CH A P. VI. 22 Of Indeterminate Problems. IT T was formerly observed, (Chap. III.) that if there are more unknown quantities in a question, than equations by which their relations are expressed, it is indetermined; or it may admit of an infinite number of answers. Other circumstances, however, may limit the number in a certain manner; and these are various, according to the nature of the problem. The contrivances by which such problems are resolved are so very different in different cases, that they cannot be comprehended in general rules. d EX HA Ε Χ Α Μ Ρ Ι Ε Ι. To divide a given square number. into two. parts, each of which all be a square number. There are two quantities fought in this question, and there is only one equation expressing their relation ; but it is required also, that they may be rational, which circumstance cannot be expressed by an equation; another condition therefore must be assumed in such a manner as to obtain a.solution in rational numbers. Let the given square be a’, let one of the squares sought be xạ, the other is a?_x Let rx--a also be a fide of the last fquare, therefore g?x?_2rxatar=a-*? By transp. gor?m? +*=2rxa Divide by x go2x+x=2ra Therefore 72 ti arha z? And rx-a= m2 +1 Let r, therefore, be assumed at pleasure, and 2ra 2 -a I جب 2 2ra and a, which must always be 72 +1° pati rational, will be the sides of the two squares required. Thus, if aż=100. Then, if r=3, the sides of the two squares are 6 and 8, for 36+64=100. 24 Also, let a’=64. Then, if r=2, the sides of the squares are 32 1024576 and 5 5 25 25 1600 =64. 25 ; and * The reason of the assumption of rxas a side of the square a2-x2, is that being squared and put equal to this last, the equation manifestly will be simple, and the root of such an equation is always rational. É X A M P L E II. To find two squate numbers whose difference is given. Let Let x2 and yü be the square numbers, and a their difference. x?+2zut 02 4 22 4 zv=x2-y2=a. IF x and y are required only to be rational, then take v at pleasure, and z= whence x and y are known. But, if x and y are required to be whole numbers, Take for z and v any two factors that produce a, and are both even or both odd numbers, and this is possible only where a is either an odd number greater than 1, or a number divifible by 4. Then and are the numbers fought. 2 For the product of two odd numbers is odd, and that of two even numbers is divisible by 4. Also, if % and v are both odd, tegers. then x= 27; and the squares are 196 and 169: or % may be 9 and v=3, and then the squares are 36 and 9. 2. If a=12, take v=2, and z=6; and the squares are 16 and 40 E X A M P L E III. To find a sum of money in pounds and fhil lings, whose half is just its reverse. a Note. The reverse of a sum of money, as 8 1. 12 s. is 12 l. 8 s. Let x be the pounds and y the shillings. j 20x + y 20x+y=40y+2x 2+=20y+* 18x=397 x : y :: (39 : 18 :: ) 13 : 6. In |