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Problem II.8 of the Arithmetica asks how a given square number is split into two other squares; in other words, for a given rational number k, find rational numbers u and v such that k 2 = u 2 + v 2. Diophantus shows how to solve this sum-of-squares problem for k = 4 (the solutions being u = 16/5 and v = 12/5). [29]
A French translation with detailed scholarly addenda and a critical edition of the Chinese text of both the book and its commentary by Karine Chemla and Shuchun Guo is Les neuf chapitres: le classique mathématique de la Chine ancienne et ses commentaires. Paris: Dunod, 2004. ISBN 978-2-10-049589-4.
For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2(y + 1) – 1, a true statement. It is also possible to take the variable y to be the unknown, and then the equation is solved by y = x – 1.
x + y = 4p 2, x − y = 4q 2. one obtains x = 2(p 2 + q 2), y = 2(p 2 − q 2) and so xy + 1 = (2p 2 − 1) 2 + 4(p 2 − q 4). For xy + 1 also to be a perfect square we must have p 2 − q 4 = 0, that is p 2 = q 4. Thus the following general solution is obtained: x = 2(q 4 + q 2), y = 2(q 4 − q 2). The value q = 2 yields the special solution ...
Konopasek's goal in inventing the TK Solver concept was to create a problem solving environment in which a given mathematical model built to solve a specific problem could be used to solve related problems (with a redistribution of input and output variables) with minimal or no additional programming required: once a user enters an equation, TK ...
For example, when d=4, the hash table for two occurrences of d would contain the key-value pair 8 and 4+4, and the one for three occurrences, the key-value pair 2 and (4+4)/4 (strings shown in bold). The task is then reduced to recursively computing these hash tables for increasing n , starting from n=1 and continuing up to e.g. n=4.
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On the other hand, if such a tiling uses exactly k of the 2 × 1 tiles, then it uses n − 2k of the 1 × 1 tiles, and so uses n − k tiles total. There are ( n − k k ) {\displaystyle {\tbinom {n-k}{k}}} ways to order these tiles, and so summing this coefficient over all possible values of k gives the identity.