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I don't know where to start. Let $2k$ be an even integers and $a^2 - b^2$...

hjhjhj57
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Aze Aicrag
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  • This may be useful http://math.stackexchange.com/questions/263101/prove-every-odd-integer-is-the-difference-of-two-squares – hjhjhj57 Mar 06 '15 at 07:53

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I do not believe this question is correct as stated. A counterexample is the number $2$. $2$ is not the difference of squares. If it were, say $a^2 - b^2 = 2$, then $(a-b)$ would divide $2$. So $(a-b)$ must one of $1$ or $2$, since $a-b$ is positive. $a+b$ must then be $2$ or $1$ respectively. In either case, we are forced to have that $(a+b) + (a-b) = 2a = 3$. But $2$ does not divide $3$.

What is true, however, is that every integer of the form $4k$ for some integer $k$ is the difference of squares. We may assume $k$ is positive. If $k = 0$, then $a=b=0$ works. If $k$ is negative, then assuming the above, there are $a$ and $b$ so that $a^2 - b^2 = -k$. Then $b^2 - a^2 = k$.

Now, simply notice that $$(k+1)^2 - (k-1)^2 = k^2 + 2k + 1 - k^2 + 2k - 1 = 4k,$$ so $4k$ is the difference of squares.

BigMathTimes
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I think we are lacking informations...

Trivial in R If a and b belong to R then if if k positive a = sqrt(2k) and b = 0. If k negative a = 0 b = sqrt (-2k)

False in N Starting point is the fact that every odd number is the difference if two squares.

Jdffdffttt
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  • In these types of problems "squares" mean "square integers". – epimorphic Mar 06 '15 at 08:06
  • I think you need to flesh this answer out a bit more. For instance why should I believe that every odd number is the difference of two squares? – Spencer Mar 06 '15 at 08:09