MR EDWARD SANG ON THE THEORY OF COMMENSURABLES. 739 



For if AB and AF have a common divisor, FB must have the same divisor. 



48. The product of two numbers each of the form o^ + ab + ¥ may be put, in 

 two ways, in the same form. 



This theorem (as is also the corresponding one of squares) is only a case of a 

 more general theorem ; instead of repeating the arithmetical proof which is to be 

 found in treatises on the Theory of Numbers, I shall give its geometrical illus- 

 tration. 



Let there be two numbers P and Q, such that lc' = a^ + ab + b'^, while 

 Q=:a'^ + a(3 + ^'^, in which we may suppose b^^a, jS^^a^ and also the quotient 



- to be greater than - . a^ ^ r — ^ r 



" ^^^^^^^r~^~ \-- \ \ \ 



From the point ^"^^-^^T"^--^^^ \ " \' \ — \w 



A in any indefinite ^~"^ ^~'~^--.-,,_^^^ \ \ // \ 



straight line, measure ^ ^~~~"----^ ^""^^'^A ^e^-'-'x^ \ .> 



off AC equal to 5, make ^"^~-~-...„^h\ ^ 



the angle ACB one- ^^o 



third of a revolution, 



and CB equal to a linear units. Join AB, then the second power of the number 

 of units in AB is equal to a^ + ab + &^ that is AB^ = P. Also, since AC is greater 

 than CB, the angle CAB is less than 30°. In AB make A7 equal to /?, the angle 

 A7/3, 120°, and cut off y(3 equal to a, join A/3. Then A^' = a' + a^ + 8' = q. Also 

 since the ratio A7 : yfS is more unequal than that of AC : CB, the angle 7A/3 is 

 less than CAB. And this disposition of the parts can always be made, unless 

 the two ratios be identic. The angle CA,5 is thus less than 60°. 



From the line A/B continued, cut off a distance AD to represent the product of 

 the roots a/P and \/Q, so that AD^ may represent the product P . Q : draw DGr 

 parallel to BC, and we shall have P . Q = AD' = DG- + DC-GA -f- GA' ; therefore the 

 product P . Q will be decomposed as asserted, provided we can show that DG and 

 GA are expressed by integer numbers. 



Draw DE parallel to (3y, EF parallel to BC, EI parallel to FG, and having 

 cut off IH equal to IE, join EH. EIH is evidently an equilateral trigon, while 

 HDE is similar to CAB. 



From the similarity of the trigons, we obtain the six proportions following, 

 with their results : — 



A/3 : iS7 : : AD : DE or v^Q : a : : \/p7Q : DE=av/P 



A|8: A7: : AD: AE VQ : /? : : \/PVQ : AE=/3v/P 



AB : BC : : AE : EF >JF : a : : (Ss/F : E¥=a(3 



AB : AC: : AE : AF sfP : b : : (3^F : AF = &/3 



AB : BC : : DE : EH ^P : a : : aVF : EK=aa 



AB : AC : : DE : DH VF : b : : aVF : DH- ha . 



