Conference rusure::math

     Are you including numbers like 1 = 000001 = (000 + 001)^2
     in the search space?  Or only integers n such that
     100000 <= n <= 999999?
     
     One way to limit the search is to consider only squares of
     numbers <= sqrt(999999) < 1000.  So only examine n^2 for
     1 <= n <= 999 (or 317 <= n <= 999 if leading zeroes don't
     count).  Also the more significant half of the number must
     be even.
     
     Dan

I don't know what you had in mind, but here's one way to approach it.
Answer follows form feed...

Let the 6 (or less?) digit number be X Y where X and Y are each three
digit numbers.  Then we can write the number as X*1000 + Y.
The problem statement gives us that (X + Y)^2 = X*1000 + Y

Solve for X gives:

X^2 + X*(2*Y - 1000) + (Y^2 - Y) = 0

Quadratic formula gives:

	X = 500 - Y +- sqrt( 500^2 -999*Y)

In order to have integer X, the Sqrt( 500^2 - 999*Y) must be an integer.
Call it N.  Then 

	N^2 = 500^2 -999*Y

A little reordering gives:

	(N+500)*(N-500) = -999*Y

But 999 has factors 3*3*3*37.  Therefore either N+500 or N-500 must be
divisible by 37, and there must be three factors of 3 in N+500 and N-500.

So we try all the multiples of 37 between 1 and 1000 (there are 27 such
numbers).  There are two values of N of interest, N = 203 and N = 499.

	(703 = 37*19, 297 = 27*11)
	(999 = 37*27, 1 = 1)

The first gives Y = 19*11 =  209 with solutions X = 88 and 494 (due to +-).
The second gives Y = 1 with solutions X = 998 and X = 0.

Therefore the solutions are:

	998001 (given)
	000001 (not really 6 digits)
	 88209 (ditto)
	494209 - a real second solution!

Dave

    Okay, how about:
    
    	abcdefgh = (abcd+efgh)^2
    Or:
    	abcdefghij = (abcde+fghij)^2
    
    Or:
    	abcdef = (ab+cd+ef)^3
    Or:
    	abcdefghi = (abc+def+ghi)^3
    

     re .3
     
>>     Okay, how about:
       ^^^^
     
     Only "Okay"?  I thought .2 was very good.
     
     Dan

Well done! This is another of the problems from "The Puzzled Programmer" 
and shows once again that a little analysis ahead of time can save a lot of 
computing.

Lynn Yarbrough 

>     Only "Okay"?  I thought .2 was very good.
    
    Sorry.  Dave dispatched that one so easily, I was anxious to see how
    slightly harder forms of the problem would fare.


    Let's consider (abcd+efgh)^2 = abcdefgh.  As in .2, (X+Y)^2 = X*10000+Y,
    so

	X^2 + X*(2*Y - 10000) + (Y^2-Y) = 0,

	X = 5000 - Y +- sqrt ( 5000^2 - 9999*Y )

	N^2 = 5000^2 - 9999*Y

	(N + 5000) * (N - 5000) = -9999 * Y = -3*3*11*101*Y

    We can partition these factors amoung N+5000 and N-5000 in eight ways:

	N+5000 = u,		N-5000 = 3*3*11*101*v
	N+5000 = 3*3*u,		N-5000 = 11*101*v
	N+5000 = 11*u,		N-5000 = 3*3*101*v
	N+5000 = 3*3*11*u,	N-5000 = 101*v
	N+5000 = 101*u,		N-5000 = 3*3*11*v
	N+5000 = 3*3*101*u,	N-5000 = 11*v
	N+5000 = 11*101*u,	N-5000 = 3*3*v
	N+5000 = 3*3*11*101*u,	N-5000 = v

    In any case, we have N = f*u - 5000 = g*v + 5000, where f*g = 9999. 
    The solution to this is N = 9999*z + k, where z is a free variable
    integer, and k satisfies: k == 5000 (mod g), and k == -5000 (mod f).

    Eight applications of Euclid's algorithm and the Chinese remainder theorem
    yield the solutions for N:

	N = 9999*z - 5000*1*1    + 5000*9999*0	
	N = 9999*z - 5000*9*247  + 5000*1111*7	(9*247 = 1 (mod 1111))
	N = 9999*z - 5000*11*248 + 5000*909*8	(11*248 = 1 (mod 909))
	N = 9999*z - 5000*99*50  + 5000*101*50		(etc)
	N = 9999*z - 5000*101*50 + 5000*99*50	
	N = 9999*z - 5000*909*8  + 5000*11*248	
	N = 9999*z - 5000*1111*7 + 5000*9*247	(1111*7 = 1 (mod 9))
	N = 9999*z - 5000*9999*0 + 5000*1*1	

    And now, expanding the above folding multiples of 9999 into the z), we
    simply the expressions for N, and solve for Y = (5000^2 - N^2)/9999.

	N = 9999*z + 4999	Y =    1 - z*(2+9999*z)
	N = 9999*z + 2777	Y = 1729 - z*(2+9999*z)
	N = 9999*z + 2272	Y = 1984 - z*(2+9999*z)
	N = 9999*z +   50	Y = 2500 - z*(2+9999*z)
	N =-9999*z -   50	Y = 2500 - z*(2+9999*z)
	N =-9999*z - 2272	Y = 1984 - z*(2+9999*z)
	N =-9999*z - 2777	Y = 1729 - z*(2+9999*z)
	N =-9999*z - 4999	Y =    1 - z*(2+9999*z)

    (or, just four equations, realizing that N = (+-)(9999*z+...))

    Now X = 5000 - Y +- N.  For convenience, we drop the z stuff (note that
    |z| >= 1 would subtract at least 9997 from Y), and have:

	 N'	Y'	X'	X''
	 4999	0001	9998	0000
	 2777	1729	6048	0494
	 2272	1984	5288	0744
	 0050	2500	2550	2450

    And so we have the eight solutions:

	(0000+0001)^2 = 00000001
	(0494+1729)^2 = 04941729
	(0744+1984)^2 = 07441984
	(2450+2500)^2 = 24502500
	(2550+2500)^2 = 25502500
	(5288+1984)^2 = 52881984
	(6048+1729)^2 = 60481729
	(9998+0001)^2 = 99980001

     re .6
     
     It's too bad that the number 1729 shows up in your
     solutions.  It seems like such a dull, boring number.
     
     Dan

>     It's too bad that the number 1729 shows up in your
>     solutions.  It seems like such a dull, boring number.
    
    Oh, I don't know.  I once rode in a cab with that number...

re .7, .8; In case someone missed the point of this exchange, it's an "in" 
joke of sorts. The mathematician Hardy once visited the m'n Ramanujan in 
the hospital. H commented that the number of the taxi that delivered him 
was 1729. R remarked that that was a very interesting number, the smallest
integer that can be expressed as the sum of two cubes in two distinct ways
(1729 = 10^3+9^3 = 1^3+12^3). Hardy's comment later was that R seemed to 
have a personal aquaintance with each of the positive integers.

Lynn