[math-fun] Diophantine
a^2 - b^2 = c^2 - d^2 = e^2 - f^2 a^2 + b^2 = c^2 + f^2 = e^2 + d^2 Any solutions in nonzero integers?
On 11/4/2014 9:18 PM, David Wilson wrote:
a^2 - b^2 = c^2 - d^2 = e^2 - f^2
a^2 + b^2 = c^2 + f^2 = e^2 + d^2
Any solutions in nonzero integers?
For any nonzero N, you could take a = b = c = d = e = f = N. A little more interesting: If X, Y, Z are integers whose squares are in arithmetic progression (such as X = 1, Y = 5, Z = 7), then a = b = Y, c = d = Z, e = f = X will work. -- Fred W. Helenius fredh@ix.netcom.com
On 11/4/2014 9:18 PM, David Wilson wrote:
a^2 - b^2 = c^2 - d^2 = e^2 - f^2
a^2 + b^2 = c^2 + f^2 = e^2 + d^2
Any solutions in nonzero integers?
Given two triples of integers with their squares in arithmetic progression, with both progressions having the same common difference (e.g., 1, 29, 41 and 23, 37, 47), you can construct a solution with distinct nonzero integers, such as a = 37, b = 29, c = 23, d = 1, e = 47, f = 41. -- Fred W. Helenius fheleni@emory.edu
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Fred W. Helenius