436 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xiv 



numbers. Their squares are in A. P. and their sums by twos will be squares 

 if 2a 2 +2& 2 = D, which is known to hold if a, b = 2mn(m 2 n 2 }. The same 

 problem was treated by A. Cunningham and F. Phillips. 9 A. E. Jones 10 

 started with any three numbers 



x m 2 -\-2mn-\-n 2 , y = m 2 -{-n 2 , z = m 2 -\-2mn n 2 



whose squares are in A. P., and called P, Q, R the values obtained from them 

 by replacing ra by x 2 , n by z 2 . Then P, Q, R are the desired numbers since 



P+Q = 2z 2 (x 2 +z 2 ) = 4zV, P+R=4x 2 z 2 , Q+R = 2x*(x*+z 2 ) =4x*y z - 



C. Campbell 11 treated the similar problem to find three numbers x, y, z 

 the difference of any two of which is a square and whose squares are in A. P. 

 Let xy = m 2 , xz = n 2 , yz = p 2 . Then n 2 p z = m 2 . Take n-\-p = ms, 

 np = m/s. Since x 2 -\-z 2 = 2y 2 gives x, we get y and z in terms of m, n. 



J. Cunliffe 12 treated the problem to find 3 squares in A. P. such that the 

 sum of each and its root shall be a square. 



J. Wright 12a found three squares x 2 , y 2 , z 2 in harmonical progression 

 such that each exceeds its root by a square. If a, b = 2rs (r 2 s 2 ), c = r 2 +s 2 , 

 a 2 +6 2 = 2c 2 and the squares of x = n 2 /d, y = bn 2 /(cd), z = bn 2 /(ad) are in har- 

 monical progression. For d = m(2n 7ri), x 2 =D. Also, y 2 y=O if 

 b 2 n 2 -bcd=n = (bn-pm) 2 , which holds if m = 2bn(c-p)}(bc-p 2 ). Then 

 z 2 2=D if (be p 2 ) 2 ap(b p)(c p} = D = (bc+2ap p 2 ) 2 , which gives 

 p = 2bc/(b+c-a). 



J. Ivory 126 found two sets a 2 , b 2 , c 2 and of, 6f, c\ of three squares in A. P. 

 having the same sum. The conditions are a 2 -\-c 2 = 2b 2 = 2bl = a\-\-c\, or 

 4& 2 = S(ad=c) 2 = 2(aiCi) 2 . Hence we require a square which is a sum of two 

 squares in two ways. The least numbers are obtained from 



To find three numbers whose sum is 117 and whose squares are in 

 A. P., S. Jones 13 took x, 5x, 7x as the numbers, whence x = 9. S. Ryley took 

 2mn(m 2 n 2 ), m 2 -{-n 2 as the numbers. Then (n+2m) 2 = 117+3m 2 = D 

 f or m = 3 ; the resulting numbers 9, 45, 63 are said to give the only solution 

 in positive integers. 



R. Adrain 14 used the squares u 2 y = (up} 2 , u 2 , u 2 +y = (u+q) 2 , whence 

 2pu y = p 2 , y 2qu = q 2 . Solving, we get u = (p 2 +q 2 )/{2(p q) }. There 

 results Frenicle's 5 solution. 



J. Surtees 15 noted that (a n) 2 , a 2 +n 2 , (a+n) 2 are in A. P. and 

 o 2 +n 2 = D if a r 2 1, n = 2r. 



J. R. Young 16 found three squares in A. P. such that the roots increased 



9 Math. Quest. Educ. Times, 24, 1913, 107. 



10 Math. Quest, and Solutions, 5, 1918, 62-3. 



"The Gentleman's Diary, or Math. Repository, London, No. 65, 1805, 40-1, Quest. 873. 



12 Math. Repository (ed., Leybourn), London, 3, 1804, 97-106, Prob. 7. 

 12a New Series of Math. Repository (ed., T. Leybourn), 1, 1806, I, 99. 

 126 Ibid., 121-3. 



13 The Gentleman's Math. Companion, London, 2, No. 9, 1806, 15-17. 

 " The Math. Correspondent, New York, 2, 1807, 14. 



15 Ladies' Diary, 1811, 39, Quest. 1217; Leybourn's Math. Quest. L. D., 4, 1817, 139. 



16 Algebra, 1816; Amer. ed., 1832, 333^ (329-31). 



