caledonia Posted February 16, 2014 Share 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 ? Link to comment Share on other sites More sharing options...
daniton Posted February 16, 2014 Share Posted February 16, 2014 Why don't you try substituting m=n+d where d is some constant and solve the resulting equation. Link to comment Share on other sites More sharing options...
studiot Posted February 17, 2014 Share 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 Link to comment Share on other sites More sharing options...
Olinguito Posted March 4, 2014 Share 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 Link to comment Share on other sites More sharing options...
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