244 ^^ POLYCvONAL NUMBEftS. 



number, but not every natural number a square integer. 

 On the CQntrary, s is unlimited in the aggregates of poly- 

 gons, as well as in natural numbers; therefore, my oppo- 

 nent's second parody is equally unsuccessful with his first; 

 because its imaginary resemblance to my syllogism has 

 been shown to be spurious. In the conclusion of this ob- 

 jection it is remarked, that the corollary under consideration 

 might be assumed as a postulate, i. e. as a self evident 

 problem; but far from treating it either as postulate or 

 axiom, he agrees with me in giving it the importance of a 

 theorem, and demonstrates it accordingly. 



5i objection Mr. Barlow's third objection is occasioned by an obvious 



"^ ' mistake on his part : I have said, that, if e — y -\- t can be 

 resolved into m — / polygons, e^\-fm^y be resolved into 

 m polygons of the same denomination. My critic puts a 

 construction on this expression, which makes me suppose, 

 that y 4- * can be resolved into m — f polygons of the de- 

 nomination OT, in all cases. This is evidently a misconcep- 

 tion ; for, had my opinion agreed with Mr. Barlow's inter- 

 pretation of it, why have 1 attempted to demonstrate the 

 theorem of Fermat; the truth of which I am supposed to 

 assert without demonstration in the preceding quotation 

 from my essay? The genuine meaning of the passage is 

 obviously this; if e ~ ^ -}- f can be resolved into m — /' po- 

 lygons in any one case; e -k- f may be resolved into m such 

 polygons in the same case ; which construction of the ex- 

 pression refutes this part of the criticism. 



The proposi- All ray opponent's objections have now been controverted ; 



be universal. ^^^ ^^ farther remarks, that there is a dilVereace betwixt 

 doing a thing, and proving that it may be done in all cases. 

 The justice of this observation obliges me to show, that 

 my essay also contains the principles of the latter demon- 

 stration. For this purpose let the reader look at example 2, 

 prop. 7, which will assist him in the following reasoning. 

 If t =: y + m — 1, it may be resolved into m polygons (by 

 CQr. prop. 6} : again, if e zz-y -\- m, it inoy be resolved into 

 two polygons, which are less than m ; here /"— m — 2, and 

 e -{-/=z y -\- 2 m — 2 is resolved into m polygons, (cor. 

 prop. 6) : moreover, if <? — y -j- 2 m, it consists of three 

 polygons (by prop. 6, and cor. 1, prop. 2), but three is the 



least 



