**Problem 2000-4/A** (Open problem, by Frits Beukers)

*
Let a, b, c be integers such that the symmetric matrix
0 a b
a 0 c
b c 0
has three integer eigenvalues. Prove, or give a counter example to, the following statement: either abc = 0 or
(a*

__Solution:__

The proposed statement is false. Here follows how I found a counter example.

The characteristic polynomial is -x^{3} + x (a^{2} + b^{2} + c^{2}) + 2abc =
-(x - x_{1})(x - x_{2})(x - x_{3}), where x_{1}, x_{2}, x_{3} are its zeroes.

So x_{1} + x_{2} + x_{3} = 0, x_{1}x_{2} + x_{1}x_{3} +
x_{2}x_{3} = -(a^{2} + b^{2} + c^{2}),
x_{1}x_{2}x_{3} = 2abc.

I wrote and ran a Pascal program that seeks integers x1, x2, x3, a, b, c (in the range from -200 to 200)
that satisfy these three equations.

I found x1 = -190, x2 = 55, x3 = 135, a = -125, b = -99, c = -57.

**Problem 23, september 2001** (Open problem)

*Does there exist, for any positive integer N, a positive integer k whose square x is the sum of m ≥ N distinct
powers of 3?
*

__Partial solution:__

To explore the case, I wrote a Pascal program to find such k for m up to 20. I found for m = 20:

1876138^{2} = 1 + 3 + 3^{2} + 3^{4} + 3^{6} + 3^{7}
+ 3^{8} + 3^{9} + 3^{10} + 3^{13} + 3^{14} + 3^{16}
+ 3^{17} + 3^{18} + 3^{19} + 3^{20} + 3^{22} + 3^{23}
+ 3^{25} + 3^{26}.

Furthermore I found: 3 solutions for m = 19, 7 solutions for m = 18, 17 solutions for m = 17, 23 solutions for m = 16,
18 solutions for m = 15, 24 solutions for m = 14, 90 solutions for m = 13, 47 solutions for m = 12,
27 solutions for m = 11, 89 solutions for m = 10, 63 solutions for m = 9, 48 solutions for m = 8,
18 solutions for m = 7, 0 solutions for m = 6, 187 solutions for m = 5, 19 solutions for m = 4,
0 solutions for m = 3, 13 solutions for m = 2, 14 solutions for m = 1.

These are all 708 solutions that can be found with a home computer.

If we ignore the solutions x that can be divided by 9
and thus reduced to x/9, we have: 1 solution for m = 20,
2 solutions for m = 19, 5 solutions for m = 18, 11 solutions for m = 17, 15 solutions for m = 16,
9 solutions for m = 15, 9 solutions for m = 14, 31 solutions for m = 13, 17 solutions for m = 12,
9 solutions for m = 11, 22 solutions for m = 10, 14 solutions for m = 9, 10 solutions for m = 8,
4 solutions for m = 7, 0 solutions for m = 6, 28 solutions for m = 5, 2 solutions for m = 4,
0 solutions for m = 3, 1 solutions for m = 2, 1 solutions for m = 1.

So these are the 191 primitive solutions within the capacity of the
computer.

For instance, for m = 4 we have 16^{2} = 1 + 3 + 3^{2} + 3^{5} and
3788^{2} = 1 + 3^{2} + 3^{3} + 3^{15}.

In general, there are 3^{n} distinct squares that are at most 3^{2n}, and 2^{2n} sums of
distinct powers of 3 that are at most 3^{2n}.

So I thought the number of solutions that are at most 3^{2n} must be (4/3)^{n}.
Then for n = 14 we get 56, so 708 and 191 are quite a bit more than expected.

Moreover, I found a solution with m = 20.

Therefore I think the answer to the open problem is 'yes'.

**Star problem september 2003**

* Determine the maximal area of a rectangle that can be covered by six circular disks of unit diameter.*

__Numerical solution:__

The rectangle with side lengths 3√2/2 and 2√2/2 can be partitioned into six squares with side length √2/2 whose circumscribing circular disks have unit diameter and cover this rectangle with area 3.

To find a rectangle with an even greater area, I wrote and ran a Pascal program. The listing is as follows:

program cover;

var m,n:integer;x,y,x1,x2,x3,x4,x5,x6,y1,y2,y3,y4,y5,y6:real; gevonden:boolean;

begin

for m:=0 to 300 do

repeat

x1:=1*sqrt(2)/4+(3*m/10000)*(-1+2*random); y1:=1*sqrt(2)/4+(2*m/10000)*(-1+2*random);

x2:=3*sqrt(2)/4+(3*m/10000)*(-1+2*random); y2:=1*sqrt(2)/4+(2*m/10000)*(-1+2*random);

x3:=5*sqrt(2)/4+(3*m/10000)*(-1+2*random); y3:=1*sqrt(2)/4+(2*m/10000)*(-1+2*random);

x4:=1*sqrt(2)/4+(3*m/10000)*(-1+2*random); y4:=3*sqrt(2)/4+(2*m/10000)*(-1+2*random);

x5:=3*sqrt(2)/4+(3*m/10000)*(-1+2*random); y5:=3*sqrt(2)/4+(2*m/10000)*(-1+2*random);

x6:=5*sqrt(2)/4+(3*m/10000)*(-1+2*random); y6:=3*sqrt(2)/4+(2*m/10000)*(-1+2*random);

n:=0;

repeat

n:=n+1; gevonden:=false;

x:=(3*sqrt(2)/2+2*m/10000)*random; y:=(2*sqrt(2)/2+3*m/10000)*random;

if ((x-x1)*(x-x1) + (y-y1)*(y-y1)) > 1/4 then

if ((x-x2)*(x-x2) + (y-y2)*(y-y2)) > 1/4 then

if ((x-x3)*(x-x3) + (y-y3)*(y-y3)) > 1/4 then

if ((x-x4)*(x-x4) + (y-y4)*(y-y4)) > 1/4 then

if ((x-x5)*(x-x5) + (y-y5)*(y-y5)) > 1/4 then

if ((x-x6)*(x-x6) + (y-y6)*(y-y6)) > 1/4 then gevonden:=true;

until gevonden or (n=1000000);

if not gevonden then

begin

writeln(m:5);

writeln(x1:13:10,y1:13:10,' ',x2:13:10,y2:13:10,' ',x3:13:10,y3:13:10);

writeln(x4:13:10,y4:13:10,' ',x5:13:10,y5:13:10,' ',x6:13:10,y6:13:10);

writeln; readln

end

until not gevonden

end.

This yields a rectangle with (rounded) side lengths 2.1233 and 1.4172 and (rounded) area 3.0092.

Then I tried to extend the rectangle further in the direction (1,0).

This yields a rectangle with (rounded) side lengths 2.1235 and 1.4172 and (rounded) area 3.0095.

The (rounded) centers of six covering circles are:

(0.3560,1.0614), (1.0638,1.0623), (1.7703,1.0596),

(0.3508,0.3525), (1.0589,0.3523), (1.7683,0.3518).

**Star problem september 2002**

* Let x=(x1,x2,...,xn) be a random variable such that for each i holds P(xi=1) = P(xi=-1) = 1/2.
Let v be a vector of unity in R ^{n}.
Calculate the probability that the inner product x.v is smaller than 1.*

__Numerical solution:__

I wrote and ran a Pascal program for n = 1, 2, ... , 6 wherein a random vector of unity and a random vector x as above
are generated, and the inner product x.v calculated, one million times.

I found the probabilities are:

For n = 1 prob = 0.5, for n = 2 prob = 0.75, for n = 3 prob = 0.7916, for n = 4 prob = 0.8028, for n = 5 prob = 0.8073,
for n = 6 prob = 0.8095, for n = 7 prob = 0.8099, for n = 8,9,10 prob = 0.8100.

I think for larger n we always have prob = 0.8100 (rounded in 4 decimals).