Let x2 be one of the parts; then 100-x3 will be the other part; which is also to be a square number.

Assume the side of this second square=2x-10, Then will 100-x2=(2x — 10)2 =4x2 −40x+100 ; And, consequently, by reduction, x=8, and 2x-10=6, Therefore 64 and 36 are the parts required.

Or, the same may be done generally, thus:

Let a2= given square number, x2= one of its parts, and a2x2= the other; which is also to be a square number.

Assume the side of this second square=rx-α,
Then will aa—x2=(rx—a)3±r3x2—2arx+a2 ;

quired; where a and r may be any numbers, taken at pleasure.

2. To divide a given number (13) consisting of twe known square numbers (9 and 4) into two other square numbers. (r)

equation would have been a −20x+100=100-x2; in which case, x, the side of the first square, would have been found = 10, and x-10, or the side of the second square =0; for which reason the substitution x-10 was avoided; but 3x-10, 4x-10, or any other quantity of the same kind, would have succeeded as well as the former, though the results would have been less simple.

(r) To this we may add the following useful property.

If sand r be any two unequal numbers, of which is the greater,

For the side of the first square sought, put rx-3; and

for the side of the second, sznumber, and s the less.

2; r being the greater

Then will (rx-3)2 + (sx − 2)2 = (r2x2 −6rx+9)+(s2 x2-4sx+4)=(r2+s2)x2 - (6r+4s)x+13=13, or (r2+ s2)x2=(6r+4s)x.

So that if r be taken` =2, and s=1, we shall have

it can then be readily shown, from the nature of the problem, that 2r8, s2-r2 and s2+ra

will be the perpendicular, base, and hypothenuse of a right-angled triangle.

From which expressions, two square numbers may be found, whose sum or difference shall be square numbers; for (2rs)2+(s2r2)2=(s2+r2)2, and (s2 +r2)2 — (2rs)2 —(s2 — r2 )2, or (62+r2 )2— (832)2=(2rsj2; where s and r may be any numbers whatever.

The question in the text, is considered by Diophantus as a very important one, being made the foundation of many of his other problems. In the solution of it given above, the values of r and s may be taken at pleasure, provided the proportion of them be not the same as that of 3 (a) to 2(b), or 3+2(a+b) to 3—2(a−b); the reason of which restriction is, that if r and s were so taken, the sides of the squares sought would come out the same as the sides of the known squares which compose the given number, and therefore the operation would be useless.

The excellent old Kersey, after amplifying and illustrating this problem in a variety of ways, concludes his chapter thus: "For a further account of this rare speculation, see Andersonus, Theorem 2, of Vieta's mysterious Doctrine of Angular Sections; and likewise Herigonius, at the latter end of the first tome of his Cursus Mathematicus.

3124 Ars-3s2 17: 6rs-2r2+2826

and

5

of the squares, in numbers, as was required.

for the sides

And if a2+b2 be put equal to the number to be divided, the general solution may be given in the same manner.

3. To find two square numbers, whose difference shall be equal to any given number.

Let the difference d be resolved into any two unequal factors a and b ; a being the greater and 6 the less.

Also put x for the side of the less square sought, and x+b= for the side of greater.

Then (x+6)—x2 = x2+2bx+b2x2=2bx+b2=d ab.

And if this be divided by b, we shall have 2x+b=a.. a-b

Whence, x= = the side of the least square sought,

and x+b=

2

So that by putting d=60, and a Xb=2X30, we shall

(16)2=256, for the squares in numbers; and so for any difference or factors whatever.

4. To find two numbers such, that, if either of them be added to the square of the other, the sum shall be a square number.

Let the numbers sought be x and y.

Then x+y, and y2+x=0,

And, if r-x be assumed for the side of the first square x2+y, we shall have x2+y=r2-2rx+x2, or y=r3 -2rx.

Therefore, by reduction, 2rx=r2 −y, or x=

3. To find a number, which, being divided by 6, shall leave the remainder 2, and when divided by 13, shall leave the remainder 3. Ans. 68

4. It is required to find a number, which being divided by 7, shall leave 5 for a remainder, and if divided by 9, the remainder shall be 2. Ans. 110

5. It is required to find the least whole number, which, being divided by 39, shall leave the remainder 16, and when divided by 56; the remainder shall be 27.

Ans. 1147

6. It is required to find the least whole number, which, being divided by 7, 8, and 9, respectively, shall leave the remainders 5, 7, and 8.

Ans. 1727

7. It is required to find the least whole number, which, being divided by each of the nine digits, 1, 2, 3, 4, 5, 6, 7, 8, 9, shall leave no remainders. Ans. 2520

8. A person receiving a box of oranges, observed, that, when he told them out by 2, 3, 4, 5, and 6 at a time, he had none remaining; but when he told them out by 7 at a time, there remained 5; how many oranges were there in the box? Ans. 180

OF THE

DIOPHANTINE ANALYSIS.

This branch of Algebra, which is so called from its inventor, Diophantus, a Greek mathematician of Alexandria in Egypt, who flourished in or about the third century after Christ, relates chiefly to the finding of square and cube numbers, or to the rendering certain compound expressions free from surds; the method of doing which is by making such substitutions for the unknown quantity, as will reduce the resulting equation to

a simple one, and then finding the value of that quantity in terms of the rest. (p)

These questions are so exceedingly curious and abstruse, that nothing less than the most refined Algebra, applied with the utmost skill and judgment, can surmount the difficulties which attend them. And, in this respect, no one has extended the limits of the analytic art further than Diophantus, or discovered greater knowledge and penetration in the application of it.

When we consider his work with attention, we are

(p) That Diophantus was not the inventor of Algebra, as has been generally imagined, is obvious; since his method of applying it is such, as could only have been used in a very advanced state of the science; besides which, he no where speaks of the fundamental rules and principles, as an inventor certainly would have done, but treats of it as an art already sufficiently known; and seems to intend, not so much to teach it, as to cultivate and improve it, by solving such questions as, before his time, had been thought too difficult to be surmounted.

It is highly probable, therefore, that Algebra was known among the Greeks, long before the time of Diophantus; but that the works of preceding writers have been destroyed by the ravages of time, or the depredations of war and barbarism.

His Arithmetical Questions, out of which these problems were mostly collected, consisted originally of thirteen books; but the first six only are now extant; the best edition of which is that published at Paris, by Bachet, in the year 1670, with Notes by Fermat. In this work, the subject is so skilfully handled, that the moderns, notwithstanding their other improvements, have been able to do little more than explain and illustrate his method. Those who have succeeded best in this respect, are Vieta, Kersey, De Billy, Ozanam, Prestet, Saunderson, Fermat, and Euler; the last of whom, in particular, has amplified and illustrated the Diophantine Algebra in as clear and satisfactory a manner as the subject seems to admit of.

The reader will find a methodical abstract of the several methods made use by these writers, with a variety of examples to illustrate them, in the first and second volumes of my Treatise of Algebra, before quoted.