724 MR EDWARD SANG ON THE 



If now b be also indivisible by 3 its square must be of the same form, so that 

 a^ + y^ would be of the form 3w + 2, which cannot possibly be that of a square 

 number. And thus one of the two must be divisible by 3. 



9. Of a right-angled trigon in integers, one of the two sides is divisible by 4. 

 The square of every odd number is of the form 8??-l- 1, wherefore if « and h 



were both odd numbers, a^ + Jf would be of the form 8^2 + 2, which cannot be that 

 of a square ; thus one or other of the two must be even. Now if a be even, h and 

 c must be both odd, wherefore a^ being the difference of two odd squares, viz., 

 c^—V^ must be divisible by 8 ; but no square can be divisible by 8 unless its root 

 be divisible by 4, wherefore the even side a must be divisible by 4. 



10. Of a right-angled trigon in integers, one of the three sides is divisible by 5. 

 All numbers not divisible by 5 are of the forms 5?z ± 1, bn =!= 2, the squares of 



which are contained in the forms 5w + 1, 6n—\. If neither a nor h be divisible 

 by 5, their squares cannot be both of the form 5w4- 1, nor both of the form 5?^— 1, 

 for then a^ + If would be of the form on + 2, or bn—2, neither of which belongs to 

 a square. Wherefore, if a be of one of the forms 5^=*= 1, h must be of one of the 

 forms 5w±2, in which case a^ + If is of the form bn, and therefore may be a 

 square. 



Again, if the hypotenuse c be not divisible by 5, c^v!?=,5?2± 1 ; and similarly, ifb 

 be not divisible by 5, &^v?=45?^ ±1. If now If'is^^.on + 1 while r vs^^.ow— 1, the difference 

 c^—lf would be of the form bn—2 ; or if 5^v?=457z— 1 whilec^'a?=i.5w4- 1, c^—lf would 

 be of the form 5?^ -f 2 ; neither of these can be square. The other two combina- 

 tions H^Hs^bn + l, c^'fsp\57i + l, and 6^Vi?=i5n— 1, c^v;=45w— I, both give c'—b'^^5n 

 w^hich can be a square, and then a must be divisible by 5. 



11. The hypotenuse of a right-angled trigon in its lowest terms can never be 

 divisible by 7. 



For if c were divisible by 7, neither a nor b can be so, as then all three would 

 be divisible by 7. Hence a and b must be of some of the forms 7?^=*=l, 7n^2, 

 7n=^3, and therefore their squares must belong to some of the forms 77i + 1, 7?i + 4, 

 7n + 2. Now no two of the remainders 1, 1 ; 4, 4 ; 2, 2 can make up 7, and there- 

 fore or + F' can never be divisible by 7. 



The same argument may be used to show that 19, 23, 31, &c., cannot be divi- 

 sors of the hypotenuse of a right-angled trigon in its lowest terms. 



12. Lemma. If two numbers be each the sum of two squares, their product 

 may be decomposed into two squares at least in two ways. 



Let d—s'^ + f, and = 11^ + 1)^, then, on multiplying we obtain 



which may be put under either of the forms, 



de — s'^u- + 2stuv + <^w^ + s^w^ — ^atuv + i'^ur ; 

 de = s^u^ — 2stuv + fv^ + s^v^ + 2stuv + t^u- ; 



