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 Sophie Squares (Posted on 2005-07-13)
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.)

 See The Solution Submitted by Gamer Rating: 3.2500 (4 votes)

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 No Subject | Comment 6 of 9 |

(a + c + d ) = b --------> (c + d) = (b - a)

a^2 + c^2 + d^2 + 2 * (ac +ad + cd) + acd = b^2

b^2 = 2a^2 + 1

d^2 = 2c^2 + 1

a^2 + c^2 + (2c^2 + 1) +  2a(c+d) + acd + 2a = (2a^2 + 1)

3c^2 - a^2 + 2a(b - a) + acd + 2a = 0

3c^2 - a^2 + 2ab - 2a^2 + acd + 2a = 0

3c^2 - 3a^2 + 2ab + acd + 2a  = 0

3(c+a)(c-a) + a(2b + cd + 2) = 0

Since a, b, c, and d are nonnegative integers, the first term must be negative, so c < a.

Making a = c + r ------> c = (a - r), r > 0.

2ab + acd + 2a = 3(c+a)(c-a)

2ab + a(a-r)d + 2a = 3(2a-r)(-r)

2ab + da(a-r) + 2a = - 6ar + 3r^2

d = (- 6ar + 3r^2 - 2ab - 2a)/a(a-r)

3r^2 - 6ar - 2a(b+1) >=0

r = [6a +- sqrt (36a^2 + 24a(b+1))]/6

So, it works only for

36a^2 + 24ab + 24 >=0 and a square, and that makes r integer.

With the value of r, we found the value of c = a - r, and d = b - a - c = b - a - a + r = b - 2a + r.

ThereŽs some mistake in my accounts, but I think that reasoning like this we find the answer.

Where I made a mistake ? IŽll post this and review my calculus.

 Posted by pcbouhid on 2005-07-13 18:23:54

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