Prove that for all nonnegative integers a and b, such that 2a² + 1 = b², there are two nonnegative integers c and d such that 2c² + 1 = d² and a + c + d = b, or give a counterexample.
(For example if a = 0 and b = 1, or a = 2 and b = 3 then c=0 and d = 1.)
From the example,
a b c d
---------------
0 1 0 1
2 3 0 1
Let a > 2
let c = 3a - 2b and d = 3b - 4a.
3a < 2b => 9a^2 < 4b^2 => 9a^2 < 4(2a^2 + 1) => a^2 < 4 =><=
Therefore, c >= 0.
3b < 4a => 9b^2 < 16a^2 => 9(2a^2 + 1) < 16a^2 => 2a^2 < -9 =><=
Therefore, d >= 0.
b = a + (3a - 2b) + (3b - 4a) = a + c + d.
d^2 = (3b - 4a)^2 = 9b^2 - 24ab + 16a^2
= (18a^2 - 24ab + 8b^2) + 1 + (b^2 - 2a^2 - 1)
= 2(3a - 2b)^2 + 1 + (0)
= 2c^2 + 1.
Edited on July 13, 2005, 6:12 pm
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Posted by Bractals
on 2005-07-13 18:09:21 |
Solution
