The following Diophantine Problem was proposed for solution some months ago in a Periodical Journal, which has since been discontinued. To those who are interested in speculations of this nature, we presume that the following solution, forwarded by Professor Strong, of Hamilton College, will not be unacceptable.
Problem.
To find three positive rational Numbers, x, y, and z, such that x2 – y, x2 – z, y2 – x, and y2 – z may all be squares.
Assume x – ay for the root of the square x2 – y: then x2 – y = (x – ay)2, whence x = a2y + 1 2a . In like manner, by assuming x – bz for the root of the square x2 – z, we find z = 2bx – 1 b2 . But y2 – x = y2 – a2y + 1 2a , (since x = a2y + 1 2a ); and as this is to be made a square, assume y – c ( a2y + 1 2a ) for its root; whence, by proceeding as before, we find y = 2a + c2 4ca – a2c2 . But x = a2y + 1 2a = (by substituting for y its value) a2 + 2c 4ca – c2a2 . Again z = 2bx – 1 b2 = (by substituting for x its value)
2b ( a2 + 2c 4ca – c2a2 ) – 1 ; hence
b2
y2 – z = ( 2a + c2 4ca – c2a2 )2 × b2 – 2b ( a2 + 2c 4ca – c2a2 ) + 1
b2
(by substituting for y and z their values;) and as this also is to be made a square, assume for its root be – 1 b . Then we shall have
( 2a + c2 4ca – c2a2 )2 × b2 – 2b ( a2 + 2c 4ca – c2a2 ) + 1 = (be – 1)2 ; from which, by reduction,
b = 2 × e(4ca – c2a2)2 – (a2 + 2c)(4ca – c2a2) e2(4ca – c2a2)2 – (2a + c2)2 .
Hence the values of the required numbers are as follows: z = 2bx – 1 b2 , (in which the value of b is to be found from the last equation,) x = a2 + 2c 4ca – c2a2 , and y = 2a + c2 4ca – c2a2 .
The numbers a, c, and e, are to be so assumed that x, y, and z may come out positive. If a = 1, c = 2, and e = 2, then will x = 5/4, y = 3/2, and z = 14/9, which numbers will be found upon trial to satisfy the question. It may also be observed that c and a being positive, ca must not exceed 4; but the form of the above expressions for x, y, z, will be sufficient to direct us how a, c, and e, are to be assumed.