Here is the problem as stated by Heinrich Dörrie in his book 100 Great Problems of Elementary Mathematics: Their History and Solution translated by David Antin:
The sun god had a herd of cattle consisting of bulls and cows, one part of which was white, a second black, a third spotted, and a fourth brown.This is (almost) the problem as stated in the first 30 lines of the Greek poem. Symbolically, this can be restated in the following set of seven equations in eight unknowns (symbols defined below):Among the bulls, the number of white ones was one half plus one third the number of the black greater than the brown; the number of the black, one quarter plus one fifth the number of the spotted greater than the brown; the number of the spotted, one sixth plus one seventh the number of the white greater than the brown.
Among the cows, the number of white ones was one third plus one quarter of the total black cattle; the number of the black, one quarter plus one fifth the total of the spotted cattle; the number of the spotted, one fifth plus one sixth the total of the brown cattle; the number of the brown, one sixth plus one seventh the total of the white cattle.
What was the composition of the herd?
(1) A-C = 5/6 B (4) a = 7/12 (B+b)
(2) B-C = 9/20 D (5) b = 9/20 (D+d)
(3) D-C = 13/42 A (6) c = 13/42 (A+a)
These 7 equations can be solved (as outlined in Dörrie to obtain:
bulls cows bulls cows
---------- --------- ----- ----
White 10,366,482 7,206,360 A a
Black 7,460,514 4,893,246 B b
Brown 4,149,387 5,439,213 C c
Spotted 7,358,060 3,515,820 D d
Total 50,389,082
as the solution in smallest terms. Any integer multiple of this is also a
solution. This solution multiplied by 80 was provided with the Greek
manuscript in 1773.The last 14 lines of the poem imposes the additional constraints that (8) the white and black bulls together form a square and that (9) the brown and spotted bulls together form a triangle. Symbolically, this can be expressed as:
(8) A" + B" = Z2 (9) C" + D" = 1/2 n(n+1)
where Z and n are integers.Throughout the 19th century, mathematicians attempted to obtain reasonable solutions by relaxing the two additional constraints or ignoring them altogether. Wurm completed solved the problem in 1830 by treating the sum of the black and white bulls as a square figure, that is, the product of two numbers. Another group attested that Archimedes had no knowledge of triangular numbers and thus solved the problem ignoring this condition.
Subject to both these conditions, the smallest solution to the problem was computed by Dr. A. Amthor in 1880 to be 7.766 x 10206544 cattle. To obtain this the Fermat equation: X2-DY2=1 with D=4729494 had to be solved for integer values of X and Y.
Starting in 1889, A. H. Bell with two associatates spent nearly four years working out the solution to 30 digits of accuracy on the left and 12 on the right without finding the intermediate ones. They solved the Fermat equation X2-DY2=1 with D=410286423278424.
Total cattle: 7 760 271 406 486 818 269 530 232 833 209 - with 68,834 intervening periods of three digits - 719 455 081 800.
G. Herdmann and K. B. Mollweide are both on record for stating that Karl Gauss completely solved the cattle problem in about 1830. This had been highly disputed ever since. Could Gauss have solved the cattle problem exactly?
This was the state of the problem when I was introduced to it in February 1974. As I was assimilating the solution in Dörrie's book, I was appalled by a seeming logical error. In this book, before the two constraints, Dörrie presents a poetic translation of the Greek poem as found here.
Try as I would I could not get his original statement of the problem out of this poem. Instead of his equations 1-3, I obtained the following:
(1') (A+a)-(C+c) = 5/6 (B+b)
(2') (B+b)-(C+c) = 9/20(D+d)
(3') (D+d)-(C+c) = 13/42(A+a)
with equations 4-7 remaining the same.Although this was purely a sematic problem, I proceeded to solve the cattle problem as I understood it because of the tremendous change it could elicit in its solution. Using simple algebra, one can obtain the following solution in smallest terms:
bulls cows total
---------- --------- -----
White 12,915 9,345 22,260
Black 8,910 7,110 16,020
Brown 2,020 6,890 8,910
Spotted 12,533 3,267 15,800
Total 62,990
which is three orders of magnitude smaller than the historic solution!I then proceeded to solve the problem subject to the two additional constraints: (8) and (9).
A + B = 21825 = 32×52×97 C + D = 14553 = 33×72×11From this it can be seen that A + B can be made a square by multiplying it by 97. Thus all but (9) can be simply satisfied.
To also comply with constraint (9) we must solve:
97(C + D)Z2 = 1/2 n(n + 1) where Z and N are integerss
1411641 Z2 = 1/2 n(n + 1)
n2 + n = 2823282 Z2
4n2+4n+1 = 4x2823282Z2 + 1
(2n + 1)2 = 2823282(2Z)2 + 1
X2 - DY2 = 1 X= 2n + 1; Y = 2Z; D=2823282
which is equivalent to solving for smallest integer terms. The smallest
solution to this Fermat equation is: X = 1049067577601 and Y = 624347440
hence: n = 524533788800 and Z = 312173720.Thus multiplying our initial solution by 97Z2 which is 945288585148792480 we obtain the following complete solution:
white bulls 122 084 020 771 966 548 792 000
cows 88 337 218 282 154 657 256 000
cattle 210 421 239 054 121 206 048 000
black bulls 84 225 212 936 757 409 968 000
cows 67 210 018 404 079 145 328 000
cattle 151 435 231 340 836 555 296 000
brown bulls 19 094 829 420 005 608 096 000
cows 65 130 383 516 751 801 872 000
cattle 84 225 212 936 757 409 968 000
spotted bulls 118 473 018 376 698 161 518 400
cows 30 882 578 076 811 050 321 600
cattle 149 355 596 453 509 211 840 000
total cattle 595 437 279 785 224 383 152 000
This many cattle could not graze on the island of Sicily, but if each
were given a three meter cube, the cattle would fit within our sun!This solution leaves three remaining unanswered questions:
References