In the above diagram, we note the following dimensions:

Length = 720
Width = 132
Height = 85

These three dimensions are all integers. So far, there are no problems finding our “all integers” solution.

   We can use the Pythagorean Theorem to calculate the external diagonals involved since these diagonals are all equivalent to the hypotenuse of a right triangle.

For the outside surface at the left end we have:
Hypotenuse (the diagonal) = Sqrt(Heigth^2 + Width^2)
Using our example gives:
Diagonal = Sqrt(85^2 + 132^2)
= 157

A similar calculation for the front, right surface yields:
Hypotenuse (the diagonal) = Sqrt(Height^2 + Length^2)
Diagonal = Sqrt( 85^2 + 720^2)
= 725

And finally we can find the diagonal across the top surface:
Hypotenuse (the diagonal) = Sqrt(Width^2 + Length^2)
Diagonal = Sqrt(132^2 + 720^2)
= 732

   So far we have integer length-width-height dimensions for the brick and all of its external diagonals. For most brick dimensions the external diagonals will not be integers. However, an infinite number of “bricks” exist that do have integer external diagonals. Some examples include:

Height   Width    Length
85        132       720
117        44       240
231       160       792
275       240       252
351       132       720

As an exercise in using the Pythagorean Theorem, you may want to use these combinations to verify that they all have integer external diagonals.



   The internal diagonal that runs from any given corner through the center of the rectangular solid and terminates at the far corner can also be calculated using the Pythagorean Theorem. One combination for the two sides to this triangle would be “height” and “top-surface diagonal”. In our example we would use the height of 85 and the top surface diagonal which we previously calculated and found to be 732. (Note: Other dimension/diagonal combinations are also valid. Alternately, you could calculate the size of the center diagonal using:
Center Diagonal = Sqrt( Height^2 + Width^2 + Length^2).)

The calculation here would be:
Hypotenuse (center diagonal) = Sqrt(Height^2 + Top-surface-diagonal^2)
Center Diagonal = Sqrt (85^2 + 732^2)
Note that Sqrt(85^2 + 132^2 + 720^2) will produce the same result.

   Unfortunately this ends up equaling 736.91858+ which is not an integer. It starts getting worse as none of the other integer dimension examples (shown above) work either. In fact there is no known combination of dimensions that work. On the other hand, no one has proved that there are no solution combinations. Thus it is an unsolved problem in mathematics.



   If you could find a dimension combination such that all diagonals are integers, including the central diagonal, the world might reward you with fame, gifts, money, etc. for solving the unsolved problem. (This might be a “slight exaggeration”, but at least your name would be recorded as “the solver”). In this pursuit you might try writing a computer program to try a large number of combinations to see if a solution could be found. A possible algorithm might be:

For (Length equals 1 to billions - or more)      //  Try a whole bunch of possible lengths
   For (Width equals 1 to billions – or more)    //  Same for width
      For (Height equals 1 to billions, etc)     //  Hey, just do more of the same
             Do all the length^2, height^2, width^2, Sqrt() calculations
             reviewed previously to see if anything works.
      Repeat this loop for a whole bunch of possible “heights”
   Repeat this loop for a whole bunch of possible “widths”
Repeat for a whole bunch of possible “lengths”

   Technically, if a solution exists, this algorithm would find it. In reality, the number of calculations required is mind boggling. You and the universe will die of old age before your search gets very far.



   Since you are going to let a computer program try to find a solution, you should at least give it a good set of rules to work with. You still might not find a solution, but you can greatly increase the search area for the dimensions of “the brick”.

   In the inefficient algorithm given above, we just tried “a whole bunch of numbers” and then applied the calculations of the Pythagorean Theorem to see if the diagonals were integers. It is much faster to use algebraic expressions derived from the Pythagorean Theorem to directly generate right triangles where the hypotenuse is also an integer.

   First, we note that all the “brick” examples given earlier had one dimension that was “odd”, while the other two dimensions were “even”. If a solution exists for the integer brick problem, then either this solution is a “primitive” solution, or it is a multiple of a primitive solution. If a primitive solution exists, exactly one of its dimensions will be “odd” while the other two dimensions will be “even”. (If all three dimensions were even, then everything could be repeatedly divided by 2 until one of the dimensions became odd. There can’t be two sides that are “odd” as the “Y” side of an integer Pythagorean Triangle must be even. See calculations below.)

If we represent the sides of a right triangle by the variables “X”, “Y”, and “Z” (where “Z” is the hypotenuse), then right, integer triangles can be generated using the following equations:
X = (P^2 – Q^2)*K        (We will let “X” = an odd number. Alternately:  X = (P-Q)(P+Q)K)
Y = 2PQK                     (Since either P or Q must be even, Y is always divisible by 4)
Z = (P^2 + Q^2)K          (We won’t use this equation in the actual computer algorithm)
P, Q, and K can be any integers (also perform multiplications for adjacent letters P, Q, K)

For example, if we let P = 2, Q = 1, and K = 1, then we get X = 3, Y = 4, and Z = 5. This is the smallest possible integer right triangle.

   We note that the “X” term above has three components: (P-Q), (P+Q), and K. If we are going to generate “odd” numbers for our trial bricks, then each of these components must be odd. For example, if we want to generate a trial brick where the odd side is equal to 85, then the 85 is separated into three components whose product equals 85. Here, the only two possibilities are 1 times 5 times 17 and 1 times 1 times 85. (These triplets can be in any order.) We will use these triplets to solve for “P”, “Q”, and “K”. In turn, we will use the results to generate the 85, 132, 720 example given earlier that “almost” solves the integer brick problem.

   In the multiplication of 1 x 5 x 17 = 85, the 1, 5, and 17 factors can be in an order. However, the terms (P-Q)(P+Q)K require (P+Q) to be larger than (P-Q). Thus, for our multiplication, we only use permutations where the second term is larger than the first. We now have:
1 x 5 x 17 = 85
1 x 17 x 5 = 85
5 x 17 x 1 = 85

Each of these can be solved to get a “P”, “Q”, “K” triplet

If we let (P-Q) = 1 and (P+Q) = 5, then we get P = 3, Q = 2, and K = 17
Substituting these in Y = 2PQK yields Y = 204   (In practice, this result won’t be useful)

If we let (P-Q) = 1 and (P+Q) = 17, then we get P = 9, Q = 8, and K = 5
Substituting these in Y = 2PQK yields Y = 720  (One of the sides in our brick)

Finally, if we let (P-Q) = 5 and (P+Q) = 17, then we get P = 11 , Q = 6, and K = 1
Substituting these in Y = 2PQK yields Y = 132  (The other side that we used in the brick)

The only other valid multiplication is:
1 x 85 x 1 = 85

If we let (P-Q) = 1 and (P+Q) = 85, then we get P = 43, Q = 42, and K = 1
Substituting these in Y = 2PQK yields Y = 3612.

Thus we have found the four right triangles that can have one side equal to 85. The other side can equal 132, 204, 720, or 3612. (These are the only four. There are no others.)

   These four results are then tried in various combinations for trial sides of the integer brick. You can reduce the trial combinations a great deal further by observing that if a solution exists, the “top-surface” diagonal would be one of the members of the above trial list.

   In the above example, there are only 4 divisors of 85 (1, 5, 17, and 85). For other odd “X” sides, there may be many possible triplets. (e.g. If the odd number is a product of many small odd prime numbers.) Our more efficient algorithm thus becomes:

For “odd side” equals 3, 5, 7, etc.
  Generate all possible triplets that can be multiplied to produce the odd number
  For each valid permutation, solve for P, Q, K and the second side of a right triangle
    Make a list of all of these trial “Side 2’s”
    Search the list to see if any combination yields a solution
Repeat the above for the next odd number.

   Finally, we have not mentioned an efficient way to find the “triplet” trial divisors. You could of course try dividing the “odd side” by 3, 5, 7, etc. up to Sqrt(odd number). This is not efficient.

A much faster way of finding trial divisors for the “triplets” can be found using a “sieve” algorithm. In the table below, note where the “1’s” show up.

Trial       < - - - - Trial “Odd Numbers” - - - - >
Divs.  3  5  7  9 11 13 15 17 19 21 23 25 27 29 31 33 35..
       - - - - - - - - - - - - - - - - - - - - - - - - -
  3    1  0  0  1  0  0  1  0  0  1  0  0  1  0  0  1  0
  5    0  1  0  0  0  0  1  0  0  0  0  1  0  0  0  0  1
  7    0  0  1  0  0  0  0  0  0  1  0  0  0  0  0  0  1
  9    0  0  0  1  0  0  0  0  0  0  0  0  1  0  0  0  0
Etc.

   The top row lists the trial odd numbers that will be used for the odd side of the brick. The left edge lists all odd numbers which potentially might be evenly divisible into the trial odd number. We know that 3 can be divided by 3. Then we simply go three columns to the right of this number to find another trial odd number that is also divisible by 3. We can continue going right 3 columns at a time as far as needed.

   A similar process works for Trial Divisor = 5. We just mark a “1” in every 5th column. The process can be extended for as many rows as needed. If you use this “sieve” method to find trial divisors, all you have to do is start a list for each column. Then process each row by adding a new divisor to the appropriate list each time a “1” would be encountered. (For computer science students, think “Link List”). Periodically, a large block of the table can be generated which will generate lists for large numbers of trial “Odd Numbers”.



   The author wrote a computer program based on the algorithm outlined above. (Actually, a couple of different versions were used.) All odd numbers out to 21 billion were tried. (The search was stopped at this point as there was minimal prospect of finding a solution at larger numbers.) No solutions were found. The program was run on a 3GHz Pentium 4 under Windows XP using the lcc-win32 C compiler system. (http://www.cs.virginia.edu/~lcc-win32/) It took less than 1 second to search all odd combinations out to 200,000 and about two weeks to reach 20 billion. The possibility exists that the computer program was faulty, and a solution actually exists somewhere in this range. Viewers are encouraged to launch their own search efforts in case a solution actually exists.

   At “odd side” = 9,704,539,845 (3x3x5x7x11x13x17x19x23x29) the program generated 16,402 right triangles that shared the same common odd edge. Even with this many candidates for the other dimensions of the brick, there were no combinations that produced a solution.

   The program could theoretically continue running out to 10^12 (One trillion). All entries in the following list of “Likely suspects” were also tried – again no solutions. The number of right triangles that share the “Candidate” (odd side for the brick) can be found by subtracting MinDiv4 from MaxNot4 and adding 1. As a note regarding numerical precision problems, the other side (and the hypotenuse) of some of these triangles has a length exceeding one million light years as measured in inches. Then of course these numbers are squared for the X^2 + Y^2 = Z^2 calculations.

Candidate                       MinDiv4    MaxNot4
15x3x5x7x11x13x17x19x23x29       18,162     45,498
17x3x5x7x11x13x17x19x23x29       20,893     37,294
19x3x5x7x11x13x17x19x23x29       20,904     37,305
21x3x5x7x11x13x17x19x23x29       18,171     45,507
23x3x5x7x11x13x17x19x23x29       20,895     37,296
25x3x5x7x11x13x17x19x23x29       19,257     42,219
27x3x5x7x11x13x17x19x23x29       17,622     47,145
29x3x5x7x11x13x17x19x23x29       20,904     37,305
31x3x5x7x11x13x17x19x23x29       17,607     47,130
33x3x5x7x11x13x17x19x23x29       18,162     45,498
35x3x5x7x11x13x17x19x23x29       18,171     45,507
37x3x5x7x11x13x17x19x23x29       17,622     47,145
39x3x5x7x11x13x17x19x23x29       18,162     45,498
41x3x5x7x11x13x17x19x23x29       17,610     47,133
43x3x5x7x11x13x17x19x23x29       17,622     47,145
45x3x5x7x11x13x17x19x23x29       15,435     53,706
47x3x5x7x11x13x17x19x23x29       17,607     47,130
49x3x5x7x11x13x17x19x23x29       19,251     42,213
51x3x5x7x11x13x17x19x23x29       18,173     45,509
53x3x5x7x11x13x17x19x23x29       17,622     47,145
55x3x5x7x11x13x17x19x23x29       18,162     45,498
57x3x5x7x11x13x17x19x23x29       18,162     45,498
59x3x5x7x11x13x17x19x23x29       17,622     47,145
61x3x5x7x11x13x17x19x23x29       17,622     47,145
3x3x5x7x11x13x17x19x23x29x31     12,711     61,917
5x3x5x7x11x13x17x19x23x29x31     12,711     61,917
7x3x5x7x11x13x17x19x23x29x31     12,684     61,890
9x3x5x7x11x13x17x19x23x29x31      7,770     76,659
3x3x5x7x11x13x17x19x23x29x33     15,543     53,706
5x3x5x7x11x13x17x19x23x29x33     13,614     59,175
7x3x5x7x11x13x17x19x23x29x33     13,605     59,166
9x3x5x7x11x13x17x19x23x29x33     12,696     61,902
3x3x5x7x11x13x17x19x23x29x35     13,605     59,166
5x3x5x7x11x13x17x19x23x29x35     15,426     53,697
7x3x5x7x11x13x17x19x23x29x35     15,441     53,712
3x3x5x7x11x13x17x19x23x29x37     12,696     61,902
5x3x5x7x11x13x17x19x23x29x37     12,696     61,902
7x3x5x7x11x13x17x19x23x29x37     12,705     61,911




   An external solution consists of integer edges and integer external diagonals. One version of the program generated a disk file that contained all the external solutions with the 3 sides less than 1,000,000. There were 9,011 of these which were subsequently read into a spreadsheet so that they could be “played with”. (Click here if you would like to view/copy this list.) Various experiments were tried using Greatest Common Divisor, Mod 2 arithmetic, Mod 3 arithmetic, etc. (e.g. The product of the 3 dimensions is always divisible by 95,040.) No reliable pattern was recognized that would prevent a solution that would also include an integer internal diagonal.

   If you use one inch as a unit of measure, then at 21 billion for the “odd side”, if a solution exists, this odd side will have a dimension in excess of 331,439 miles. It’s looking grim that any solution exists. Unfortunately, this still doesn’t answer the question of the existence of a solution to the integer brick problem. Viewers/readers are encouraged to pursue further research.



For additional information about the Euler Brick Problem please see:
Eric Weisstein’s "Perfect Cuboid"  web page:      http://mathworld.wolfram.com/PerfectCuboid.html
and
Fred Curtis’s "Primitive Euler Bricks"  web page:     http://f2.org/maths/peb.html



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