Popular Science Monthly/Volume 67/November 1905/The Cattle Problem of Archimedes

by Mansfield Merriman

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THE CATTLE PROBLEM OF ARCHIMEDES.

By Professor MANSFIELD MERRIMAN,

LEHIGH UNIVERSITY.

THE sleepy town of Wolfenbüttel in northern Germany is the proud possessor of a library containing about 240,000 books and 10,000 manuscripts, many of the latter being Greek and Latin writings of interest and value. Lessing, the poet and philosopher, was appointed its librarian in 1769, and a few years later he published translations of some of the unique manuscripts with commentaries thereon. One of these was a Greek poem of forty-four lines which states an arithmetical problem that has since attracted much attention on account of the difficulty of its solution and the enormous numbers required to fulfil its conditions. The name of Archimedes appears in the title of the poem, it being said that he sent it in a letter to Eratosthenes, the Cyrean, to be investigated by the mathematicians of Alexandria. Opinions differ as to the truth of this statement, and it may well be doubted if Archimedes was the real author, particularly as no mention of the problem has been found in the writings of the Greek mathematicians.

The following statement of the cattle problem has been abridged from the German translations published by Nesselmann in 1842, and by Krumbiegel in 1880:

Compute, O friend, the number of the cattle of the sun which once grazed upon the plains of Sicily, divided according to color into four herds, one milk-white, one black, one dappled and one yellow. The number of bulls is greater than the number of cows, and the relations between them are as follows:

${\begin{aligned}&{\text{White bulls}}=({\tfrac {1}{2}}+{\tfrac {1}{2}})\ {\text{black bulls}}+{\text{yellow bulls}},\\&{\text{Black bulls}}=({\tfrac {1}{4}}+{\tfrac {1}{5}})\ {\text{dappled bulls}}+{\text{yellow bulls}},\\&{\text{Dappled bulls}}=({\tfrac {1}{6}}+{\tfrac {1}{7}})\ {\text{white bulls}}+{\text{yellow bulls}},\\&{\text{White cows}}=({\tfrac {1}{3}}+{\tfrac {1}{4}})\ {\text{black herd}},\\&{\text{Black cows}}=({\tfrac {1}{4}}+{\tfrac {1}{5}})\ {\text{dappled heard}},\\&{\text{Dappled cows}}=({\tfrac {1}{3}}+{\tfrac {1}{6}})\ {\text{yellow herd}},\\&{\text{Yellow cows}}=({\tfrac {1}{6}}+{\tfrac {1}{7}})\ {\text{white heard}}.\end{aligned}}$

If thou canst give, O friend, the number of each kind of bulls and cows, thou art no novice in numbers, yet can not be regarded as of high skill. Consider, however, the following additional relations between the bulls of the sun:

${\begin{aligned}&{\text{White bulls}}+{\text{black bulls}}={\text{a square number}},\\&{\text{Dappled bulls}}+{\text{yellow bulls}}={\text{a triangular number}}.\end{aligned}}$

If thou hast, computed these also, O friend, and found the total number of cattle, then exult as a conqueror, for thou hast proved thyself most skilled in numbers.

The Wolfenbüttel manuscript has an appendix, also in Greek, giving numbers in answer to the problem, the total number of cattle being stated as 4,031,126,560, but the results satisfy only the first seven conditions. Lessing also gives results computed by Leiste, a clergyman of Wolfenbüttel, whose solution satisfies these seven conditions and likewise the eighth one. In 1821 J. and K. L. Struve published a critical and mathematical discussion, which was followed in 1828 by another from G. Hermann. The latter makes the interesting remark that Gauss had arrived at a complete solution; but, if so, no further information regarding it has been obtained. Many other attempts at solution were made, but the large numbers required to satisfy the nine conditions discouraged many investigators. Some critics thought that the original problem of Archimedes included only the first seven conditions and that the two others had been added by a later writer. It is, of course, clearly seen that the exercise as stated includes two problems, the first to find integral numbers that satisfy the first seven conditions, and the second to find integral numbers that satisfy all the nine conditions. As the poem says, the first problem may be solved by those of moderate proficiency in numbers, while the second can only be done by those of the highest skill.

The first problem is an easy one for a boy in the high school. Let W, B, D, Y represent the number of white, black, dappled and yellow bulls and let w, b, d, y represent the number of white, black, dappled and yellow cows. The seven conditions then give the seven equations:

${\begin{aligned}W&={\tfrac {5}{6}}\;B+Y,&(1)&\qquad \qquad \qquad &w&={\tfrac {7}{12}}(B+b),&(4)\\B&={\tfrac {9}{26}}D+Y,&(2)&&b&={\tfrac {9}{26}}(D+d),&(5)\\D&={\tfrac {12}{42}}W+Y,&(3)&&d&={\tfrac {11}{36}}(Y+y),&(6)\\&&&&y&={\tfrac {13}{42}}(W+w),&(7)\\\end{aligned}}$

and these contain eight unknown quantities. The problem, therefore, is of the kind called indeterminate, for many sets of numbers may be found to satisfy the seven equations. That set having the smallest numbers is the one required, for any other set may be found by multiplying these numbers by the same integer. If B and W are eliminated from equations (1), (2), (3), there will be found the single equation $891\,D=1{,}580\,Y,$ and hence $T=891$ and $D=1{,}580$ are the smallest integral numbers satisfying it; from these are found $B=1{,}602$ and $W=2{,}226.$ These values, before insertion in equations (4) to (7), are to be multiplied by a factor m, the value of which is later to be determined, so that the number of cows in each herd shall be an integer. Proceeding with the elimination, the values of w, b, d, y are deduced in terms of m, and it is then seen that 4,657 is the least value of m which will make the results integers. It is thus easily found that

${\begin{aligned}B&=\;\;7{,}460{,}514,\qquad \qquad &b&=4{,}893{,}246,\\W&=10{,}366{,}482,\qquad \qquad &w&=7{,}206{,}360,\\D&=\;\;7{,}358{,}060,\qquad \qquad &d&=3{,}515{,}820,\\Y&=\;\;4{,}149{,}387,\qquad \qquad &y&=5{,}439{,}213,\\\end{aligned}}$

are the least numbers satisfying the conditions of the first problem. The total number of cattle is 50,389,082, not too great to graze upon the island Sicily, the area of which is about 7,000,000 acres. If the above numbers are multiplied by 80 they give the results stated in the appendix to the Wolfenbüttel manuscript.

The second or complete problem includes the determination of numbers which satisfy not only equations (1) to (7), but also

W	$+$	B	$=$	a square number,	(8)
D	$+$	Y	$=$	a triangular number,	(9)

and this is to be done by finding an integer N to multiply into each of the results of the first problem. Since W + B is 17,826,996 and D + Y is 11,507,447, these equations become

17,826,996	N	=	a square number,
11,507,447	N	=	a triangular number.

A number N that will satisfy one of these conditions can be found without difficulty, but to determine one that will satisfy both is a task requiring an enormous amount of labor and patience. In fact, this required number N has never been completely computed.

It has been claimed by some critics that the ninth condition should be rejected altogether, for they asserted that there is no evidence that Archimedes or the Greek mathematicians had the idea of a triangular number. On this hypothesis the solution is easy. Since W + B is 17,826,966 N or 4 $\times$ 4,456,749 N, and since 4,456,749 contains no number that is a perfect square, it is plain that N must be 4,456,749. Accordingly, each of the numbers found in the first solution must be multiplied by 4,456,749 in order to satisfy equations (1) to (8) inclusive; the number W + B is then 79,450,446,596,004, which is a perfect square, but the number D + Y is 51,285,802,909,803, which is not a triangular number. This solution is identical with that of Leiste as published by Lessing in 1773.

For the benefit of those who are neither novices nor of high skill in numbers, it is now time to explain what is meant by a triangular number. The number 10 is triangular because ten dots can be arranged in rows in the form of a triangle, there being one dot in the first row, two in the second, three in the third and four in the fourth. The next higher triangular number is 15 and the next 21, and in general ¹⁄₂ n (n + 1) is a triangular number whenever n is an integer, n being the number of rows parallel to one side of the triangle. The proof that 51,285,802,909,803 is not a triangular number consists in equating it to 1/2 n (n $+$ 1) and computing the value of n by the solution of the quadratic equation; this value of n is found to be not an integer.

Some mathematicians who desired to solve the cattle problem have claimed that W $+$ B is not required to be a perfect square, because the statement of the eighth condition in the Greek manuscript does not use the term square number, but mentions 'a square figure.' Since the length of a bovine animal is greater than its breadth, it was maintained by Wurm, about 1830, that W $+$ B is required to be a rectangular number, that is, a number having two factors. On this hypothesis he made a solution which gave the total number of cattle as 5,916,837,175,686, and the number of white and black bulls as 2,093,299,351,328, which has the factors 704,538 $\times$ 2,971,166, as well as many others, while the number of dappled and yellow bulls is 1,351,238,949,081, which is a triangular number, so that these bulls could be arranged in a triangle with 3,287,843 rows.

The consensus of opinion regarding the eighth and ninth conditions is expressed, however, in the statement of the problem as given above, namely, that the terms 'square figure' and 'triangular figure' should be understood to mean square number and triangular number. Since 51,285,802,909,803 is the number of dappled and yellow bulls which results from a solution that satisfies conditions (1) to (8) inclusive, it is plain that the ninth condition may be expressed by

51,285,802,909,803, x² $=$ 1/2 n (n $+$ 1),

in which $x$ and n are to be integers. When $x^{2}$ has been found, each of the numbers of the first solution is to be multiplied by 4,456,749 $x^{2}$, in order to give the number of bulls and cows in each herd, satisfying the nine imposed conditions.

These numbers were readily seen to be so great that the island of Sicily could not contain all the cattle, as the problem seems to demand. This requirement, however, was understood to be only figurative, and mathematicians agreed that the numbers, though very large, could be found, but that no useful purpose would be attained by computing them. Thus the question rested until 1880, when Amthor undertook to determine how many figures were required to express one of the numbers. His lengthy investigation demonstrates that 206,545 figures are needed to express the total number of cattle. He further computed that 766 are the first three figures of this number, so that 766 $\times$ 10^206,542 is the approximate number of cattle. This is an enormous number, and it is easy to show that a sphere having the diameter of the milky way, across which light takes ten thousand years to travel, could contain only a part of this great number of animals, even if the size of each is that of the smallest bacterium.

It would be thought that, after this investigation by Amthor, the question of solving the cattle problem would have been finally dropped, but such was not the case. The certainty that numbers could be found satisfying all nine conditions existed, and until they had been actually computed the challenge of the author of the problem still remained open. The way to solve it was well understood from the theory of indeterminate analysis. Let the preceding equation be multiplied by 8 and unity added to each member, and let 2 n $+$ 1 be called y; then it reduces to

$y^{2}-410,280,423,278,424x^{2}=1$

which is of the form $y^{2}-Ax^{2}=1$, and it is known that when A is an integer there can always be found integral values of y and x which satisfy the equation. The method of obtaining such values of x and y can not well be explained here, but such a method was devised many years ago by Pell and by Fermat, and it is well known to those skilled in higher arithmetic. For example, take the simple case where $A=19$, or $y^{2}-19x^{2}-=1$, then the smallest integral values of y and x which satisfy this equation may be found to be 170 and 39, so that the square of 170 minus 19 times the square of 39 equals unity.

In 1889 A. H. Bell, a surveyor and civil engineer of Hillsboro, Illinois, began the work of solution. He formed the Hillsboro Mathematical Club, consisting of Edmund Fish, George H. Richards and himself, and nearly four years were spent on the work. They computed thirty of the left-hand figures and twelve of the right-hand figures of the value of $x^{2}$ without finding the intermediate ones. This value is

$x^{2}=$34,555,906,354,559,370 . . . . . . . . . 252,058,980,100

in which the dots indicate fifteen computed figures which it is here unnecessary to give and 206,487 uncomputed ones; the total number of figures in this number is 206,531. The final step is to multiply each of the numbers of the first solution by 4,456,749 and by this value of $x^{2}$, and thus are obtained

White bulls	$=$	1,596,510	341.800
Black bulls	$=$	1,148,971	178,600
Dappled bulls	$=$	1,133.192	894,000
Yellow bulls	$=$	639,034	026,300
White cows	$=$	1,109,829	564,000
Black cows	$=$	735,594	645,400
Dappled cows	$=$	541,460	318,000
Yellow cows	$=$	837.676	113,700
Total cattle		7,760,271	081,800

in which the dots represent 206,532 figures, the total number of figures in each line being either 206,545 or 206,544. In each of these lines there are omitted twenty-four figures at the left-end and six at the right-end which were computed by the Hillsboro Mathematical Club.

This solution is published in the American Mathematical Magazine for May, 1895, where Bell remarks that each of these enormous numbers is 'one half mile long.' A clearer idea of its length may be obtained from considering the space it would take to print it. Each page of this Monthly contains 45 lines and in each line about 50 figures may be set, so that one page would permit a number of about 2,250 figures to be printed. To print a number containing 206,545 figures there would be required 92 pages, and to print the nine numbers indicated above a volume of about 830 large octavo pages in this size of page and type would be needed.

It is known that Archimedes speculated regarding large numbers, for his book called Arenarius is devoted to showing that a number may be written that will express the number of grains of sand in a sphere of the size of the earth. It can not be proved that Archimedes was, or was not, the author of the cattle problem, but, as Amthor remarks, the enormous numbers required in its solution render it worthy of his genius and proper to bear his name. Its closing challenge still remains open, for the complete solution has not yet been made. Moreover, it is practically impossible that the long numbers can ever be computed, since the investigations of Bell show that this would require the work of a thousand men for thousand years. The little prairie town of Hillsboro may, however, well exult as a conqueror, for its mathematical club has made the most complete of all solutions of the cattle problem and has proved itself to be highly skilled in numbers.