 4n}

{a,b,c} = {n, m(p+1), (m+1)(p+n+1)},  where p = mn+1  

{a,b,c} = {n, m(p+1),  4p(p+n)(mp+m+p)},  where p = mn+1

 

by Kevin Brown, Richard Saunderson, and Jim Buddenhagen, respectively.  The first two obey,

 

(a+b+c)2 - 4(ab+ac+bc) = 4

 

but the third does not.  For a further discussion (but without a mention of Fibonacci numbers), see Brown's article.

 

 

(Update, 7/16/10):  In 2000, Shuwen found the first soln to the multigrade (k.6.6),

 

x1k+x2k+x3k+x4k+x5k+x6k = y1k+y2k+y3k+y4k+y5k+y6k        (eq.1)

 

for k = 1,3,5,7,9, as,

 

[313, 311, 193, 29, 59, 247]k = [323, 289, 269, 173, 7, 91]k

 

after two months(!) of computer time.  Nowadays, it can be done faster but still only six solns are known.  (See below.)  The thing about odd powers is that you can transpose or move terms around, and in fact there is a transposition of the above with six terms to a side, not necessarily positive, such that it will be good for k = 2 also.  However, Wroblewski found that by exchanging the pairs in blue, you can get a three-way (k.6.6.8) chain,

 

 [313, 311, 193, 29, -7, -91]k = [323, 289, 269, 173, -59, -247]k = [373, 341, 253, 251, 89, -31, -151, -377]k

 

which is the only known known so far.  Wroblewski also found the 6th soln to eq.1 as,

 

[899, 821, 713, 331, 319]k =  [67, 73, 83, 449, 671, 863, 877]k

 

after transposing one term to the LHS, though this one has no (k.6.6) arrangement such that it is good for k = 2 as well.

 

 

(Update, 7/16/10):  Paul Cheffers gave a simple soln to,

 

x2+y2 = 2z2+2t2     (eq.1)

 

as {x,y} = {z+t, z-t}.  A four-variable soln by Euler is {x,y,z,t} = {2(pr-qs),  2(ps+qr),  (p+q)r-(p-q)s,  (p-q)r+(p+q)s}.  Note that eq.1, also known as the parallelogram law, can be transformed to the form p2+q2 = x2+y2 which has a complete soln in four variables.  See also Sums of Two Squares, form 8.

 

 

(Update, 7/15/10):  Alain Verghote gave,

 

(3n+1)2 + (4n+3)2 = (5n+3)2 + 1 

 

It can be showed that some Pythagorean triples {a,b,c} can be used to solve the Diophantine eqn,

 

x2+y2 = z2+1      (eq.1)

 

using the form,

 

(an+1)2 + (bn+a)2 = (cn+a)2 + 1

 

for free variable n, and where {b,c} are consecutive integers, or a leg-hypotenuse Twin Pythagorean triple.  These can be completely parameterized by,

 

{a,b,c} = {2m+1, 2m2+2m, 2m2+2m+1}

 

thus the first leg is any odd integer a > 1 with the first few as {3,4,5}, {5,12,13}, {7,24,25} and so on.  For other ways to solve eq.1, such as using Pell equations, see Sums of Two Squares, form 4.

 

 

(Update, 7/13/10):  Solns to the special Pell equation x2-2y2 = -1,

 

{x,y} = {1, 1}, {7, 5}, {41, 29}, {239, 169}, {1393, 985},…

 

appear in a variety of Diophantine eqns.  In the examples below, [1] and [2] will involve second and fourth powers of the same sums; likewise for [4] and [3].

 

0. In Pythorean-like triples a2 + b2 = c2+1 (sufficient condition):

 

74 + (2*5)2 = (2*52)2 + 1

414 + (2*29)2 = (2*292)2 + 1

 

General rule:  x4 + (2y)2 = (2y2)2 + 1

 

1. In Pythagorean triples where the legs differ by 1 (necessary condition):

 

32 + 42 = 52

202 + 212 = 292

1192 + 1202 = 1692

 

General rule: ((x-1)/2)2 + ((x+1)/2)2 = y2

 

2. In Pythagorean triples where the sum is a fourth power (sufficient condition):

 

72 + (52-1)2 = 54

412 + (292-1)2 = 294

2392 + (1692-1)2 = 1694

 

General rule: x2 + (y2-1)2 = y4

 

3. The sum of the first x2 cubes equal to a 4th power (necessary condition):

 

13 + 23 + 33 + … + (72)3 = (7*5)4

13 + 23 + 33 + … + (412)3 = (41*29)4

13 + 23 + 33 + … + (2392)3 = (239*169)4

 

General rule: 13 + 23 + 33 + … + (x2)3 = (xy)4

 

Proof:  The sum of the first consecutive n cubes is the square of the nth triangular number T(n) = n(n+1)/2,

 

13 + 23 + 33 + … + n3 = (1 + 2+ 3+ ... + n)2 = T2(n) = (n(n+1)/2)2

 

If we require as well that,

 

T(n) = n(n+1)/2 = m2

 

or a square triangular number, this can be partially solved by letting {n,m} = {x2, xy}, where x2-2y = -1.  (Note:  The complete soln is {n,m} = {(u-1)/2, v/2}, where u2-2v = 1, since this will include the ones above.)

 

4. The sum of the first y odd cubes equal to a square (necessary condition):

 

13 + 33 + 53 + … + (2*5-1)3 = (7*5)2

13 + 33 + 53 + … + (2*29-1)3 = (41*29)2

13 + 33 + 53 + … + (2*169-1)3 = (239*169)2

 

General rule:  13 + 33 + 53 + … + (2y-1)3 =  (xy)2

 

Proof:  Since the formula F for sums of cubes in arithmetic progression d, or F(d,a,n) := a3 + (a+d)3 + (a+2d)3 + ….  + (a+dn-d)3 is,

 

F(d,a,n) := (n/4)(2a-d+dn)(2a2-2ad+2adn-d2n+d2n2

 

then for {d,a,n} = {2,1,y}, this becomes,

 

F(2,1,y) := 13 + 33 + 53 + … + (2y-1)3 =  y2(-1+2y2)

 

and making it a square reduces to the Pell eqn -1+2y2 = x2.

 

P.S.  Incidentally, the sums of the first odd cubes, stopping at diferent points, can also express all perfect numbers other than the first one,

 

28   = 13 + 33

496  = 13 + 33 + 53 + 73

8128 = 13 + 33 + 53 + 73 + … + 153

 

and so on.  Q: Any other special Diophantine eqn where this Pell eqn is a necessary condition?

 

 

(Update, 7/12/10): Alain Verghote gave a variety of identities for squares as n squares with some terms consecutive,

 

n = 3:  a2 + (a+1)2 + (ab)2 = (ab+1)2,   where b = a+1

 

n = 4:  (2(a+b))2 + (2(a-b))2 + (2c-1)2 + (2cd)2 = (2cd+1)2,   where {c,d} = {a2+b2, a2+b2+1}

 

n = 4:  (a+b)2 + (a-b)2 + (c+d)2 + (a2+b2+c2+d2-1)2 = (a2+b2+c2+d2)2,   where d = c+1

 

n = 4:  (ab-c2)2 + (ac-b2)2 + (bc-a2)2 + (ab+ac+bc)2  = (a2+b2+c2)2

 

Note that the first three have one addend and sum as consecutive {n, n+1}, while the last has a nice form with terms that cycles its variable {a,b,c}.  

 

 

(Update, 7/12/10):  Using the (k.5.5) Letac-Sinha identity and Theorem 5, Seiji Tomita found a (k.8.8) identity for k = 1,3,5,7,9,

 

[-2c+a, -c+3b+d, -c+3b-d, -c+4a, -2c-a, -c-3b-d, -c-3b+d, -c-4a]k =

[-2c+3a, -c+b+d, -c+b-d, -c+4b, -2c-3a, -c-b-d, -c-b+d, -c-4b]k,

 

where {a,b,c,d} must satisfy a2+12b2 = c2, and 12a2+b2 = d2.

 

Likewise, using the (k.7.7) identity by Wroblewski and Piezas, Tomita found another (k.12.12) for k = 1,3,5,7,9,11,

 

[a+8b, -4a-4b, 6a-4b, 2a-2d, a-12b+4c, a+12b+4c, a-8b, 6a+4b, -4a+4b, 2a+2d, a+12b-4c, a-12b-4c]k

[a-16b, -3a+8b, 5a+8b, 2a+4c, -2a-2d,  4a-2d,  a+16b,  5a-8b,  -3a-8b,  2a-4c,  4a+2d,  -2a+2d]k

 

where a2+b2 = c2, and a2+52b2 = d2.

 

 

(Update, 7/3/10):  There is this unusual eqn,

 

 7(214+94+134+174-34-214-234-14)(219+99-139-179+39+219-239-19) =

12(216+96+136+176-36-216-236-16)(217+97-137-177+37+217-237-17)

 

and an infinite more like it.  Note that 21k+9k-13k-17k = -3k-21k+23k+1k,  for k = 1,3,5, and 21+9-13-17 = 3+21-23-1 = 0.  This is just one instance of a more general theorem found by this author:

 

If ak+bk+ck+dk = ek+fk+gk+hk, for k = 1,3,5 where a+b+c+d = e+f+g+h = 0, call this system M, then,

 

 7(a4+b4+c4+d4-e4-f4-g4-h4)(a9+b9+c9+d9-e9-f9-g9-h9) =

12(a6+b6+c6+d6-e6-f6-g6-h6)(a7+b7+c7+d7-e7-f7-g7-h7)

 

There are many known parametrizations to M.

 

 

(Update, 7/1/10):  Given the triple of integers {44, 117, 240}, then,

 

442 + 1172 = 1252

442 + 2402 = 2442

1172 + 2402 = 2672

 

Such a triple can be considered as the length, width, height {a,b,c}of a rational cuboid, also known as a Euler brick,

 

a2 + b2 = x2

a2 + c2 = y2

b2 + c2 = z2

 

where the face diagonals {x,y,z} are rationals.  If the space diagonal d,

 

a2 + b2 + c2 = d2

 

is rational as well, then this is a perfect cuboid, or a perfect Euler brick, though none have yet been found.  The first polynomial soln was given by Nicholas Saunderson in 1740.  (Saunderson, who was Lucasian professor of mathematics, was blind from infancy1.)  His soln was later re-discovered by Euler.  This is,

 

{a,b,c} = {(4p2-r2)q,  4pqr,  (4q2-r2)p}

 

where {p,q,r} is the Pythagorean triple, {m2-n2, 2mn, m2+n2}.  Explicitly,

 

{a,b,c} = {2mn(3m4-10m2n2+3n4),  8mn(m4-n4),  m6-15m4n2+15m2n4-n6}

 

though these are factorable polynomials.  This soln has the smallest degree known2, and it is unknown if there others with deg ≤ 6.  (Note that the middle diagonal a2+c2 = y2 is a perfect 6th power.)  More details at Form 8, Simultaneous Polynomials to be made Squares.

 

Note 1:  As Kevin Brown points out, "...in nearly the only personal comment of the entire 3-volume History of the Theory of Numbers, Dickson notes parenthetically that Saunderson was blind from infancy."

Note 2:  Bremner’s 1988 paper, The rational cuboid and a quartic surface gives 8th deg parametrizations, and points out that evidence suggests these can be found for every even deg ≥ 6.

 

 

You can email the author at tpiezas@gmail.com. 

 

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