All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 5 of 9 |
` `
`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
 Posted by Bractals on 2005-07-13 18:09:21

 Search: Search body:
Forums (0)