CHAPTER XX. 



QUADRATIC FORM MADE AN NTH POWER. 



BINARY QUADRATIC FORM MADE A CUBE. 



Diophantus, VI, 19, to find a right triangle the sum of whose area x 

 and hypotenuse h is a square and perimeter is a cube, took 2 and x as the 

 legs and h+x = 25, noting that the square 25 when increased by 2 becomes 

 the cube of 3. Then h 2 =x 2 +2 2 gives 3 = 621/50. 



Jordanus 35 of Ch. XII noted that x(x-\-l) is never a cube. 



C. G. Bachet 1 noted that from 5 2 +2 = 3 3 we can find other [rational] 

 numbers x making x 2 +2 a cube. Let x = 5N. To make 27 WN+N 2 

 the cube of 3 z, equate the second term 272 of its cube to WN, whence 

 z = ION/27. We now get N. In VI, 20, we have 17 = 2 3 +3 2 and seek a 

 cube which increased by 17 gives a square; take N 2 and 3+Z as the sides 

 of the cube and square, and equate the second terms 122V and Qt of the 

 expansions, whence N = 10, i 20. 



Fermat 2 stated that he could give a rigorous proof that 25 is the only 

 integral square which is less than a cube by 2. 



Fermat 3 stated elsewhere this result on 25 and the fact that 4 and 121 

 are the only integral squares which when increased by 4 give a cube. 



L. Euler 4 proved that z 3 -|-l = n has no [positive] rational solution 

 except x = 2. To show that, for a and b relatively prime, a 3 6+6 4 = D 

 only when a = 2, 6 = 1, set a+b = c. The condition becomes bcg=H, 

 g=c 2 3&C+36 2 . First, let c be not divisible by 3. Then 6, c, g are rela- 

 tively prime and hence each is a square. Set g=(bmln c} 2 and solve for 

 6/c. If m is not divisible by 3, c = (ra 2 - 3rc 2 ) , 6 = (2mn - 3?i 2 ) . For the 

 lower sign, c is not a square. Hence c = m z 3n 2 =[3 = (m np/q} 2 , 

 m/n = (p z +3q*)f(2pq). Then 6/n 2 = G/(pg), G = p 2 -3pq+3q 2 . Thus 

 pqG=\3, so that the method of descent applies. Next, for m = 3k, 

 b : c = ri L c Zkn : n? 3k 2 . As before, 



c = n 2 -3k 2 =(n-kp/qY, b/n 2 = (p 2 +3g 2 -4pg)/(3g 2 +p 2 ). 



Hence (3q 2 +p 2 )(p-q}(p-3q} = D. Let p-q = t, p-3q = u. Then 



and the method of descent applies. Finally, let c = 3d. Then 



is of the initial type with the former 6, c replaced by d, 6. Since 6 is prime 

 to 3, the descent applies. It is stated that a like proof shows that x 5 1 4= D . 



1 Diophanti Alex. Arith., 1621, 423-5. 



2 Oeuvres, I, 333-4; French transl., Ill, 269. 



3 Oeuvres, II, 345, 434, letters to Digby, Aug., 1657, and to Carcavi, Aug., 1659. E. Brassinne, 



Precis dee Oeuvres math, de P. Fermat et de 1' Arith. de Diophante, Me"m. Acad. Sc. 

 Toulouse, (4), 3, 1853, 122, 164. 



4 Comm. Acad. Petrop., 10, 1738, 145; Comm. Arith. Coll., I, 33-34; Opera Omnia, (1), II, 



56-58. Proof republished by E. Waring, Medit. Algebr., ed. 3, 1782, 374-5. 



533 



