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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 



aa! sin»m&r=- 


^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)<]. 


The velocity with which the string leaves the pulley 

Pressure on pulley at time , = ^' (»+| +ljft=g = to< «+«+l)(— *) 


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, 

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.