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This is a problem from Discrete Mathematics and its Applications. enter image description here

I understand the basic ideas of the direct proof. Basically a proof is a conclusion from a series of steps to establish the truth of a mathematical statement. And a direct proof starts with an assumption and then works through steps to a conclusion that can be reached from that assumption. Correct me if any of this is wrong by the way, this is just my interpretation from https://courses.cs.washington.edu/courses/cse311/14au/slides/lecture07-filled.pdf, slide 8

Here is my work so far enter image description here

I only introduce the variables y and r if the integer to be checked(x) is odd. y and r are just used to find one possibility of the difference of two squares. I did factoring with y^2 - x^2 and ended up with 2k + 1 = (y + r)(y-r). My question is where do i go from here equality between the terms? The structure of the problem really makes me want to solve this like a quadratic but there are three variables involved, not just one. What is the next step?

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You can try the following: 2k+1=(k+a)^2-(k+b)^2 Expand the right hand side and compare with the left hand side, you will get a pair of simultaneous equations. Solve the simultaneous equations and you will have a=1, b=0. Therefore 2k+1=(k+1)^2-k^2, a difference of two squares.

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  • $\begingroup$ where did you get k+a and k+b from? aren't the squares are supposed to be one some arbitrary number, not in relationship to k? $\endgroup$ Jan 11 '15 at 9:41
  • $\begingroup$ Like how would you get to k+a and k+b from what i had, y and r? $\endgroup$ Jan 11 '15 at 9:42
  • $\begingroup$ As of the question is about existence of two squares such that their difference is odd. I just try one and that works. Your proof should start from assuming a form of a general form of an odd number, which you did, 2k+1. Then you try to find out two numbers that satisfy the following conditions. Your numbers chosen should be related to the odd number that you initial set, which is 2k+1. $\endgroup$
    – Novice
    Jan 11 '15 at 9:46
  • $\begingroup$ The arbitrariness of the two numbers is not an issue here. The question that is given to you is given an odd number, find two numbers such that the difference in their squares is that odd number. $\endgroup$
    – Novice
    Jan 11 '15 at 9:48
  • $\begingroup$ like in terms of a proof though, how would i explain that i came up with k + a and k + b or the idea of having some distance from k? cause my initial quantifier made sense, some y and r $\endgroup$ Jan 11 '15 at 9:50

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