Conference rusure::math

>	(m'+2)(n'+2) = (m+2)(n+2)

I think the only positive-integer solutions of this are {m', n'} = {m, n}.
We can solve the original problem, however; two solutions are
	{m, n} = {4, 6} and
	{m, n} = {3, 10}.

In each case the number of blue and white tiles are the same.

    	.1 lists the solutions where the number of white tiles equals the
    number of blue ones.   Where the number of white tiles *exceeds* the
    number of blue ones, imagine the blue ones as a perimeter for the white
    ones.   It's only a vey limited number of rectangles for which the
    perimeter can be greater than the area, and this breaks the back of
    the problem.   To be specific, we can show that m, say = 1,2,3 or 4.
    
    	The rest is just symbolic manipulation.
    
    Cheers,
    Andrew.

        Where is it stated that the border width is one?
        
        Dan

Not here, certainly, although I suppose there may be aesthetic arguments 
for a border 1 unit wide. For borders of k units wide, there are:

	k = 1: {m, n} = {3, 10}*
			{4, 6}*
	k = 2: {m, n} = {5, 36}*
			{6, 20}
			{8, 12}
	k = 3; {m, n} = {7, 78}*
			{8, 42}*
			{9, 30}
			{10, 24}*
			{12, 18}
	k = 4; {m, n} = {9, 136}*
			{10, 72}
			{12, 40}
			{16, 24}
	k = 5; {m, n} = {11, 210}*
			{12, 110}*
			{14, 60}*
			{15, 50}
			{18, 35}*
			{20, 30}

etc. The ones marked * are not multiples of smaller ones, i.e. 'primes' in 
some sense.

    Some other solutions:
      {3, 18} and {2, 23}
      {4, 10} and {2, 16}
      {6, 6}  and {2, 14}
    
    I'm pretty sure there aren't any others involving 2.
    The reasoning is simple if you note that the border is always 8 greater
    than the area when one of the dimensions of the rectangle is 2.
    
    The formula eventually comes down to n = (2m+12)/(m-2).  All the
    positive integer solutions of this are listed on the left side above.
    
    Jon

    > That's mn white tiles and 2m+2n+4 blue tiles.  If m' and n' are
    > such that m'n'=2m+2n+4 and 2m'+2n'+4=mn, we have a solution.
    >
    > Observe that 2m+2n+4 = (m+2)(n+2)-mn, and similarly for m' and n'.  So:
    > 
    >         m'n' = (m+2)(n+2) - mn.
    >         mn = (m'+2)(n'+2) - m'n'.
    
    Fine up until here.
    
    > Substitute the latter into the former:
    > 
    >         m'n' = (m+2)(n+2) - (m'+2)(n'+2) + m'n'.
    
    Substituting here doesn't eliminate the need to satisfy at least one of
    the equations above.  This is why the simplified equation:
    
    >        (m'+2)(n'+2) = (m+2)(n+2).
    
    has so many more solutions than the actual problem.  Some solutions are 
    {m',n'} = {m,n} = any pair of positive integers, as mentioned in .1,
    but these aren't the only ones.  For instance, we know that 3*8 == 4*6, 
    so we can choose {1,6} and {2,4} as pairs that satisfy the simplified
    equation, but which don't satisfy the pair of equations at the top.
    
    To see why there are so many more solutions after the substitution, we
    need only plug in values like {1,2} and {1,2} (clearly not a solution
    to the original problem).  The original pair of equations becomes
    
    	 2 = m'n' = (m+2)(n+2) - mn     = 3*4 - 2 = 10
    and	 2 = mn   = (m'+2)(n'+2) - m'n' = 3*4 - 2 = 10
    
    which are clearly both wrong, but when the right hand side (call it
    "rhs") of the bottom "equation" (which equals 10) is substituted for the
    left hand side of the bottom "equation" (mn, which equals 2) in the
    top "equation", you get
    
        2 = m'n'  = (m+2)(n+2) - rhs    = 3*4 - 10 = 2
    
    The problem is that we tend to think we're eliminating an equation from
    a system of equations when we substitute, when actually we're not.  We
    have to keep the equation we used for the substitution in the system of
    equations to be satisfied.  
    
    What we usually do when we use substitution is to simplify one equation
    in a system of equations to a point where the left hand side is a
    single variable, then we substitute the rhs of the equation for that
    variable in every other equation in the system, so that the variable
    now only appears in the left hand side of a single equation.  That
    equation then becomes a "definition" of the value of that variable in
    terms of the other variables, and we forget about that equation and
    that variable until we've solved for the other variables.  We haven't
    eliminated the need to satisfy that "definition" equation, we just tend
    not to look at it as part of the problem anymore.  Satisfying it has
    become a trivial matter, once we have solved the other equations for
    the other variables.
    
    I'll post what I think is a complete list of answers to this problem in
    my next reply.
    
    Jon

    Take the equations presented in .1 (m', n', m and n must be positive
    integers):
    
    (1)   m'n' = 2m + 2n + 4
    (2)   mn = 2m' + 2n' + 4
    
    divide both sides of (1) by m' and substitute:
    
    (1.1)    n' = (2m+2n+4)/m'
    (2.1)    mn = 2m' + 2(2m+2n+4)/m' + 4
    
    We can leave the solution of 1.1 for the end and work on 2.1
    
    (2.2)    (m - 4/m')n = (2m'� + 4m + 8 + 4m')/m'
    
    (2.3)              n = (4m + 8 + 4m' + 2m'�)/(m'm - 4)
    
    Although we could continue this general formula by setting z=m'm-4
    and substituting to get rid of the m, it gets very messy and you still
    end up with lots of m' terms, so let's just substitute some small
    integers for m' instead.
    
    Let m' = 1.  Then we get
    
              n = (4m+14)/(m-4)
    substituting z = m-4, you get
    
              n = (4z+30)/z
    which has integer solutions for z = 1,2,3,5,6,10,15 and 30.
    
    These yield the following pairs for {m,n}:
    {5,34}, {6,19}, {7,14}, {9,10}, {10,9}, {14,7}, {19,6} and {34,5}, with
    the repetition being due to the symmetry of the original equations.
    
    Going back to 1.1 to get n', we get the following solutions:
    
    {1,82} & {5,34}
    {1,54} & {6,19}
    {1,46} & {7,14}
    {1,42} & {9,10}
    
    
    Going back to (2.3) and letting m' = 2, then 3, then 4, then 5  (3 and
    5 get a little hairy, with z=m'm-4 and modulo's and such), you get the
    following solutions, which have all been mentioned in previous notes:
    
    m'=2   {2,23} & {3,18}  (note .5)
           {2,16} & {4,10}  (note .5)
           {2,14} & {6,6}   (note .5)
    
    m'=3   {3,18} & {2,23}  (repeat of above)
           {3,10} & {3,10}  (note .1)
    
    m'=4   {4,10} & {2,16}  (repeat of above)
           {4,6}  & {4,6}   (note .1)
    
    m'=5   {5,34} & {1,82}  (repeat)
    
    It seems that continuing in this way won't give us anything more than
    the pairs already listed, so I'm pretty sure this is an exhaustive list
    of solutions.  That means there are 9 unique solutions for border = 1.
    
    Jon

    Re .6:
    
    I didn't mean to imply the substituion yielded a complete result.  I
    should have said that .0 was just a start.  I saved the Usenet posting,
    diddled with a bit, and eventually just dropped the file in here.
    
    
    Re .7:
    
    Hmm, I'm not comfortable that there aren't more solutions.  Can anybody
    prove there aren't?
    
    
    				-- edp

>    Re .7:
>    
>    Hmm, I'm not comfortable that there aren't more solutions.  Can anybody
>    prove there aren't?

	It's clear.   If we have A white tiles, a blue tiles, with A >= a.
We can state:

	A = MN = 2m + 2n + 2
	a = mn = 2M + 2N + 2

this will get us MN + mn = (m+2)(n+2).

	Use the fact that A >= a to get

	2mn =< (m+2)(n+2), and since wlog m =< n, we can derive m < 5 from
this.

	I think the most useful approach for the rest is to eliminate n and
get:
	(mN-4)(mN-4) = 2m� + 4m� + 8m + 16

which we can solve for m = 1,2,3 & 4 to get the 9 solutions described.

	The only other curiosity I can see to mention is that *two* of these
solutions have a = 46, with A = 54 or 98.

Regards,
Andrew.