caledonia Posted February 16, 2014 Posted February 16, 2014 m^2 = n^2 * 2 has no solution for integers m and n because root two is irrational. But m^2 = n^2 * 2 -1 does have solutions, the first of these being 7&5, 41&29, 239&169, 1393&985, 8119&5741. I believe that there are an infinite number of solutions, in other words for all N, there exists a solution with m and n both greater than N. Can anyone give me a proof ?
daniton Posted February 16, 2014 Posted February 16, 2014 Why don't you try substituting m=n+d where d is some constant and solve the resulting equation.
studiot Posted February 17, 2014 Posted February 17, 2014 (edited) I don't have a full solution, but try the substitution [math]n = \sqrt 2 p[/math]Which leads to the condition [math]p = \sqrt {\frac{{1 + {m^2}}}{4}} [/math] Which should be easier to handle. Edited February 17, 2014 by studiot
Olinguito Posted March 4, 2014 Posted March 4, 2014 (edited) But m^2 = n^2 * 2 -1 does have solutions, the first of these being 7&5, 41&29, 239&169, 1393&985, 8119&5741. I believe that there are an infinite number of solutions, in other words for all N, there exists a solution with m and n both greater than N. Can anyone give me a proof ? This is an example of Pell’s equation. Edited March 4, 2014 by Olinguito 1
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