Numbers Expressible As (a^2 - 1)(b^2 - 1)
Some time ago I mentioned that the number 588107520 is expressible
in the form (X^2 - 1)(Y^2 - 1) (where X,Y are integers) in five
distinct ways, and asked if anyone knew a 6-way expressible number.
So far, no 6-way expressible number has been found, although such
a number has not been proved impossible.
Regarding 5-way numbers, Dean Hickerson and Fred Helenius both
independently found five more, so as of now the complete list of
5-way expressible numbers is
588107520
67270694400
546939993600
2128050512640
37400697734400
5566067918611200
No one seems to know if there are infinitely many such numbers, or
even if there are any more beyond this list.
There are several possible approaches to constructing numbers of
this kind. One way is to notice that if a,b,c,d are four integers
such that the product of any two of them is a "shy square", i.e.,
ab = x^2 - 1 ac = y^2 - 1 ad = z^2 - 1
bc = u^2 - 1 bd = v^2 - 1 cd = w^2 - 1 (1)
then clearly (abcd) can be expressed in at least three ways, namely
abcd = (ab)(cd) = (ac)(bd) = (ad)(bc) (2)
There are some nice parametric formulas for sets of four numbers
with property (1). For example, take
a = n
b = q(qn+2)
c = (q+1)((q+1)n+2)
d = 4abc+2(a+b+c) (3)
where n is an integer and q is any rational number such that b
and c are integers. The product abcd is
F(q,n) = 4qn(qn+1)(qn+2)(q+1)(qn+n+1)(qn+n+2)(qn(q+1)+2q+1)
The values of F(q,n) are sure to have at least three representations,
but they may have more, and obviously any number that has more than
three must have three, so these are good numbers to check. In fact
we find that each of the numbers
F(3,4) = 588107520
F(5,5) = 67270694400
F(10,3) = 546939993600
F(12,2) = 2128050512640
F(4,-20) = 37400697734400
F(4/3,156) = 5566067918611200
has five distinct representations, as mentioned previously. Since
each of these numbers splits into two shy squares in five ways, each
of them is 3-way expressible in 10 different ways (i.e., there are
10 ways of choosing 3 out of 5 factorizations). The fact that each
of these numbers is of the form F(q,n) implies that at least one of
the 10 3-way sets must be of the form (2).
Not all 3-way solutions are of the form (2). The most general 3-way
set would consist of eight components a,b,c,d,e,f,g,h, with the three
factorizations
abcdefgh = (abcd)(efgh) = (abef)(cdgh) = (aceg)(bdfh) (4)
Similarly, the most general 4-way set would consist of 16 components,
and the most general 5-way set would consist of 32 components. To
see why an N-way set may require 2^N components, notice that the
components are bisected N times, and in each bisection a component
either is or isn't in the same segment as, say, component "a".
Thus we can encode each component's behavior in an N-bit binary
number, where of course a={111...1}. Every other N-bit number
can occur, so there may be 2^N different types of components.
Anyway, another approach to constructing multi-expressible numbers is
to look at integers of the form c^2-1 that can be expressed as a
product of two other integers of the same form. Define
f(a,b,...,z) = (a^2 - 1)(b^2 - 1)...(z^2 - 1)
and then find three integers a,b,c such that f(a,b), f(a,c) and
f(b,c) are all shy squares. Then the integer
N = f(a,b,c)
has the three representations
N = f(a,b)f(c) = f(a,c)f(b) = f(b,c)f(a)
Noticed that for any integer u the integers v such that f(u,v)
is a shy square occur in a second-order linear recurring sequence.
For example, the integers v such that f(5,v) is a shy square are
..., 3821, 386, 39, 4, 1, 6, 59, 584, 5781, ...
where the terms satisfy the recurrence s[n] = 10*s[n-1] - s[n-2]. In
general, for any integer u, the sequence has the central values
..., u-1, 1, u+1, ...
and it satisfies the recurrence s[n] = 2u*s[n-1] - s[n-2]. Also,
notice that for any integers u and j the integer f(s[u;j],s[u;j+1])
is a shy square. Therefore, the three numbers u, s[u;j], and s[u;j+1]
satisfy the stated condition, so the number
N = h(u,s[u;j],s[u;j+1])
has three distinct representations as a product of two shy squares for
any choice of the integers u and j. To illustrate, with u=5 and j=2 we
have
s[5;2]=59 s[5;3]=584
so the number f(5,59,584) is expressible as a product of two shy squares
in three distinct ways, as follows
f(5,59,584) = f(5,34451) = f(59,2861) = f(584,289)
Thus, the s-sequences together constitute a 2-parameter family of 3-way
expressible numbers.
Furthermore, for any given integer u you can generate a sequence of
this type from any integer 'a' such that
(u^2-1)(a^2-1) = (x^2-1) (5)
because if we define b = x + au we have
(u^2-1)(b^2-1) = (y^2-1)
where y = bu - a. Repeating this process we can define c = y + bu and
then we have
(u^2-1)(c^2-1) = (z^2-1)
where z = cu - b, and so on. Substituting y = bu - a into the equation
c = y + bu gives the recurrence
c = 2ub - a
so the numbers a,b,c,... constitute a sequence of solutions. To prove
that f(s[j],s[j+1]) is itself a shy square for any of these s-sequences,
notice that putting x = b - au in (5) and expanding the terms gives
(ua)^2 - u^2 - a^2 + 1 = b^2 - 2abu + (au)^2 - 1
Adding (b^2-u^2)(a^2-1) to both sides gives
(ab)^2 - a^2 - b^2 + 1 = (ab)^2 - 2abu + u^2 - 1
which factors as
(a^2-1)(b^2-1) = (ab-u)^2 - 1
showing that any two consecutive elements of the sequence a,b,c,...
satisfy a relation of this form.
Yet another approach was taken by Dean Hickerson and (independently)
Fred Helenius:
Define f(a,b) = (a^2 - 1)(b^2 - 1)
and
4mn(m^2 - 1)(n^2 - 1)(m n - 1)
g(m,n) = ------------------------------
(m-n)^2
If m and n are integers such that n-m divides m^2-1, then g(m,n)
is an integer and has three representations:
/ 2 m n^2 - m - n \ / 2 n m^2 - m - n \
g(m,n) = f( m, --------------- ) = f( n, --------------- )
\ n - m / \ n - m /
/ mn - 1 \
= f( ------, 2mn - 1 )
\ n - m /
Furthermore, for certain values of m and n there are additional
representations. In particular, if we set n = m+3 where m is not
divisible by 3, we have a four-way expressible integer
/ 2m^3 + 12m^2 + 16m - 3 \
g(m,m+3) = f( m, ----------------------- )
\ 3 /
/ 2m^3 + 6m^2 - 2m - 3 \
= f( m+3, --------------------- )
\ 3 /
/ m^2 + 3m - 1 \
= f( ------------, 2m^2 + 6m - 1 )
\ 3 /
/ 2m^2 + 6m - 5 \
= f( -------------, m^2 + 3m + 1 )
\ 3 /
This proves there are infinitely many four-way expressible numbers.
By examining lots of these you sometimes find one that has a 5th
representation. For example, with m=31 or 37 we have
g(31,34) = 546939993600 = f(31, 23869) = f(34, 21761) = f(271, 2729)
= f(351, 2107) = f(701, 1055)
g(37,40) = 2128050512640 = f(9, 163097) = f(37, 39441) = f(40, 36481)
= f(493, 2959) = f(985, 1481)
Another interesting case is to set m = 2r^2 - r - 2 and n = 2r^2 - 1,
which gives the four-way expressible numbers
g(m,n) = f( 2r^2 - r - 2 , 16r^5 - 24r^4 - 8r^3 + 16r^2 - 1)
= f( 2r^2 - 1 , 16r^5 - 32r^4 - 4r^3 + 28r^2 - 2r - 5)
= f((2r - 1)(2r^2 - 2r - 1) , (2r + 1)(4r^3 - 4r^2 - 4r - 3))
= f(4r^3 - 2r^2 - 4r + 1 , 8r^4 - 12r^3 - 4r^2 + 6r + 1)
Examining some of these reveals that with r = 3 or 4 there's a 5'th
representation as well:
g(13,17) = 588107520 = f(13, 1871) = f(17, 1429) = f(55, 441)
= f(79, 307) = f(129, 188)
g(26,31) = 67270694400 = f(26, 9983) = f(31, 8371) = f(161, 1611)
= f(209, 1241) = f(433, 599)
Another family of 4-way expressible numbers is given by setting
m = r(r^2 - 3)/2 n = (r^3 + r^2 - 4r - 2)/2
We also have the miscellaneous result
g(209,365) = F(4/3,156) = 5566067918611200
which doesn't seem to be part of an algebraic family of 4-way numbers.
In summary, the basic building block of multi-representable numbers
is the 2-parameter formula for three-way expressible numbers, which
can be algebraically specialized to 1-parameter families of four-way
expressible numbers. Some of these four-way expressibles also have a
fifth representation, but it isn't clear whether there is an algebraic
specialization to give these, or they are just numerical accidents.
There also appear to be some 5-way numbers that are not members of an
algebraic 4-way family. It's unknown if there exist any 6-way
expressible numbers.
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