And when paid ev'ry way 'twill admit, the amount 8. DIOPHANTINE PROBLEMS. Unlimited problems relating to square and cube numbers, right angled triangles, &c. were first and chiefly treated of by Diophantus of Alexandria, and from that circumstance they are usually named Diophantine Problems'. These problems, if not duly ordered, will frequently bring out answers in irrational quantities; but with proper management this inconvenience may in many cases be avoided, and the answers obtained in commensurable numbers. The intricate nature and almost endless variety of problems of this kind, render it impossible to lay down a general rule for their solution, or to give rules for an innumerable variety of particular cases which may occur. The following rules will, perhaps, be found among the best and most generally applicable of any that have been proposed. RULE I. Substitute one or more letters for the required root of the given square, cube, &c. so that, when involved, either the given number, or the highest power of the unknown quantity, may be exterminated from the given equation. f Diophantus has been considered by some writers as the inventor of Algebra; others have ascribed to him the invention of unlimited problems: but the difficult nature of the latter, and the masterly and elegant solutions he has given to most of them, plainly indicate that both opinions are erroneous. Diophantus flourished, according to some, before the Christian æra; some place him in the second century after Christ, others in the fourth, and others in the eighth or ninth. His Arithmetics, (out of which have been extracted most of the curious problems of the kind at present extant,) consisted originally of thirteen books, six of which, with the imperfect seventh, were published at Basil in 1575, by Xylander; this fragment is the only work on Algebra, which has descended to us from the ancients: the remaining books have never been discovered. See Vol. I. p. 327. Of those who have written on, and successfully cultivated, the Diophantine Algebra, the chief are, Bachet de Mezeriac, Brancker, Bernoulli, Bonnycastle, De Billy, Euler, Fermat, Kersey, Ozanam, Prestet, Saunderson, Vieta, and Wolfius. II. If, after this operation, the unknown quantity be of but one dimension, reduce the equation, and the answer will be found. III. But if the unknown quantity be still a square, cube, &c. substitute some new letter or letters for the root, and proceed as before directed. IV. Repeat the operation until the unknown quantity is reduced to one dimension; its value will then readily be found, from whence the values of all the other quantities will likewise be known. 1. To divide a given square number into two parts, so that each may be a square number. ANALYSIS. Let a2=the given square number, x2=one of the parts; then will a2x2-the other part, which, by the problem, must likewise be a square. Let rx-a=the side of the latter square, then will (rx—a2=) r2x2-2 arx+a2=a2-x2, whence x= the side of the first square, and rx—a=(; 2 ar2 r2+1 = 2 ar r2+1 ar2-a r2+1 the side of the second square; wherefore 2 ar r2+11 2 and a2rt-2 a2r2+a2 2 ar ar2. a ar2 -a r2+1 2 are the parts required; where a and r may be any numbers taken at pleasure, provided r be greater or less than unity'. Q. E. I. SYNTHESIS. First, 4 a2r2 |2 = (- + r4+2 r2+1.a2 = a2, which is ar2. -a r2+1 the first condition. Secondly, 2 and 2 ar r2+1 2 are evidently both squares, which is the second condition. Q. E. D. EXAMPLES.-Let the square number 100 be proposed to be divided into two parts, which will be squares. Mr. Bonnycastle, in his solution of the problem, (Algebra, third Edit. p. 143.) has omitted this restriction, which is evidently necessary; for if r be supposed =1, then will the numerator of the fraction ar2. -α vanish, and the +1 solution become nugatory. VOL. II. Here a2=100, and a=10. First, assume r=2, then will 40 2 ar = =) 8=the side of the first square, and rx—a=6= r2+1 5 the side of the second square; for 8)2+6)2=(64+36=)100, as was required. Secondly, assume r=3, then will x=(==)6, and rx-a=8, 60 10 To divide 25 into two square numbers. Ans. 16 and 9. 2. To find two square numbers having a given difference. Let d the given difference, axb=d, whereof ab, and let x=the side of the less square, and x+b=the side of the greater ; then will x+b2-x2 = (x2+2 bx+b2x2=) 2 bx+b2=ab; divide the side of the less square; this by b, and 2x+b=a, ': x= a-b =the side of the greater square: a2+2ab+b2 wherefore = 2 the greater square required, 4 by hypothesis ab=d. Q. E. D. EXAMPLES.-To find two square numbers, whereof the greater exceeds the less by 11. 2 2 Whence 6)2=36, and 5)2=25, are the squares required. To find two square numbers differing by 6. Here d=6(=3×2), a=3, b=2. Ans. 64 To find two squares, whose difference is 15. and 49. To find two squares differing by 24. 3. To find two numbers, whose sum and difference will be both squares. Let x=one of the numbers, x2 -x=the other; then will their sum (x+x2 —x=) x2, evidently be a square number. And since (x? —x—x=) x2 -2x=their difference, must likewise be a square; let its side be assumed=x-r, then will (x =)x2—2 xr+r2 =x2 —2 x, or 2 xr—2 x=r2, '.' x: 2 T-2' 2 and до в 2r-2 h If 2 be substituted in this example for r, both numbers will come out=2; that is, their sum will be 4, and difference 0; wherefore r must not only be greater than 1, (as is asserted in Bonnycastle's Algebra, p. 146.) but greater: than 2. Let r=5, to find the numbers. 4. To divide a given number, which is the sum of two known squares, into two other squares. 2 Let a2+b2=the number given, rx-a=the side of the first required square, sx-b=the side of the second, where r > s. Then will rx-a2 + sx—b2 = = (r2x2—2 arx+a2+s2x2 — 2 bsx +b2=) r2+s3.x2-2 ar+2 bs.x+a2+b2=a2+b2 r2+s2.x2 -2 ar+2 bs.x=o, or r2+s2.x2=2 ar+2 bs.x; ... dividing by x, EXAMPLES.-Let a=6, b=4, r=5, s=3; then will x= 42 17' Let a=4, b=3, r=2, and s=1, be given. 5. To find two numbers, of which the sum is equal to the square of the least. Ans. 6 and 3. 6. To divide the number 30 into two parts, such that their product will be a square number. Ans. 27 and 3. 7. To divide the number 129 into two parts, the difference of which will be a square number. Ans. 105 and 24. 8. What two numbers are those, whose product added to the sum of their squares, will make a square? Ans. 5 and 3. 9. To find two squares, such that their sum added to their 16 product may likewise make a square. Ans. and 10. To find two numbers, one of which their product, the remainder will be a cube. 9 being taken from Ans. 3 and 108. 11. To find two numbers, such that either of them being added to the square of the other, the sum will be a square. An 12. To find three numbers, such that their sum, and likewise the sum of every two of them, will each be a square number. Ans. 42, 684, and 22. |