Solution of a Diophantine Problem. 295 



Art. XI. — An easy solution of a Diophantine Problem ; by A. D. 

 Wheeler, Principal of the Latin Grammar School, Salem, Mass. 



Problem. — To find two squares, whose sum shall be a square ; or 

 in other words, to find rational values for the legs and hypothenuse 

 of a right angled triangle. 



Rule. — Take any two numbers, of which the difference is 2. 

 Their sum will be the root of one square ; their product, that of the 

 other. Add 2 to the product, just found, and you obtain the root of 

 the sum of the squares, or thevalue of the hypothenuse. 



Example. — Take the numbers 10 and 12; then 104-12=22 = 

 the root of one square, and 10X12 = 120-— that of the other. 

 Furthermore, 120+2 = 122= the root of their sum; since (22) ^ 

 + (120)=' = (122)^. These quantities maybe multiplied by any 

 number whatever, and their products, when squared, will still answer 

 to the conditions of the problem. 



Demonstration. ^Lie\. z represent the smaller number, and z-^2, 

 the larger. Their sum z-\-{z-\-2)=2z-}-2 [A], and their product, 

 z.{z+2)=z^- +2z [B]. Now {2z-{-2y -\-{z^ -\-2zY =z^ +42^3 + 

 82; 2 -[-82 4- 4= a square, whose root is z^ -\-2z-\-2 — [z^ -\-2z)-{- 

 2 [C] = the product [B] -\-2. Multiplying the expressions [A], 

 [B], and [C], by m, and squaring, we have m^ {2z -\-2)^ -\-m^ [z'^ -{- 

 2zY=m'- (;s4-]-4^^ + 8^=^4-8^+4)=m2 {z--\-2z-{-2Y, whatever 

 be the values of m and z. Q. E. D. 



When m=0, and z is an odd number, the quantities [A], [B], 

 [C], diVe prime to each other. But when ;r is even, these quantities 

 may be divided by 2, and by no other number. For, if we suppose 

 the quantity [B] to be divisible by n^ then, when w is a prime number, 

 either z or z-\-2 must be divisible by it. Because, if a prime num- 

 ber will divide neither factor, it cannot divide the product. (Euler, 

 App. 10). But since the difference between z and z-}-2 is only 2, 

 the number n, when greater than 2, cannot divide them both ; and, 

 consequently, cannot divide the quantity [A]. (Bonnycasde, p. 145). 

 Again if [B] and [C] are divisible by n, then the parts of [C], {z^ + 

 2z) and 2 may be divided by n. But, it is plain, that this cannot be 

 done, unless, as before, n—2. 



If n be not a prime number, it is necessary to remark, that all 

 compound numbers may be resolved into prime factors, each of 



