Conference rusure::math

Bypass this if you want to try it yourself...

   4      2                 2       2
  x  - 13x  + 36 <= 0 --> (x  - 4)(x  - 9) <= 0 --> (x-3)(x-2)(x+2)(x+3) <= 0

This will be true when any of the terms are 0 or when an odd number of them
are negative; i.e. the set of real numbers that satisfies the inequality is

			[-3,-2] U [2,3].

		   3	      2	
The derivative of x  - 3x = 3x  - 3 which is positive throughout both
subranges, so the maximum must be at an righthand endpoint; trying -2 and 3
shows the max is at 3 and it is 18.

Bypass this if you want to try it yourself...


    20000     100          19900            -100
  10     / (10    + 3) = 10      / (1 + 3*10    )

     19900         1    -100    2     -200    3     -300
 = 10      * (1 - 3 * 10     + 3  * 10     - 3  * 10     + ...

	       198     -19800    199     -19900    200     -20000
	... + 3    * 10       - 3    * 10       + 3    * 10       - ...)

		      100    199     200      100
 = (some integer) * 10    - 3    + (3    / (10    + 3))

				    200    100     100
Now, the last term is less than 1 (3    = 9    < 10   ), so the rightmost
digit of the quotient will be expressed as

		 199		     49
	nnnn0 - 3     = nnnn0 - ((81)   * 27) = nnnnn0 - mmmmm7 = 3

Bypass this if you want to try it...


			  -1  2		  -1
Lemma: sigma(i=0 to n) cot  (i +i+1) = cot  (1/(n+1))

						  -1	     -1
Proof by induction: 1)	its true for n=0, i.e. cot  (1) = cot  (1)

2)	If its true for i=0 through n-1, then

			   -1  2	   -1           -1  2	
	sigma(i=0 to n) cot  (i +i+1) = cot  (1/n) + cot  (n +n+1)

	Using the formula cot(x+y) = (cot(x)cot(y)-1)/(cot(x)+cot(y))
	this yields the desired result and we have a lemma.

Therefore,
			   -1  2			     -1
 sigma(i=0 to infinity) cot  (i +i+1) = lim(n-->infinity) cot  (1/(n+1))

			= cot (0) = pi/2

    finally, one I CAN solve! (it IS the easiest one of the lot)  
    
              .
             / \
        ............. ____
        |  /     \  |
        | /   +   \ |   h
    	|/_________\| ____

                    
    	"+" :== center of circle
    
    when the height of the triangle that sticks up above the rectangle
    is equal to "h", the two areas will be the same.
    
    thus h + (1/2)h = r = 1
    
    		h = 2/3
    
                                                   
                  /
                 (  ___
                  ) ///
                 /
    

Bypass this if you want to try it yourself. Its not a very pretty
solution - is there a better one?


Since each transversal contains exactly one element from each row and column,
if you pick two elements a(i,j) and a(k,l) in a transversal and substitute
the "other corners" a(i,l) and a(k,j) for them you still have a transversal.
If the sum of every transversal in A is the same, it follows that

	a(i,j) + a(k,l) = a(i,l) + a(k,j) for all i,j,k and l

which implies a(i,j) - a(k,j) = a(i,l) - a(k,l), which implies that each
row of the matrix A differs from each other row, and specifically from the
first row, by a constant; i.e.

(1)	a(i,j) = a(1,j) + d(i) for all i and j.

Conversely, any matrix that obeys formula (1) will have the sum of all
its transversals equal to sigma(j=1 to n)a(1,j) + sigma(i=2 to n)d(i), and
is thus the kind of matrix we are looking for. So, the set of matrices we
are looking for is exactly the set of matrices that obey (1).

Now, how many matrices of order n with elements in {-1, 0, 1} satisfy (1)?

Well, there are 3^n possible "first rows" of such a matrix. They fall into
three categories:

(a) First rows where max(a(1,j))-min(a(1,j)) = 0 - there are three such first
	rows, namely all 0, all 1, and all -1.

(b) First rows where max(a(1,j))-min(a(1,j)) = 1 - there are 2^(n+1)-4 such
	first rows, namely all rows consisting of just {0,1} except for all 0
	and all 1 (2^n-2 of these), and all rows consisting of just {0, -1}
	except for all 0 and all -1 (another 2^n-2 of these).

(c) First rows where max(a(1,j))-min(a(1,j)) = 2 - all the remaining first
	rows are of this type, namely 3^n-2^(n+1)+1.

Now:
For first rows of type (a) there are 3 choices of d(i) for every other row
For first rows of type (b) there are 2 choices of d(i) for every other row
For first rows of type (c) there are 1 choices of d(i) for every other row

So, the total number of matrices of order n is:

3 * 3^(n-1)  +  (2^(n+1)-4) * 2^(n-1)  +  (3^n-2^(n+1)+1) * 1^(n-1)

which in the form needed by the problem is:

	2 * 3^n  +  1 * 4^n  +  (-4) * 2^n  +  1 * 1^n

Whew!

Problem A-2:

�	What is the units (i.e. rightmost) digit of [10^20000 / (10^100 + 3)]?
� Here [x] is the greatest integer <= x.

    Expanding the denominator in a power series we get

	10^19900 * [ 1 - 3*10^-100 + 9*10^-200 - 27*10^-300 + ... ]

    Deep in the bowels of the parenthesised series we find a term like

	(-1)^k * 3^k * 10^(100*k) * (1 - 3 * 10^-100) +

    Set k to 199 and combine with 10^19900 and get

	- 3^199 * (1 - 3 * 10^-100)

    For powers of 3 the least significant decimal digit runs thru
    {1, 3, 9, 7} as k runs thru the integers, so we find the least
    digit to be 7 since 199 mod 4 = 3 and the term is

	- ...6.9999999999

    So the least digit must be 3.

    - Jim

Problem B-1:


�	 Inscribe a rectangle of base b and height h and an isosceles triangle
� of base b in a circle of radius 1 in such a way that the base of the triangle
� coincides with a base of the rectangle.  For what value of h do the rectangle
� and triangle have equal areas.

    Let the angle from the center of the circle subtended by the base
    of the triangle be t.  Then the area of the rectangle will be
    4*cos(t)*sin(t), and the area of the isosceles triangle will be
    (1+cos(t))*sin(t).

    By the stated equality,

	4*cos(t)*sin(t) = (1+cos(t))*sin(t),

	cos(t) = 1/3.

    And the height of the triangle will be 1+cos(t) = 1+1/3.

    Also, the angle of the isosceles triangle subtended by its base will
    be t/2.

    - Jim 

    B-4 solution:
    
Show that G goes to zero by picking a particular m and n for each r.

 **  **  **  **  **  **  **  **  **  **  **  **  **  **  **  ** 

Let  H(r)  =  | r - sqrt(M^2 + 2*N^2) |

where   M = floor (r)
and     N = floor ( sqrt( (r-M)*M ) )

Then G(r) <= H(r), and H(r) ---> zero

(1)  Example:	r = 1234.5
		M = 1234
		sqrt (617) = 24.8
		N = 24
		sqrt (1234^2 + 2 * 24^2) = sqrt (1523908) = 1234.467
		H(1234.5) = | 1234.5 - 1234.467 | = .033

                | r^2 - M^2 - 2*N^2 |         | r^2 - M^2 - 2*N^2 |
(2)  H(r) =  ---------------------------  <  -----------------------
              | r + sqrt(M^2 + 2*N^2) |                 r

(3)  Work on the numerator:

	Let   X = r - M   and   Y = sqrt( (r-M)*M ) - N

	| r^2 - M^2 - 2*N^2 |

	=  | r^2 - (r-X)^2 - 2*(sqrt( (r-M)*M - Y)^2 |

	=  | r^2 - r^2 + 2*r*X - X^2 - 2*(sqrt( X*M - Y)^2 |

	=  | 2*r*X - X^2 - 2*( X*M - 2*Y*sqrt( X*M ) + Y^2 ) |

	=  | 2*r*X - X^2 - 2*X*M + 4*Y*sqrt( X*M ) - 2*Y^2 |

	=  | 2*r*X - X^2 - 2*X*(r-X) + 4*Y*sqrt( X*M ) - 2*Y^2 |

	=  | 2*r*X - X^2 - 2*r*X + 2*X^2 + 4*Y*sqrt( X*M ) - 2*Y^2 |

	=  | X^2 + 4*Y*sqrt( X*M ) - 2*Y^2 |

	<= | X^2 | + | 4*Y*sqrt( X*M ) | + | 2*Y^2 |

	<= 1  +  4 * sqrt (M)  +  2

	<= 4 * sqrt(r)  +  3

                  4 * sqrt(r)  +  3
(4)  So H(r)  <  -------------------
                          r

	which ---> zero as r ---> infinity.
    
    
    John

    re .7:
    
    So what is "h"? 
                                                   
                  /
                 (  ___
                  ) ///
                 /
    

    Re:             -< after all that, what's the answer? >-

    Oh well, maybe I'll learn to read someday.

    - Jim

Problem B-2:

�	Prove that there are only a finite number of possibilities for the
� ordered triple T = (x-y, y-z, z-x) where x, y, and z are complex numbers
� satisfying the simultaneous equations
� 
�		x (x-1) + 2 y z = y (y-1) + 2 z x = z (z-1) + 2 x y,
� 
� and list all such triples T.

    First assume the numbers distinct.  With no loss in generality,
    translate so x is at the origin giving:

	2*y*z = y*(y-1) = z*(z-1)

    This is effectively a first order system, giving y = -1, z = -1,
    and x assumed to be zero.

    The triple is translation invariant, giving the only triple,

	(1, 0, -1).

    Now suppose x and y are equal, and translate so they are zero:

	z*(z-1) = 0,

    There are two roots and two triples:

	(0, 0, 0),
	(0, 1, 1)

    If all 3 are equal the equations are homogenous and the only triple
    is

	(0, 0, 0)

    - Jim

Solution for B-2 follows


The simultaneous equations

		x (x-1) + 2 y z = y (y-1) + 2 z x = z (z-1) + 2 x y,

are equivalent to

		(x-y) (x+y-1-2z) = 0
		(y-z) (y+z-1-2x) = 0

The solutions to these are:

		((x = y) or (x+y = 1+2z)) and ((y = z) or (y+z = 1+2x))

Going through the 4 possibilities, the solutions are:

		x = y = z
		x = y, z = 1+x
		y = z, x = 1+y
		z = x, y = 1+z

In these cases, the ordered triple T = (x-y, y-z, z-x) takes values:

		( 0,  0,  0)
		( 0, -1,  1)
		( 1,  0, -1)
		(-1,  1,  0)

I agree with .12.

The (0,1,1) triple can't be right because the components
have to sum to (x-y)+(y-z)+(z-x) = 0.

I 'found' what appears to be a solution to problem A-6, but as yet have
no idea of why it turns out the way it does, or even whether it is the
general solution.  It follows the <FF>.


	Let a(1), ..., a(n) be real numbers, and let b(1), ..., b(n) be
distinct positive integers.  Suppose there is a polynomial f(x) satisfying 
the identity (1 - x)^n f(x) = 1 + sum(a(i) x^b(i)), where the sum is taken
from 1 to n.  Find a simple expression (not involving sums) for f(1) in
terms of b(1), ..., b(n) (but independent of a(1), ..., a(n).)

	f(1) = product(b(i)) / n!, where the product is taken from 1 to n.

� The (0,1,1) triple can't be right because the components
� have to sum to (x-y)+(y-z)+(z-x) = 0.

  That was sort of a typo, I really meant (0,-1,1) but still missed one
  triple.

  Two of the easiest ones out of 3 wrong, I should just give up on this
  nonsense...

  - Jim

I can't resist trying yet another one.

    Instead of summing the angles we can consider instead the problem
    of rotating in the complex plane and recover the resulting angle.

    arccot(k^2+k+1) = the included angle of the complex number (k^2+k+1) + j.

    Form the product PROD (k>=0) [(k^2+k+1) + j] = (1+j)*(3+j)*(7+j)*...

    The first few partial products z[k] = x[k] + j y[k] are

	z[0] = 1 + j
	z[1] = 2 + j 4
	z[2] = 10 + j 30
	z[3] = 100 + j 400

    Now assume that in the k'th product y[k-1] = k*x[k+1]
    and write out the recurrance

	x[k] = x[k-1]*(k^2+k+1) - y[k-1] = x[k-1]*(k^2+k+1) - k*x[k-1]
	y[k] = y[k-1]*(k^2+k+1) + x[k-1] = k*x[k-1]*(k^2+k+1) + x[k-1]

	x[k] = x[k-1]*(k^2+1)
	y[k] = x[k-1]*(k^3+k^2+k+1)

    But k^3+k^2+k+1 = (k+1)*(k^2+1) and so y[k] = (k+1)*x[k], and the
    included angle must therefore approach pi/2 as k approaches oo.

    - Jim

>	f(1) = product(b(i)) / n!, where the product is taken from 1 to n.

Yeah, I suspected that as the answer by plugging in

		f(x) = 1 + x + x^2 + ... + x^n

which makes the a(i) and b(i) terms easier to figure.

One approach that looks real promising, but has me pretty well bogged down
in the details, is to make use of the following properties of polynomials
with repeated roots, which can be derived by expressing the polynomial in
product-of-roots form and differentiating it repeatedly:

1)	If a polynomial P has an n'th order repeated root r, then

			     2			n-1
		dP       ,  d P       ,  ...   d   P
	P(r),	---(r)	    ---(r)	       -----(r) are all zero
		dx	      2			 n-1
			    dx		       dx

2)	for that same polynomial P, the value of the reduced polynomial
	Q = P / (r-x)^n at r is given by

	    n	n
	(-1)   d P
	---- * ---(r)
		 n
	 n!    dx

In this case, property (1) with r=1 gives a set of n simultaneous equations
in the a(i) and b(i), which can be used to express the a(i) in terms of the
b(i), and property (2) gives the value of f(1).

The term prod(b(i)) showed up real early when I tried to solve the set of
equations, but the complexity quickly became more than I could afford.....

There must be an easier way!
							Richie

Currently I haven't solved A5.   Of the others, I think A3,A6,B4 and B6 are 
quite nice.

-------------------------------------------------------------------------------

Problem A-1.
	Find, with explanation, the maximum value of f(x) = x^3 - 3x on
the set of real numbers x satisfying x^4 + 36 <= 13x^2.

Answer A-1.

	Extrema of f are either at stationary points, or at boundaries of
valid intervals which x can belong to.

	x^4 + 36 =< 13x^2
<=>	(x+3)(x+2)(x-2)(x-3) =< 0
<=>	x is in [-3,-2] or [2,3]

	Where are the stationary points of f(x)?
	df/dx = 3x^2 - 3 = 0
<=>	x = � 1
	Neither of these lies within the feasible intervals, so the maxima
must be at a boundary of the valid intervals.   x = �2 or �3, with f(x) being
�2 or �18.   Therefore, maximum is:

===>>> 18

-------------------------------------------------------------------------------
 
Problem A-2.
	What is the units (i.e. rightmost) digit of [10^20000 / (10^100 + 3)]?
Here [x] is the greatest integer <= x.

Answer A-2.
	[x/y] == [(x + 10zy)/y] (mod 10)

So, mod 10
a =	[10^20000 / (10^100 + 3)]
==	[(10^20000 - 10^19900(10^100+3)) / (10^100+3)]
==	[-10^19900.3 / (10^100+3)]
...
==	[-10^100.3^199 / (10^100+3)]

Now we want to add 3^199(10^100+3)/(10^100+3)
This = 3^199, which is == 7 mod10

a =	[3^200 / (10^100+3)] + 3

But the term in brackets is less than 1, and so a =

===>>> 3

-------------------------------------------------------------------------------

Problem A-3.
	Evaluate sum(arccot(n^2 + n + 1)) as n runs over non-negative integers.
Here arccot(t) for t >= 0 denotes the number theta in the interval 
0 < theta < pi/2 with cot(theta) = t.

Answer A-3.
I like this one.
		   1 + cot(a)cot(b)
	cot(a-b) = ----------------
		    cot(b) - cot(a)

	So let cot(a) = 1/(n+1), cot(b) = 1/n.   Now:
	cot(a-b) = n^2 + n + 1
=>	arccot(n^2+n+1) = arccot(1/(n+1)) - arccot(1/n)
=>	sum(n=0 to N)(arccot(n^2+n+1)) 
   =	sum(n=1 to N)(arccot(n^2+n+1)) + arccot(1) 
   = arccot(1/(N+1)) - arccot(1) + arccot(1)
The limit as N goes to infinity of arccot(1/(N+1)) is pi/2, so
	sum(arccot(n^2+n+1)) =

===>>> pi/2

-------------------------------------------------------------------------------
 
Problem A-4.
	A transversal of an n x n matrix A consists of n entries of A, no two
in the same row or column.  Let f(n) be the number of n x n matrices A
satisfying the following two conditions:
 
	a)  Each entry a(i,j) of A is in the set {-1, 0, 1}.
	b)  The sum of a transversal is the same for all transversals of A.
 
An example of such a matrix is
 
				[-1  0 -1 ]
			    A = [ 0  1  0 ].
				[ 0  1  0 ]
 
Determine with proof a formula for f(n) of the form
 
		f(n) = a(1)b(1)^n + a(2)b(2)^n + a(3)b(3)^n + a(4),
 
where the a(i)'s and b(i)'s are rational rational numbers.
 
Answer A-4.
(R1)	a(i,j) is in {-1, 0, +1} for all i,j
(R2)	sum(i=1 to n)(a(i,p(i)) = K, for all perms p of the set {1,...,n}	

Suppose that p is a perm taking i to j, and i' to j', and q is the 
transposition (j,j').   Then have
	sum(a(i,p(i)) = sum(a(i,pq(i))
=>	a(i,j)+a(i',j') = a(i,j')+a(i',j)
=>	a(i,j) = a(i,j')+a(i',j)-a(i',j')
So in particular if i'=j'=1, then a(i,1) & a(1,j) for i,j = 1 to n, determine
all the rest, and have:
	a(i,j) = a(i,1)+a(1,j)-a(1)
This is sufficient to satisfy R(2) since
	sum(i=1 to n)(a(i,p(i))
   =	sum(i=1 to n)(a(i,1) + a(1,p(i) - a(1,1))
   =	sum(i=1 to n)(a(i,1) + a(1,i) - a(1,1))
which is independent of the perm chosen.

	It remains to satisfy (R1) in our choice of a(i,1) & a(1,j).
If we pick the 2n-1 values from the set {-1,0,1}, then we must also ensure that
for all i,j > 1:
	a(i,1)+a(j,1)-a(1,1) = -1,0 or 1.
If a(i,1) = a(1,1) for all i, then there are no constraints on a(1,j) for j>1
Thus we have 3^(n-1) possibilities.   Similarly, if a(1,j) = a(1,1) for all j, 
then we have 3^(n-1) possibilities, of which 3 have been counted already.
	If a(i,1)=1, and a(i',1)=-1, then all a(1,j) must be the same, else 
we have a(1,j) > a(1,j') => a(i,j)-a(i',j')>2, which is impossible.   Thus
we have no new cases.
	If a(1,1)=�1 then, all new cases must have a(i,1) & a(1,j) drawn from
{0,a(1,1)}.   All of these are legal which gives us 2.(2^(n-1)-1)^2 (must avoid
having all elements on row (or column) the same, since we've counted them
already.
	If a(1,1)=0, then must avoid having a(i,1)=a(1,j)=�1.   So draw a(i,1)
from {0,x=�1}, and a(1,j) from {0,x}.   Discounting once more the cases whre
all elements on row are the same, have 2.(2^(n-1)-1)^2.

	This gives us a grand total (which we can check for n=1,2 where
f(n) = 3,19) for f(n) =

===>>> 1.4^n + 2.3^n - 4.2^n + 1 

-------------------------------------------------------------------------------

Problem A-5.
	Suppose f1(x), f2(x), ..., fn(x) are functions of n real variables
x = (x1, ..., xn) with continuous second order partial derivatives everywhere
on R^n.  Suppose further that there are constants c(i,j) such that
 
		d fi   d fj
		---- - ---- = c(i, j) for all i, j between 1 and n.
		d xj   d xi
 
(Here d denotes partial derivative.)  Prove that there is a function g(x)
on R^n such that fi + dg / dxi is linear for all i, 1 <= i <= n.  (A linear
function is one of the form a0 + a1 x1 + a2 x2 + ... + an xn.)

Answer A-5.
??? 
Beats me.

-------------------------------------------------------------------------------
 
Problem A-6.
	Let a(1), ..., a(n) be real numbers, and let b(1), ..., b(n) be
distinct positive integers.  Suppose there is a polynomial f(x) satisfying 
the identity
 
			(1 - x)^n f(x) = 1 + sum(a(i) x^b(i)),
 
where the sum is taken from 1 to n.  Find a simple expression (not
involving sums) for f(1) in terms of b(1), ..., b(n) (but independent of
a(1), ..., a(n).)

Answer A-6.
A don't-be-frightened-of-matrices question:
	Examining the equation and its derivatives at x=1, we have that:

	sum(a(i)) = -1.

for j=1,n-1:
	sum(b(i).(b(i)-1)...(b(i)-j+1).a(i) = 0

In fact, we can clearly add these latter equations together, and derive:
for j=1,n-1:
	sum(b(i)^j.a(i)) = 0

We thus have that M.a=v, where M is an n*n matrix, and a&v are n-vectors.
	a_j = a(j)
	v_i = 1 if i=0, 0 if i=1 to n-1
	M_i_j = b(j)^i
If we can compute M^(-1).v, then we know a.

	It is clear that the determinant of M = prod(i>j)(b(i)-b(j)).   To
see this, note that the matrix is singular if b(i)=b(j), and therefore 
b(i)-b(j) must factor the determinant.   det M is a poly of degree �n(n-1),
so we only have a constant factor to calculate.   The major diagonal term
gives us the sign of det M, so we're there.

	It is nearly as easy to compute the co-factor of M_1_k, the 
determinant of the matrix without the first row and kth column, to be
�prod(i=/=k)(b(i)).prod(i>j i,j=/=k)(b(i)-b(j)).   These are the only
cofactors that we need to calculate to derive a, since v is mostly empty.
So a_k = �prod(i=/=k)(b(i)).prod(i>j i,j=/=k)(b(i)-b(j))/detM,
where the sign is -(-1)^k.

	Compute f(1) by setting x=1+y, and looking at coefficient of y^n.

f(1) = sum(k)(choose(b(k),n).a(k))
     = (1/n!).sum(k)(b(k).(b(k)-1)...(b(k)-n+1).a(k)
By the same additions as we did before:
     = (1/n!).sum(k)(b(k)^n.a(k))
     = [ (1/n!)/detM ] *
	sum(k)(�(b(k)^n).prod(i=/=k)(b(i)).prod(i>j i,j=/=k)(b(i)-b(j)))
     = [ (1/n!)/detM ] * prod(i)(b(i)) *
	sum(k)(�(b(k)^(n-1).prod(i>j i,j=/=k)(b(i)-b(j)))
But the second line is recognizable as detM, expressed as the sum of the
products of the bottom row with its cofactors.   Hence:
f(1) =

===>>> prod(i)(b(i))/n!

-------------------------------------------------------------------------------
 
Problem B-1.
	Inscribe a rectangle of base b and height h and an isosceles triangle
of base b in a circle of radius 1 in such a way that the base of the triangle
coincides with a base of the rectangle.  For what value of h do the rectangle
and triangle have equal areas.

Area of rectangle = bh
Area of triangle = �ba

	Where:
	b^2 + h^2 = 4
	a = 1 + sqrt(1-b^2/4)
hence	a = 1 + h/2

So require:	bh = �ba
=>		2h = 1 + h/2
=>		h = 2/3

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Problem B-2.
	Prove that there are only a finite number of possibilities for the
ordered triple T = (x-y, y-z, z-x) where x, y, and z are complex numbers
satisfying the simultaneous equations
 
		x (x-1) + 2 y z = y (y-1) + 2 z x = z (z-1) + 2 x y,
 
and list all such triples T.

Answer A-2.
	Let a = x-y, b=y-z, c =z-x.   Note a+b+c=0
	x^2-y^2 + x-y + 2yz-2xz = 0
<=>	(x-y).(x+y-2z+1) = 0
<=>	a(b-c+1)=0
Sim	b(c-a+1)=0
&       c(a-b+1)=0

Assume that none of a,b or c = 0.   Then b-c+1 = c-a+1 = a-c+1 = 0.   Adding
these together, get 3=0 contradiction.   So one of them is 0, wlog a.
	b(c+1) = 0
&	c(-b+1) = 0
Either b=c=0, or c+1=1-b=0.   So, remembering the arbitrary choice of a=0, the
possible ordered triples are:

===>>> (0,0,0) (0,1,-1) (1,-1,0) (-1,0,1)

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Problem B-3.
	Let Z[x] consist of all polynomials in x with integral coefficients.
For f and g in Z[x] and m a positive integer we write "f = g (mod m)" to
mean that f-g is an integral multiple of m.  Let n and p be positive integers
with p prime.  Given that f, g, h, r, and s are in Z[x] with
 
			r f + s g = 1 (mod p),
			f g = h (mod p),
 
prove there exist F and G in Z[x] with F = f (mod p), G = g (mod p), and
F G = h (mod p^n).

Answer B-3.
	Induction on n.   Given h
If f,g,r,s are in Z[x] with
		rf + sg = 1 mod p^n
		fg = h mod p^n
show that there exist F,G,R,S with
(i)		F = f mod p^n
(ii)		G = g mod p^n
(iii)		RF + SG = 1 mod p^(2n)
(iv)		FG = h mod p^(2n)
[If this can be shown, then since the initial condition are satisfied for
n=1, we can show for all p^(2^m), and a fortiori, for all p^n.

	Say that fg = h + Cp^n
		 rf + sg = 1 + Dp^n

	Define F = f -sCp^n
	       G = g -rCp^n
	       R = r + r(-D+2rsC)p^n
	       S = s + s(-D+2rsC)p^n

(i) & (ii) are trivially satisfied.   
FG = fg + (-rfC -sgC)p^n +Xp^(2n)
   = h + (C -rfC -sgC)p^n + Xp^(2n)
   = h + (C-C)p^n + (X-CD)p^(2n)
   = h + Yp^(2n)
FR+GS = (f-sCp^n)(r+r(-D+2rsC)p^n) + (g-rCp^n)(s+s(-D+2rsC)p^n) 
      = fr+gs +(-rsC +fr(-D+2rsC) -rsC +gs(-D+2rsC))p^n + Zp^(2n)
      = 1 +(D -rsC +(fr+gs)(-D+2rsC) -rsC)p^n + Zp^(2n)
      = 1 +(D -2rsC + (-D+2rsC))p^n + (Z-D+2rsC)p^(2n)
      = 1 + Wp^(2n)

===>>> So this result holds for all n.

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Problem B-4.
	For a positive real number r, let G(r) be the minimum value of
|r - sqrt(m^2 + 2n^2)|, for all integers m and n. Prove or disprove the
assertion that lim G(r) exists and is 0 as r ---> infinity.

Answer B-4.
The key is to relax the right constraint.   Don't try to calculate G(r)
precisely (there's a lot of quadratic form material that could
be dragged in here) but just serve up a crude upper bound.

	Let a = floor(r).   Then r lies in some interval:
[sqrt(a^2 + 2b^2), sqrt(a^2 + 2(b+1)^2)), where b^2 is drawn from the set
{0,1,...a}.

	Now G(r) =< f(r) = minimum value of |r - sqrt(a^2 + 2b^2)| over all
integers b.   (Note, a is a function of r).

	f(r) =< �(sqrt(a^2 + 2(b+1)^2)-sqrt(a^2 + 2b^2))

	     			2b+1
	     =	-------------------------------------
		sqrt(a^2 + 2(b+1)^2)+sqrt(a^2 + 2b^2)

		     2b+1
	     < ---------------		=< (b+�)/a	< 2/sqrt(a)
	        2sqrt(a^2+2b^2)    

but a is O(r), so f(r) -> 0 as r -> infinity.   G(r) is bounded by f(r):

===>>> as r -> infinity, lim G(r) = 0.

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Problem B-5.
	Let f(x, y, z) = x^2 + y^2 + z^2 + x y z.  Let p(x,y,z), q(x,y,z),
and r(x,y,z) be polynomials with real coefficients satisfying
 
	f(p(x,y,z), q(x,y,z), r(x,y,z)) = f(x, y, z).
 
Prove or disprove the assertion that the sequence p, q, r consists of some
permutation of (+-)x, (+-)y, (+-)z, where the number of minus signs is 0 or 2.

Answer B-5.
Is there a more elegant way of treating this?

	Let i be maximum degree of p, and let 
a = sum(i1+i2+i3=i)(d(i1,i2,i3)x^i1.y^i2.z^i3)
Similarly, define b and c, j and k for q and r respectively.

	Is a^2 = 0?   No, because for instance the unique term which
maximizes the degree of x, and then maximizes the degree of y, cannot
disappear.   Similarly b^2 and c^2?

	Is abc = 0?   No, once again, there is a unique term which maximizes
the degree of x, and then maximizes the degree of y, cannot disappear.   So
(if we assume that i >= j>= k wlog) the highest degree is 2i or i+j+k.   But
the highest degree in f(x,y,z) is 3.   So i+j+k=3, and 2i =< 3, so i=j=k=1.
	p = d(p,x)x + d(p,y)y + d(p,z)z + d(p)
	q = d(q,x)x + d(q,y)y + d(q,z)z + d(q)
	r = d(r,x)x + d(r,y)y + d(r,z)z + d(r)

	d(p,x)^2 + d(q,x)^2 + d(r,x)^2 = 1
	d(p,x)d(q,x)d(r,x) = 0
& the same for q and r.
Suppose that d(p,x)&d(q,x) are non-zero, then d(r,x) is zero.   The only
contributor to terms of form x^2y is d(p,x)d(q,x)d(r,y).   So d(r,y) = 0.
Similarly, d(r,z) would be zero.  But then k=0, contradiction.
So at least two of coeffs of x are zero, therefore exactly two, and wlog
	p = d(p,x)x + d(p)
	q = d(q,y)y + d(q)
	r = d(r,z)z + d(r)
To satisfy requirements of x^2 term, d(p,x), etc = �1, with an even number
being -, in order that xyz term has right sign.
	p = �x + d(p)
	q = �y + d(q)
	r = �z + d(r)
Now, the linear terms -2d(p)=d(q)+d(r), etc => d(p)=d(q)=d(r)=d say => d=0.

	So, bearing in mind that the values of p,q,r can be permuted, and that
any two of them can be multiplied by -1, there are

===>>> 24 solutions, being {p,q,r} = {�x,�y,�z} with 0 or 2 minus signs amongst
them.

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Problem B-6.
	Suppose A, B, C, D are n x n matrices with entries in a field F
satisfying the conditions that A T(B) and C T(D) are symmetric and
A T(D) - B T(C) = I, where T denotes transpose and I is the identity matrix.
Prove that T(A) D - T(C) B = I.

Answer A-6.
	Aha!
	     (A B)
	Let  (C D)  be a 2n x 2n matrix, built up of the smaller matrices

in the obvious way.   Inspired by the idea of matrix multiplication for 2 x 2
			  (A B)   (D~ -B~)
matrices, let's compute:  (C D) * (-C~ A~), where ~ denotes transposition.
	   (AD~-BC~ -AB~+BA~)   (I 0)
We derive  (CD~-DC~ -CB~+DA~) = (0 I).   This is deduced for the off-diagonal

terms from the symmetry of AB~ and CD~, and for the diagonal terms from the

identity of AD~-BC~ and its transpose.   So by a familiar idea, we know:
(D~ -B~)   (A B)   (I 0)              (D~A-B~C D~B-B~D)   (I 0)
(-C~ A~) * (C D) = (0 I).   We derive (A~C-C~A A~D-C~B) = (0 I), whence:

as well as the bonus result:	A~C and B~D are symmetric, we also have:

===>>> A~D - C~B = I

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Andrew Buchanan