n2+n(bc + bd — cd) + (bc + bd — cd)2 = (n + bcbd-cd 2 - b'c2d-(bcbd--cd)2 Let b -)2; then n= = acd+10=3d, and bcd+10=4th. Let ad=b+c+2p, and be + 1 = p2; then the 2d and 3d are square. ... Now bc-p-1; .. b—p—1, c = p-1, 4p a= ; by substitution in the first & third, I find, d +1=0, and (p2-1)d+1=0. The latter is a Hence 4p3 4p+1=0, which is the case when p is 1 or 6. Then a 24, b=7, c=5, and d=1. Otherways To transform into squares the 4 following formula; xyz + 1, xyv + 1, xzx + 1, and yzv + 1. Put xy-b2-c2, z=d2— e2, v=f2-g2, Then the first and 2nd formulæ are squares, if (A), be-cd=1, (B), bg — cf=±1; whence, by substitution in the third and fourth formulæ, (b2—c2) (d2—c2) ( ƒ2 — g3) y +1=0, (ď2 — e2){ƒa— g3)y+1=0. by the theory of continued fractions, if we form a series of fractions, by adding the numerators and denominators of the succeeding fractions to the preceding ones, we will satisfy the equations (A) and (B) thus:,,,,,, 18. &c., where any 3 contiguous ones may be taken for b, c, d, e, f, and g. Thus, take 3 d 4 ƒ 7 b - C 2' e 3' g Hence y=1, z=d2—c2—7, v—24, x= =5, as before. 1 37. Find three numbers whose sum shall be a biquadrate, and the sum of the 1 and 3d a square cube, and the sums of the 1st and 2d, and 2d and 3d, squares. Let 4x, x2-4x, 2x+1; then (x + 1)2 must be the 4th power; or x+1=. Also 6x+1=□,x= 30. Put r=1+s, 36+12s+6s2=36—12ts+t22; therefore t +1 s=12.1±1, if t—3, s=16, & r=17, x=48; 192, 2112, & 97. Let 192x1, 2112x, 97x, Nos. then 2892 square cube, or 1722 = cube; .. x=17, and 192∙17, 2112.17, 197-17'. See page 446, for more diophantine questions and solutions. Miscellaneous Questions. 1. What two numbers are those, whose sum is 19, and whose difference multiplied by the greater is 60? Let x, and 19-x be the numbers; .. 2x2-19x=60, and by case 1, x2—19x+(42)2=30+361-841, or 2-2018—12, or —§. 2. If the square of a certain number be taken from 40, and the square root of this difference be increased by 10, and the sum multiplied by 2, and the product divided by the number itself, the quotient will be 4. Required the number. Ans. 6. the number; ... 2/(40-2)+10}, -4, & Let =2x-10; ... 40—x2—4x2—40x+100, or x2— 8x (40-22) 12. 3. There is a field in the form of a rectangular parallelogram, whose length exceeds the breadth by 16 yards, and it contains 960 square yards. Required the length and breadth. Let x, and x-16 be the length and breadth, x2-16x=960, and 2-16x+64-1024, x-8-32, and x=40 the length and the. breadth 24 yards, Ans. 4. A person being asked his age, answered, If you add the square root of it to half of it, and subtract 12, there will remain nothing. Required his age. Ans. 16. Let x his age; then x+x-12-0, x-2/x+1=25, and x+1=5, x-4, and x16, Ans.. 5. Two casks of wine were bought for $58, one of which contained 5 gallons more than the other, and the price per gallon was $2 less than one-third of the number of gallons in the less. Required the number of gallons in each, and the price per gallon? Let 3x, 3x+5, and x-2 be the number of gallons in the less and the greater, and the price of a gallon. Then we have, ... (6x+5)(x-2)=58, and 622-7x-68. By case 2, ..-Zx+ £ 24 = 88 42 — 1681, and 2-7-11, and x = 4, and, the numbers are 12 and 17, and the price $2 per gallon. 68 6. From two places, at the distance of 320 miles, two persons, A and B, set out at the same time to meet each other. A travelled 8 miles a day more than B, and the number of days in which they met was equal to half the number of miles B went in a day. How many miles did each travel per day, and how far did each travel? Let 2x, 2x+8, and x the number B and A went per day, and number of days; .. 4x2+8x=320, and by case 1, x—8; A went 24 and B 16 miles per day, and the distances travelled by them were 128 and 192 miles. 7. The difference between the hypothenuse and base of a right angled triangle is = 6, and the difference between the hypothe nuse and the perpendicular is = 3. What are the sides? Let x, x—6, and x-3 be the hypothenuse, base and perpendicular, and x2—(x—6)2+(x—3)2—2x2—18x+45, and .. x2—18x+ 45-0, and by case 3, (x2-18x——45), or x2-18x+81 =81 – 45—36, and x=15, and 12, and 9, the sides, Ans. 8. In a parcel which contains 25 coins of silver and copper, each silver coin is worth as many pence as there are copper coins, and each copper coin is worth as many pence as there are silver coins, and the whole is worth 18 shillings. How many are there of each ? Ans. 6 of one, and 18 of the other. Let z. and 24-x be the numbers; .. 2x(24-x)=18X 12, or x2—24x——108, by case 3; ≈ = 18 or 6. 9. Bought a number of books, consisting of folios, quartos, and octavos, for $1932. Fourteen folios (which was the whole number) cost three times as much as all the quartos; and one quarto cost as many dollars as there were quartos. The number of octavos was 32, and their value was such, that 4 of them cost as much as one quarto. Required the value of each, and the number of quartos. Ans. There were 21 quartos, each folio cost $94, each quarto $21, and each octavo $5 25 cts. Let x be the number of quartos, 22 their value, and 3x2 and 8x = the value of the folios and octavos; then by the question, 4x2+8x=1932, and x-21. 10. Bought two flocks of sheep for £65, 13s., one containing 5 more than the other. Each sheep cost as many shillings as there were sheep in the flock. Required the numbers in each flock. Ans. 23, and 28. Let x, and x+5 be the numbers; then x2+(x+5)2= 1313, or 2x2+10x+25=1313, and 2-1-23, and 28, Answers. 11. A regiment of soldiers, consisting of 1066 men, is formed into two squares, one of which has four men more in a side than the other. What number of men are in a side of each of the Ans. 21, and 25. squares? Let x, and x+4 be the numbers; then x2+(x+4)=1066, and x=21 and 25 are the numbers, Ans. 12. What number is that, to which if 23 be added, and the square root of the sum extracted, this root shall be less than the original quantity by 18? Let the number; 300. Ans. 25. √(x+24)=x—18, and x2-37x= Hence x=1337—25, or 12, Ans. 13. After taking the kings, queens, and knaves out of a pack of cards, the rest were divided into three heaps. The number of pips contained in the second heap was found to be four times the square of the number in the first heap; and had the third heap contained 5 more pips than it did, the number in it would have been exactly half of what the first and second heap contained. Required the number of pips in each heap. number will be = Let x, and 4x2 be the first and second numbers, and the third 4x2+x -5, and 3(4x+x)-5=220; ... 4x2+ 2 x=150, and 4x+x+1=2401; .. 2x+1=42, and 2x=12, and x=6, 144, and 70 are the Ans. 14. A regiment of foot was ordered to send 216 men on garrison duty, each company being to furnish an equal number; but before the detachment marched, 3 of the companies were sent on another service, when it was found that each company that remained was obliged to furnish 12 additional men, in order to make up the complement 216. How many companies were there in the regiment, and what number of men was each company ordered to send at first? Let x= 216 the number; then the number each was to send was 216 15 and (x-3) (- -+12)=216, and x2—3x—54, x— + and each was to send 24 men. 2 15. A poulterer bought 15 ducks and 12 turkeys for five guinHe had two ducks more for 18 shillings than he had of turkeys for 20 shillings. What was the price of each? eas. of a turkey, and x: 1 :: 18: the number of ducks for 18s equal 18 4X4 and 7-x 9 2, or x 8 -x the number of turkeys for 20s ; .. 18 16 + -X +1; .. 63-9x=8x+7x-x, or x2-24x-63, and x=3 or 21, and the prices were 3s and 5s. 16. A and B entered into a speculation, to which B subscribed $15 more than A. After 4 months, C was admitted, who added $50 to the stock; and at the end of twelve months from C's admission they found they had gained $159; when A withdrawing, received for principal and gain. $88. What did he originally subscribe ? Ans. $40. Let x= the sum; .. x+15=B's subscription, and 16(2x+15) +12×40: 16x:: 159: 88x, or (4x+105) (88-2)=318x, and 4x2+71x+(71)2=9240301-152881, whence 2x+4=391, and 2x=80, ... x=40. 17. A wall was built round a rectangular court to a certain height, Now the length of one side of the court was two yards less than eight times the height of the wall, and the length of the adjacent side was 5 yards less than 6 times the height of the wall and the number of square yards in the court was greater than the number in the wall by 178. Required the dimensions of the court, and the height of the wall. Let x the height; ...therefore the sides are 8x-2, and 6x-5, then (8x 2)(6x-5)=2x(14x-7)+178, and 10x2-19x=84, x2-18x+(18)2=§4+263—3721; .. x—18—§1, and x=4, or the side was 30 and 19, and the height 4 yards. and 400 18. A ship containing 74 sailors, and a certain number of soldiers, beside officers, took a prize. The sailors received each onethird as many dollars as there were soldiers, and the soldiers recieved $3 apiece less, and $768 fell to the share of the officers. Had the officers however received nothing, the soldiers and sailors might have received half as many dollars per man, as there were soldiers. How many soldiers were there, and how much did each receive ? Let 3x = the number of soldiers; .. = what each sailor received, and 74x+3x2-9x+768 the value of the prize : 3x+65x+768_3x 3x+74 = 2 ; .. 3x2+92x=1536, and by case 1, x2 + 2x + ( 46 ) 2 =—1536 | 2116—6724; .. x+46—82, and x— =12. There were 36 soldiers; each soldier received $9, and each sailor $12, Ans. 19. A poulterer going to market to buy turkeys, met with four flocks. In the second were 6 more than three times the square root of double the number in the first. The third contained three times as many as the first and second; and the fourth contained 6 more than the square of one-third of the number in the third; and the whole number was 1938. How many were there in each flock? Let x = the number in the first, then 3/2x+6, 3(x+3No2x +6), and (x+3/2x+6)2 + 6 be the number in the 2d, 3d, and 4th; then.. (x+3No/2x+6)2+4(x+3No/2x+6)+4=1936, and x+3/2x+6+2=44, .. x+3;/3x=36, and 2x+6√2x+9= 72-981, .. No/2x+3=9, 2x=6, and x=18, and the numbers were 18, 24, 126, and 1770. 20. A body of men are just sufficient to form a hollow equilateral wedge, three deep; and if 597 be taken away, the remainder will form a hollow square, four deep, the front of which contains one man more than the square root of the number contained in a front of the wedge. What is the number of men? Ans. 693. Let x = the number in a side of the first triangle; .'. 3(x—1) |