1 such that n(n + 1)(n + 2) = m^r.
SaA99: Mark Saul, Titu Andreescu;
Square or not square?
Quantum July/August 1999, p45-46, 53
problem 10: n!+2000 cannot be a square
problem 11: n(n+1) cannot be a square
problem 12: n(n+2) cannot be a square
problem 14: n(n+1)(n+2)(n+3) cannot be a square
problem 15: (m+n)^2 + 3m + n + 1 cannot be a square (for n != m)
problem 16: 1!+2!+ ... +n! cannot be a square (for n > 3)
problem 17: n! cannot be a square (for n > 1)
This article contain 17 impossible square problems.
Sil65: David L. Silverman;
Mathematics Magazine 38 (Jan. 1965) 60
Product of four consecutive odd integers.
3^2 = (-3)(-1) 1 3
Tri85: Charles W. Trigg;
Mathematical Quickies,
1967 (reprinted: Dover Publications, 1985).
Problem 130: Never a Square
if n^4 + 2n^3 + 2n^2 + 2n + 1 is a square then n=0.
Problem 262: Wanted - Integer Solutions
find all integral solutions of y^2 + y = x^4 + x^3 + x^2 + x.
That is (2y+1)^2 = 4(x^4 + x^3 + x^2 + x) + 1.
==> arithmetic/consecutive.product.p <== from the rec.puzzles.archive
Prove that the product of three or more consecutive
positive integers cannot be a perfect square.
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==> arithmetic/consecutive.product.p <==
Prove that the product of three or more consecutive positive integers cannot
be a perfect square.
==> arithmetic/consecutive.product.s <==
Lemma:
If a and b are relatively prime, and ab is a square,
then a and b are squares. (This is left as an exercise.)
Three consecutive numbers:
Suppose (n - 1)n(n + 1) = k^2, where n > 1.
Then n(n^2 - 1) = k^2. But n and (n^2 - 1) are relatively prime.
Therefore n^2 - 1 is a perfect square, which is a contradiction.
Four consecutive numbers:
n(n + 1)(n + 2)(n + 3) = (n^2 + 3n + 1)^2 - 1
Five consecutive numbers:
Assume the product is a integer square, call it m.
The prime factorization of m must have even numbers of each prime factor.
For each prime factor, p, of m, p >= 5, p^2k must divide one of the
consecutive naturals in the product. (Otherwise, the difference between two
of the naturals in the product would be a positive multiple of a prime >= 5.
But in this problem, the greatest difference is 4.) So we need only consider
the primes 2 and 3.
Each of the consecutive naturals is one of:
1) a perfect square
2) 2 times a perfect square
3) 3 times a perfect square
4) 6 times a perfect square.
By the shoe box principle, two of the five consecutive numbers must fall into
the same category.
If there are two perfect squares, then their difference being less than five
limits their values to be 1 and 4. (0 is not a natural number, so 0 and 1
and 0 and 4 cannot be the perfect squares.) But 1*2*3*4*5=120!=x*x where x
is an integer.
If there are two numbers that are 2 times a perfect square, then their
difference being less than five implies that the perfect squares (which are
multiplied by 2) are less than 3 apart, and no two natural squares differ by
only 1 or 2.
A similar argument holds for two numbers which are 3 times a perfect square.
We cannot have the case that two of the 5 consecutive numbers are multiples
(much less square multiples) of 6, since their difference would be >= 6, and
our span of five consecutive numbers is only 4.
Therefore the assumption that m is a perfect square does not hold.
QED.
In general the equation:
y^2 = x(x+1)(x+2)...(x+n), n > 3
has only the solution corresponding to y = 0.
This is a theorem of Rigge [O. Rigge, ``Uber ein diophantisches Problem'',
IX Skan. Math. Kong. Helsingfors (1938)] and Erdos [P. Erdos, ``Note on
products of consecutive integers,'' J. London Math. Soc. #14 (1939),
pages 194-198].
A proof can be found on page 276 of [L. Mordell, ``Diophantine
Equations'', Academic Press 1969].
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Generalization:
Given a monic polynom p of even order.
Is there an integral solution of p(n) = k^2?
Solution:
There is a effective method to find
the finite number of solutions.
Compute the sqrt of p. Forget all term of order 1/N of higher.
This is an approximate solution.
Examples:
p(N) = N(N+1)
Sqrt(p(N)) = N + 1/2 + O(1/N)
q(N) = N + 1/2
r(N) = p(N) - q(N)^2 = -1/4
As N^2 < p(N) < (N + 1/2)^2 for N>0 it is not a square.
p(N) = N(N+1)(N+2)(N+3)
Sqrt(p(N)) = N^2 + 3N + 1 + O(1/N)
q(N) = N^2 + 3N + 1
r(N) = p(N) - q(N)^2 = -1 = O(1)
Note that r(N) is of order 0 and not of the expected order 1.
p(N) = N(N+1)(N+2)(N+3)(N+4)(N+5)
Sqrt(p(N)) = N^3 + 15/2 N^2 + 115/8 N + 75/16 + O(1/N)
q(N) = N^3 + 15/2 N^2 + 115/8 N + 75/16
r(N) = p(N) - q(N)^2 = O(N^2)
As order(r) < order(q) there can be at most finitely many solutions.
p(N) = N(N+1)(N+2)(N+3)(N+4)(N+5)(N+6)(N+7)
Sqrt(p(N)) = N^4 + 14 N^3 + 63 N^2 + 98 N + 28 + O(1/N)
q(N) = N^4 + 14 N^3 + 63 N^2 + 98 N + 28
r(N) = p(N) - q(N)^2 = (8 N + 28)^2 = O(N^2)
Note that r(N) is of order 2 and not of the expected order 3.
As order(r) < order(q) there can be at most finitely many solutions.
p(N) = N^4 + N^3 + N^2 + N + 1 [Hon78]
Sqrt(p(N)) = N^2 + N/2 + 3/8 + O(1/N)
q(N) = N^2 + N/2 + 3/8
r(N) = p(N) - q(N)^2 = 5/8 N + 55/64 = O(N)
As order(r) < order(q) there can be at most finitely many solutions.
As 2q(N) is between adjacent integers 2N^2 + N < 2q(N) < 2N^2 + N + 1
we will show that (2N^2 + N)^2 < 4p(N) < (2N^2 + N + 1)^2 except for
a finite set of integers.
First show the left side
(2N^2 + N)^2 <= 4p(N) <=> 0 <= 3N^2 + 4N + 4 = 1/3 ((3N+2)^2 + 8)
which is valid for all real N.
Second show the right side
4p(N) <= (2N^2 + N + 1)^2 <=> 0 <= N^2 - 2N - 3 = (N-3)(N+1)
which is valid for all real N > 3 and N < -1.
Checking the values N from -1 to 3 gives the integral
solutions p(-1) = p(0) = 1^2 and p(3) = 11^2.
p(N) = (N-3a)(N-a)(N+a)(N+3a) [Sil65]
Sqrt(p(N)) = N^2 - 5a^2 + O(1/N)
q(N) = N^2 - 5a^2
r(N) = p(N) - q(N)^2 = -16a^4 = O(1)
Note that r(N) is of order 0 and not of the expected order 1.
As (s-1)^2 - s^2 = -2s + 1 look for the roots of
-2(N^2 - 5a^2) + 1 = -16a^4
N = sqrt( 8a^4 + 5a^2 + 1/2 )
For a=1 we find the bounds 0<=N<4 and
for a=2 we find the bounds 0<=N<13 and
for a=3 we find the bounds 0<=N<27.
Solutions: a=1: p(0) = 3^2, p(1) = p(3) = 0^2.
Solutions: a=2: p(0) = 12^2, p(2) = p(6) = 0^2.
Solutions: a=3: p(0) = 27^2, p(3) = p(9) = 0^2.
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By partitioning the linear factors we get differences of squares.
p(N) = a1 a2 ... ak b1 b2 ... bk
= ((a1 a2 ... ak + b1 b2 ... bk)/2)^2
- ((a1 a2 ... ak - b1 b2 ... bk)/2)^2
But this approach is successfull for p4, p8, (and p2).
p2(N) = N(N+1)
= (N + 1/2)^2 - (1/2)^2
not the difference of two integers, as there are values 2 mod 4.
p4(N) = N(N+1)(N+2)(N+3)
= N(N+3) (N+1)(N+2)
= (N^2 + 3N)(N^2 + 3N + 2)
= (N^2 + 3N + 1)^2 - 1^2
Unique partition with one square of order 0.
q4(N) = N^2 + 3N + 1
p6(N) = N(N+1)(N+2)(N+3)(N+4)(N+5)
= N(N+3)(N+5) (N+1)(N+2)(N+4)
= (N^3 + 8 N^2 + 15 N)(N^3 + 7 N^2 + 14 N + 8)
= (N^3 + 15/2 N^2 + 29/2 N + 4)^2 - (1/2 N^2 + 1/2 N - 4)^2
= N(N+3)(N+4) (N+1)(N+2)(N+5)
= (N^3 + 15/2 N^2 + 29/2 N + 5)^2 - (1/2 N^2 + 5/2 N + 5)^2
= N(N+2)(N+5) (N+1)(N+3)(N+4)
= (N^3 + 15/2 N^2 + 29/2 N + 6)^2 - (1/2 N^2 + 9/2 N + 6)^2
All partitions with one square of order 2.
Note that the bigger squares are consecutive integers.
p8(N) = N(N+1)(N+2)(N+3)(N+4)(N+5)(N+6)(N+7)
= N(N+3)(N+5)(N+6) (N+1)(N+2)(N+4)(N+7)
= (N^4 + 14 N^3 + 63 N^2 + 90 N ) *
(N^4 + 14 N^3 + 63 N^2 + 106 N + 56)
= (N^4 + 14 N^3 + 63 N^2 + 98 N + 28)^2 - (8 N + 28)^2
= (q4(N) q4(N+4) - 1)^2 - (8 N + 28)^2
Unique partition with one square of order 1.
p12(N) = N(N+1)(N+2)(N+3)(N+4)(N+5)(N+6)(N+7)(N+8)(N+9)(N+10)(N+11)
= N(N+2)(N+6)(N+7)(N+8)(N+10) (N+1)(N+3)(N+4)(N+5)(N+9)(N+11)
= (N^6 + 33 N^5 + 418 N^4 + 2508 N^3 + 6952 N^2 + 6720 N ) *
(N^6 + 33 N^5 + 418 N^4 + 2574 N^3 + 8041 N^2 + 11793 N + 5940)
= (N^6 + 33 N^5 + 418 N^4 + 2541 N^3 + 14993/2 N^2 + 18513/2 N + 2970)^2
- ( 33 N^3 + 1089/2 N^2 + 5073/2 N + 2970)^2
Unique partition with one square of order 3.
p16(N) = N(N+1)(N+2)(N+3)...(N+14)(N+15)
= N(N+3)(N+5)(N+6)(N+9)(N+10)(N+12)(N+15)
(N+1)(N+2)(N+4)(N+7)(N+8)(N+11)(N+13)(N+14)
= (N^8 + 60 N^7 + 1490 N^6 + 19800 N^5 + 151569 N^4 + 665820 N^3 + 1547100 N^2 + 1458000 N ) *
(N^8 + 60 N^7 + 1490 N^6 + 19800 N^5 + 151953 N^4 + 677340 N^3 + 1671260 N^2 + 2024400 N + 896896)
= (N^8 + 60 N^7 + 1490 N^6 + 19800 N^5 + 151761 N^4 + 671580 N^3 + 1609180 N^2 + 1741200 N + 448448)^2
- ( 192 N^4 + 5760 N^3 + 62080 N^2 + 283200 N + 448448)^2
Unique partition with one square of order 4.
p20(N) = N(N+1)(N+2)(N+3)...(N+18)(N+19)
No partition with one square of order 6 possible.
---------------------------------------------------------
A classical theorem discovered independently by Sylvester and Schur,
states that the product of k consecutive integers, each greater than k,
has a prime divisor greater than k.
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Problem C
Can both x^2 + y and x + y^2 be squares for x and y natural numbers?
Solution:
No. Since x and y are both positive integers, for x^2 + y to be a
square requires y >= 2*x + 1. But similarly, for y^2 + x to be a
square, x >= 2*y + 1 is also necessary. These inequalities cannot
hold simultaneously if x and y are positive.
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Problem E:
Can p+q and p-q be squares for p and q primes?
Solution:
Let p+q = x^2 and p-q = y^2 then there difference is
(*) 2q = x^2 - y^2 = (x+y)(x-y).
Reducing mod 2 we have (x+y)(x-y) = 0 (mod 2).
As x+y = x-y (mod 2) both factors are 0 (mod 2).
That means 2q = x^2 - y^2 = (x+y)(x-y) = 0 (mod 4).
Therefore q = 0 (mod 2) which gives the solution q = 2.
Now we can substitute back into (*) finding
4 = (x+y)(x-y)
with two even factors x+y and x-y. Let both factors be positive
we find the system: x+y = 2 and x-y = 2.
Unique solution (2,2,2,0) = (p,q,x,y).
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Problem ([Trigg P262] and [Mar66])
Find all integers x, y satisfying y^2 + y = x^4 + x^3 + x^2 + x.
Solution:
Multiply both sides by 4 and add 1 gives
(2y + 1)^2 = 4x^4 + 4x^3 + 4x^2 + 4x + 1.
Let p(x) = 4x^4 + 4x^3 + 4x^2 + 4x + 1 then
Sqrt(p(x)) = 2x^2 + x + 3/4 + O(1/x)
We will show that (2x^2 + x)^2 < p(x) < (2x^2 + x + 1)^2 for all
x < -1 or x > 2.
First show the left side
(2x^2 + x)^2 <= p(x) <=> 0 <= 3x^2 + 4x + 1 = (x+1)(3x+1)
which is valid for x <= -1 or x >= -1/3.
Second show the right side
p(x) <= (2x^2 + x + 1)^2 <=> 0 <= x^2 - 2x = x(x-2)
which is valid for x <= 0 or x >= 2.
Therefore we have only to consider the cases of x in {-1, 0, 1, 2}.
Checking these values gives the six solutions
y = -1, x = -1
y = 0, x = -1
y = -1, x = 0
y = 0, x = 0
y = -6, x = 2
y = 5, x = 2
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Problem ([Hun81])
Find all integers x, y satisfying y^2 = x^4 + 4x^3 + 8x^2 + 4x + 16.
Solution:
Let p(x) = x^4 + 4x^3 + 8x^2 + 4x + 16 then
Sqrt(p(x)) = x^2 + 2x + 2 + O(1/x)
We will show that (x^2 + 2x + 1)^2 < p(x) < (x^2 + 2x + 3)^2 for all
x <= -5 or x >= 1.
First show the left side
(x^2 + 2x + 1)^2 < p(x) <=> 0 < 2x^2 + 15
which is valid for real x.
Second show the right side
p(x) < (x^2 + 2x + 3)^2 <=> 0 < 2x^2 + 8x - 7 = 2(x+5)(x-1) + 3
which is valid for x <= -5 or x >= 1.
Therefore we have only to consider the cases of x in {-4, -3, -2, -1, 0}
for exceptions. Checking these values gives the solutions
p(0) = 4^2 and p(-3) = 7^2.
We are left with the case that p(x) = (x^2 + 2x + 2)^2 the only
square in the inverval left. This is the case for x = 3 only.
So there is one further square p(3) = 17^2.
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Problem: [Tri85 Q 130]
if n^4 + 2n^3 + 2n^2 + 2n + 1 is a square then n=0.
solution Leo Moser, MM 37 (Jan. 1964) 62:
We have (n^2 + n)^2 = n^4 + 2n^3 + n^2 < n^4 + 2n^3 + 2n^2 + 2n + 1
< n^4 + 2n^3 + 3n^2 + 2n + 1 = (n^2 + n + 1)^2 for n != 0.
So the given number lies between two consecutive squares.
solution Ch. Trigg:
n^4 + 2n^3 + 2n^2 + 2n + 1 = (n^2 + 1)(n + 1)^2.
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Problem F: [SaA99 problem 17]
Show that n! cannot be a prefect square for any integer n>1.
Solution [SaA99 solution 17]
--------
Let p be the largest prime that divides n!. If we want n! to be a square,
it must contain at least one more multiple of p, namely 2p. But according
to Bertrand's postulate, between p and 2p there must be another prime.
This contradicts our assumption that p is the largest prime that divides
n!. Note that the proof doesnot apply for n=1 as no prime divides 1!.
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Problem Z: (from sci.math.research 1998-11-05)
Let P a monic polynomial in Z[X].
We suppose that :
- the degree of P is even,
- for all positive integer n, P(n) is a square in Z.
Can we say that P is a square in Z[X] ?
Solution (Mark Sapir)
--------
It can be a nice high school olympiad problem. Here is a sketch of an
elementary solution: if f(x) has degree 2n, find a polynomial g(x) such
that g(x)^2 has the same first n+1 coefficients as f(x) (it is always
possible to do); g(x) will have ratioanal coefficients, which is not a
problem, just multiply the variable x by a proper number, and you get a
polynomial with integer coefficients. Then use the fact that the
distance between two integer squares a^2-b^2 cannot be much smaller than
2*min{a,b}.
Mark Sapir msapir@math.vanderbilt.edu
Solution (Robert Israel):
--------
Yes, it must be a square. If p(z) is monic with degree 2n,
consider the Laurent series of q(z) = z^n sqrt(p(z)/z^(2n))
(taking the principal branch of the square root)
on |z| > R = max{ |r|: p(r)=0 }.
If p(z) is not a square, q(z) is not a polynomial, and we have
q(z) = r(z) + a_k z^(-k) + O(|z|^(-k-1))
where r(z) is a polynomial with rational coefficients and a_k \ne 0.
If M is the least common multiple of the denominators of the coefficients
in r(z), M (q(z)- r(z)) can't be an integer when z is a sufficiently large
integer, so neither can sqrt(p(z)).
Robert Israel israel@math.ubc.ca
Solution (Dave Rusin):
--------
Yes. When I ran into this question last year I got this information from
Bruce Reznick:
>Michael Filaseta [...] contacted Schinzel, who'd know. The
>consensus is that the original result (f: Z -> Z^2 => f = g^2) is
>apparently well-known and has been discovered several times.
I don't have any references, however.
Reznick's solution was to compute Q = sqrt(P) = F + G formally where F is
polynomial and G(n)->0 as n->oo. By assumption the values of Q are
integers so the iterated differences are too; on the other hand high-enough
differences of F also vanish. Thus the differences of G are both
integral and small, hence zero, making G itself both polynomial and small,
hence zero (as n->oo, and hence identically). So Q is a polynomial.
Here's an alternative proof Neil Dummigan and I worked out: factor
P = P1^2*P2 with P2 square-free; then y^2=P2(x) is a non-singular curve
with infinitely many integral points, hence P2 is at most quadratic.
(Obviously P2 can't be linear with all P2(n) squares unless P2=constant.)
Completing the square makes this equivalent to an equation y^2=ax^2+b
(Pell's equation) and the premise becomes that it has O(N) solutions
with x