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STOP Early Journal Content on JSTOR, Free to Anyone in the World This article is one of nearly 500,000 scholarly works digitized and made freely available to everyone in the world by JSTOR. Known as the Early Journal Content, this set of works include research articles, news, letters, and other writings published in more than 200 of the oldest leading academic journals. The works date from the mid-seventeenth to the early twentieth centuries. We encourage people to read and share the Early Journal Content openly and to tell others that this resource exists. People may post this content online or redistribute in any way for non-commercial purposes. Read more about Early Journal Content at http://about.jstor.org/participate-jstor/individuals/early- journal-content . JSTOR is a digital library of academic journals, books, and primary source objects. JSTOR helps people discover, use, and build upon a wide range of content through a powerful research and teaching platform, and preserves this content for future generations. JSTOR is part of ITHAKA, a not-for-profit organization that also includes Ithaka S+R and Portico. For more information about JSTOR, please contact support@jstor.org. 229 We must build up a value of <p,(aO satisfying the last expression. Take e-ft'sin/Sa; and e-^cos/Sa; and multiply the first by cos/M, the second by sin/SA. Subtract these and we get e-tosm§(x— X). 1 s«OD s»QO .-. (p^x) = — ) I er-»v<p(X)smP(x— X)d0dL e- o aa! sin»m&r=- a*+i ^f^ sin ^-^ = _j_L_ . 143. Proposed by W. J. GEEENSTREET, M. A., Editor of The Mathematical Gazette, Stroud, England. Beads are fastened at equal intervals on a string placed over a smooth fixed pulley. If the original position of the string is one of symmetry, find the velocity at any moment, the pressure on the pulley, and the velocity with which the string leaves the pulley. Solution by 6. B. M. ZERB, A. M., Ph. D., Professor of Chemistry and Physios, The Temple College, Phila- delphia, Pa. In this solution the weights of the pulley and string are neglected. Let 22=length of string, 2w+l=number of beads each mass m. Aoeeleration=/=— ^ — y . v , , r ^-^■=- 7i — — — g, where x can have w(2w+l) 2n+l " any value from 1 to (n— 1). Z/«=distance between two beads. now w-w-i^agt, . _ | 2?(2»+1) \ 2gl (2x-l) " * 1 ~SgK2x-l) ' v —J ( i—yl M (2w+1) - Let t be any time between the passing of the (a;+l)th and (aj-(-2)th par- ticle over the pulley, then the velocity, v lt at this time, is *=* mjte=I) | 2gl(2s + l) rsgi '•-^W n(an-l) "W n(2n+l)'-<Jn(2n + l) [ 1 +l/ 3 + ^5 + l/7+.... + 1 /(2«-l)+ 1 /(2a!+l)<]. 230 The velocity with which the string leaves the pulley Pressure on pulley at time , = ^' (»+| +ljft=g = to< «+«+l)(— *) DIOPHANTINE ANALYSIS. 92. Proposed by M. A. GBUBEB, A. II., War Department, Washington, D. G. Find the sides of integral right triangles when the difference of the legs is given. Solution by HON. JOSIAH H. DRUMMOND, LL. D., Portland, Me. Let a be the difference in the legs, and x and x-\-a=\egs. Then 2x s -\-2ax-\- 2aT» + n a s =n=(say) [px— a] 2 =p s a: 2 — 2apx + a-. Whence,a:= — L /__ J . Takep=2, 2[r4-s~\ z=3a, and the sides are 3a, 4a, 5a. Then in the formula , ~ , we have r/s ' ' r+2s =4, then p=f, and a;=20a, and the sides 20a, 21a, and 29a, and so on ad infinitum. Bemark on Problem 94 by HON. JOSIAH H. DRUMMOND, LL. D., Portland, Me. In his solution of this question, Professor Zerr gives up his general dem- onstration a little prematurely. It is true that "For integral values of m, m* + m-\-l is not a square," but fractional values of m lead to integral values of a, b, and c. The value of m which makes the expression a square is -^ — = i n which p may be any number except one. Take p=2, and »»=$. Then a=- 4 /, &—/-, and c= J ¥ 5 -. Any common multiple of a, 6, and c makes the square root of j/[a s +& 8 + c 8 ] a square and as the denominator is a square, makes abc a square, as well as these particular values. So a, b, and c may be taken=40, 24, and 15, respectively. Then a&cj/[a 8 +6 2 -|-c 8 ]=14400=49, a square number. However, the ques- tion does not call for integral values, and I had solved the question as follows from the point at which Professor Zerr leaves it. Substituting the value of m— g in the values of a, b, and c, reducing to common denominator, [p 2 — 1]» , wehavea=p[2p s +5j)+2], b=p[p+2'][p 2 — 1], and c=2p 3 =p i —2p—l, in which p may be any number except one. Hence, there is an indefinite number of ra- tional triangles whose area is a square. 96. Proposed by L. C. WALKER, A. M., Graduate Student, Leland Stanford University, Cal. (a) Find the least three integral numbers such that the difference of every two of them shall be a square number; (b) find the least three square numbers such that the dif- ference of every two af them shall be a square number.