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[lian109]

Cho x,y,z là các số nguyên dương sao cho $(xy+1)(yz+1)(zx+1) $ là số chính phương.Chứng minh rằng các số:$xy+1,yz+1,zx+1 $ đều là các số chính phương.:burnjossstick:
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[fool90]

đây là bài toán không dễ giải đâu, có thể nói là khó từ KIRAN KEDLAYA thì phải !
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[dong1919]

Hình như bài toán này đã có TQ rồi
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[n.t.tuan]

Thuổng bằng Goooogle. Ai quan tâm thì đọc nhé!

If ab+1, ac+1, and bc+1 are squares...

Given any three positive integers a,b,c such that the product of
any two is one less than an integer squared, i.e.,

ab+1 = x^2 ac+1 = y^2 bc+1 = z^2

we seek a fourth positive integer d such that its product with any
of the first three is also one less than a square, i.e.,

ad+1 = w^2 bd+1 = u^2 cd+1 = v^2

This is an old and very interesting problem. According to Dickson's
"History of the Theory of Numbers", the ancient Greek mathematician
Diophantus attacked the problem (in rational numbers) by taking x,
x+2, 4x+4 as the first three numbers, and (3x+1)^2 - 1 as the product
of the first and fourth. Thus the fourth is 9x+6. The product of the
2nd and 4th increased by unity is 9x^2 + 24x + 13. Setting this equal
to (3x-4)^2, he solved for x=1/16.

Fermat started with the three integers 1,3,8, and then the 4th number,
x, must make each of x+1, 3x+1, 8x+1 square. He solved this "triple
equation" for x=120.

Euler gave the parametric solution

a b c = a+b+2k d = 4k(a+k)(b+k)

where k^2 = ab+1. He then extended this to FIVE numbers, finding
that if we set
4r + 2u(1+s)
x = ------------
(s-1)^2

where
u = a+b+c+d r = abc+abd+acd+bcd s = abcd

then each of the numbers 1+ax, 1+bx, 1+cx, 1+dx is a rational square.
He derived this from the fact that if each of these is a square, then
their product is also a square, i.e.,

y^2 = (1+ax)(1+bx)(1+cx)(1+dx)

= 1 + ux + vx^2 + rx^3 + sx^4

where v=(ab+ac+ad+bc+bd+cd). (See Dickson, page 517 for the full
derivation.) For example, this gives the five numbers

1 3 8 120 777480 / 2879^2

Another approach to this problem, suggested by Peter Montgomery (based
on the solution of Arkin, Hoggatt, and Strauss), is to begin with the
familiar example a=1, b=3, c=8, d=120 and its permutations

ab+1 = 2^2 cd+1 = 31^2 (2)(31) = 62
ac+1 = 3^3 bd+1 = 19^2 (3)(19) = 57
bc+1 = 5^2 ad+1 = 11^2 (5)(11) = 55

and then observe that the products 62, 57, 55 are c+54, b+54, and
a+54. Also, negating one of the square roots (say replacing 19 by
-19) we get 62, -57, 55, which are 63-a, 63-d, and 63-c. This
suggests looking for a value of t such that

(a+t)^2 = (ad+1)(bc+1)
(b+t)^2 = (bd+1)(ac+1)
(c+t)^2 = (cd+1)(ab+1)

Subtract two equations (to eliminate the t^2 term) and solve for

d = a+b+c+2t

Plug this into any one of the three equations to get a quadratic
whose solution is

t = abc +- sqrt((ab+1)(ac+1)(bc+1))

Dan Cass applied Fermat's method to the triple a=6, b=8, c=28, which
means we want to make each of

6d+1 8d+1 28d+1

a square. Now 6d+1 is square exactly for d = 2p*(3p+-1), and 8d+1 is
square exactly for d = q*(2q+-1), and finally 28d+1 is square exactly
for d = r*(7r+-1). So to find d we need "only" find p,q,r such that
the following "triple equation" holds:

2p(3p+-1) = q(2q+-1) = r(7r+-1)

with one of the 8 possible choices of +- sign. Richard Pinch noted
that this triple equation reduces to the pair of simultenaeous
Pellians
4x^2 - 3y^2 = 1
7x^2 - 3z^2 = 4
where

x = 6p +- 1 y = 4q +- 1 z = 14r +- 1

Pinch also noted that Baker's method shows that any solution has
log(x) < 63.1. Using a "seiving technique", it can then be shown
that the only solutions are x = 1, y = 1, z = 1.

Jim Buddenhagen noted a connection to elliptic curves: If 6d+1,
8d+1, and 28d+1 are squares so is their product, so we have the
elliptic curve
y^2=(6d+1)(8d+1)(28d+1)

The point P=(0,1) is on the curve and can be used via the standard
chord/tangent method to get new points 2P, 3P, 4P, ... , where now
d is rational. This curve happens to be rank 1, and the point P is
a generator. Furthermore, the points P, 3P, 5P, 7P, ... each have
"d" coordinate making each of 6d+1, 8d+1, 28d+1 a rational square.
He then notes that 3P is an integer point with d-coordinate 5460.
With this value of d we get 6d+1=181^2, 8d+1=209^2, 28d+1=391^2.

In the general case, y^2=(ad+1)(bd+1)(cd+1), d and y variables.
The point P(0,1) is usually of infinite order (see Don Zagier's
survey paper on elliptic curves in Jbr. d. Dt. Math.-Verein vol
92(1990)58-76 for the exceptions). The point 3P is now has

8(a-b-c)(a+b-c)(a-b+c)
d = ---------------------------
(a^2+b^2+c^2-2ab-2ac-2bc)^2

which, if it is an integer (which is often the case) solves the
problem and in this case seems to coincide with Peter Montgomery's
solution.

Cass then asked: Given that ab+1=r^2, ac+1=s^2, bc+1=t^2 (all
integers), under what circumstances (conditions on a,b,c) is d
given by the above formula an integer? And if it is an integer, is
it necessarily equal to a+b+c+2abc+2rst ? (This was Montogmery's
solution, where the sign in front of 2rst could also be negative).

We observe that the great majority of triples (a,b,c) for which ab+1,
ac+1, and bc+1 are squares are evidently given by the two-parameter
family
a = n
b = q(qn+2)
c = (q+1)[(q+1)n + 2]

where n is an integer and q is any rational number such that b and c
are integers. This solution was given by N. Saunderson in 1740.
(In nearly the only personal comment of the entire 3-volume History,
Dickson notes parenthetically that Saunderson was blind from infancy.)
For these triples we have

ab + 1 = [ qn + 1 ]^2
ac + 1 = [ (q+1)n + 1 ]^2
bc + 1 = [ q(q+1)n + 2q + 1 ]^2

In general, for any triple a,b,c the two values of d given by
Montgomery's solution satisfy d+d' = 4abc+2(a+b+c). However, for
Saunderson's family of triples we always have d'=0, so the only
non-trivial value of d is 4abc+2(a+b+c). Substituting the values
of a,b,c gives
d = 4(nq+1)(nq+n+1)(nq^2+2q+nq+1)

so this is the 4th component of a Saunderson 4-tuple.

Of course, there are many "exceptional" triples a,b,c that are not
in Saunderson's family. Here are some examples:

(1,3,120) (2,4,420) (3,5,1008) (4,6,1980)
(1,3,1680) (2,12,420) (3,8,120) (4,12,420)
(1,8,120) (2,12,2380) (3,8,2080) (4,20,1980)
(1,8,528) (2,24,2380) (3,16,1008) etc...

These triples have two non-zero d values, and are the triples for
which Buddenhagen's solution is not an integer.

In trying to find a simple parameterization of these "exceptional"
triples, we came across this simple 1-parameter family of solution
4-tuples

a = n - 1 b = n + 1 c = 4n d = 4n(4n^2 - 1)

such that any product of 2 is one less than a square. In general,
3-tuples such as a,b,d of this form do not satisfy the condition

[(a+b+d)/2]^2 - 1 = (ab+ad+bd)

so these are among the "exceptional" 3-tuples. Buddenhagen then took
{a,b,d} of a Saunderson 4-tuple, which is an "exceptional" triple,
and determined the two possible "completions" from Montgomery's
solution. Of course, one is just c from the original Saunderson
4-tuple, but the other is d', giving the family

a = n
b = m(mn+2)
c = 4(mn+1)(mn+n+1)(m(mn+n+2)+1)
d'= (2mn+1)(2mn+3)(m(2mn+2n+3)+1)(mn(2mn+2n+5)+3n+2)

for which the product of any two is 1 less than a square.

Phil Gibbs commented that Saunderson's parameterisation is equivalent
to the class of triples (a,b,c) which are the positive integer
solutions of the equation:

a^2 + b^2 + c^2 - 2ab - 2ac - 2bc = 4

and he pointed out a simple geometric construction of this class of
triplets (not surprising, considering the similarity of these formulas
with Heron's formula for the area of a triangle). Suppose we have a
triangle in a plane whose points fall on lattice points with integer
values, and the area of that triangle is exactly one, e.g. a triangle
XYZ where

X = (0,0) Y = (2,4) Z = (5,11)

form the edge vectors

Y-X = (2,4) Z-X = (5,11) Z-Y = (3,7)

Multiplying together the coordinates of each vector, we get

8 = (2)(4) 55 = (5)(11) 21 = (3)(7)

which is a triplet. Gibbs then provided a proof by descent that
this construction gives ALL solutions of the equation. His proof
is reproduced in the Attachment. By the way, Kiran Kedlaya has
published an article in the Feb 98 issue of Mathematics Magazine, and
the main result is this proposition.
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[n.t.tuan]

Gibbs also commented that a large family of 4-tuple solutions are
given by the positive integer solutions of

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

This is a quadratic in any one of the variables, so for any one
solution we can get four more by taking the "other" root in each
variable. Repeating this, we get a tree of solutions. It can be
shown that there are exactly two disconnected trees of solutions.
Gibbs conjectures that ALL 4-tuples are solutions of this equation.
This equation can also be written in any of the three equivalent
forms

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

[(a+c)-(b+d)]^2 = 4(ac+1)(bd+1)

[(a+d)-(b+c)]^2 = 4(ad+1)(bc+1)

We also note that if we set A=sqrt(a), B=sqrt(b) and C=sqrt(c), then
the symmetrical expression for 3-tuples

a^2 + b^2 + c^2 - 2ab - 2ac - 2bc = 4

has the Heronian factorization

(A+B+C)(A+B-C)(A-B+C)(-A+B+C) = 4

and similarly the symmetrical expression for 4-tuples can be written
in either of the forms

-(A+B+C-D)(A+B-C+D)(A-B+C+D)(-A+B+C+D) = 4(1-ABCD)^2

(A+B-C-D)(A-B-C+D)(A+B-C-D)(A+B+C+D) = 4(1+ABCD)^2

where D = sqrt(d). This relates to the area of a quadralateral, just
as the previous expression relates to the area of a triangle. In both
cases we have a product of four factors equal to a square.

We also observe that Saunderson triples {a,b,c} with a < b < c
we have
a = y - x b = z - x c = y + z

which follows from the general formula

y^2 - 1 z^2 - 1
a = --------- b = ----------
c c

Since ab = x^2 - 1 we have

(cx)^2 - c^2 = (yz+1)^2 - (y+z)^2

so obviously we have a solution by setting cx = (yz+1) and c = (y+z).
Thus, for any rational values of y and z this construction gives a
rational solution triple (a,b,c) with x = (yz+1)/(y+z). This is an
integer solution if and only if y and z are integers such that (y+z)
divides (yz+1).

Of course, we can also have solutions that don't require c = y+z. For
the general rational solution we only require that c be rational for
any y and z such that ab+1 is a rational square. Thus, ALL rational
solutions are given by the following 3-parameter family for any
rational values of y, z, and q:

a = (y^2 - 1)/c
b = (z^2 - 1 )/c
c = [ (y^2-1)(z^2-1) - q^2 ] / (2q)

These formulas were found by Euler.

From this it can be shown that, given any three integers {a,b,c}, the
product of any two of them is 1 less than a square if and only if the
quantity
(abc)(abc+a+b+c) + (ab+ac+bc) (1)

is also 1 less than a square. This is equivalent to the statement that
there is an integer m such that

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

With m=0 this gives the "non-exceptional" triples.

It's interesting to review the conditions for sets of k integers such
that the product of any two is 1 less than a sqaure. In the simplest
case, k=2, we seek pairs of integers {a,b} for which:

ab = n^2 - 1 = (n-1)(n+1) (3)

The most obvious family of solutions is given by setting a=(n-1) and
b=(n+1) or, equivalently, |b-a| = 2. Squaring this gives the condition

(a+b)^2 - 4(ab) = 4 (4)

so these might be called the "non-exceptional" doubles. Of course,
this does not cover all doubles, because the quantity (n-1)(n+1) can
have other factorizations besides the algebraic one.

Proceding to triples {a,b,c}, we've seen that a large family of
triples (the "non-exceptional" ones) are given by

(a+b+c)^2 - 4(ab+ac+bc) = 4 (5)

but again this does not cover all possible triples.

Going on to 4-tuples, we have Gibbs' interesting conjecture that

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

covers all possible 4-tuples. Clearly the conjecture does not hold
if we allow zero values. For example, {a,b,c,0} is, strictly speaking,
a 4-tuple iff {a,b,c} is a triple. But if we set d=0 in the preceding
equation, it reduces to the equation for just the "non-exceptional"
triples, and we know there are exceptional triples. Thus we have
4-tuples of the degenerate form {a,b,c,0} that do not satisfy the
conjecture. Therefore, Gibbs' conjecture must stipulate positive
integers.

The technique of considering degenerate cases could have been used to
show that the non-exceptional equation for triples cannot be complete,
because it reduces to the non-exceptional double equation if we set
c=0, and we know there are exceptional doubles. The correct condition
is given by setting c=0 in the COMPLETE equation (2) for triples given
above, which then yields the true basic doubles requirement: ab+1
equals a square.

On a related question, Gibbs asks if it's true that given any four
integers {a,b,c,d} the product of any two of them plus 1 is a square
if and only if the quantity

(ab+1)(ac+1)(bc+1)(ad+1)(bd+1)(cd+1)

is a square? Montgomery replied that the answer is no. For example,
let a = b = c = d. Or let a = 1, c = b, d = b^2, with b arbitrary.
Here are some solutions (a, b, c, d, x) where a < b < c < d < 100
and x < 2^31 (x is the square root of the big product):

1 3 13 23 33600
1 5 11 19 40320
2 16 33 46 11035101
5 11 46 49 44324280

We may also consider a different generalization; given any four
integers {a,b,c,d}, is it true that the quantity

(abc+1)(abd+1)(acd+1)(bcd+1)

is a square iff (abc+1), (abd+1), (acd+1), and (bcd+1) are each
squares? There are certainly 4-tuples of this kind. Two examples
are {1,5,7,24} and {2,4,15,28}.

Notice that if this construction is extrapolated in reverse, it would
lead to the conjecture that (a+1)(b+1) is a square iff (a+1) and (b+1)
are each squares, which is obviously false. So, in an effort to
devise a fully general proposition, we conjecture that, given any n
positive integers x1,x2,...,xn, the quantity

n / Q \
PROD ( ----- + 1 ) where Q = (x1)(x2)...(xn)
i=1 \ xi /

is a (n-1)th power iff each of the n factors (Q/xi)+1 is a (n-1)th
power. This is trivially true for n=2, and it certainly seems to be
true for n=3.

Incidentally, triples {a,b,c} such that

ab+1 = 2x^2 ac+1 = 2y^2 bc+1 = 2z^2

seem to be much more rare than the simple square triples. The only
known example in integers greater than 1 is a=49, b=79, c=943. It
is evidently possible for the product (ab+1)(ac+1)(bc+1) to be of
the form 2m^2 while the individual factors are not.
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[n.t.tuan]

Bài của Kiran Kedlaya đây!
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