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Show that the solution for the Diophantine equation $x^2 - y^2 = N$ is unique if and only if $\mid N \mid$ or $\frac{\mid N \mid}{4}$, respectively, is $1$ or prime.

I have an idea of how to show $\mid N \mid$ or $\frac{\mid N \mid}{4}$ is $1$ or prime $\Rightarrow$ the solution for the Diophantine equation is unique.

We can show this by going through the separate cases $ \mid N \mid = 1$ or $p$ and $\frac{\mid N \mid}{4} = 1 $ or $p$. For example, if $\mid N \mid = 1$, then $x^2 - y^2 = (x -y)(x+y) = 1$ or $(x -y)(x+y) = -1 $. For $(x -y)(x+y) = 1$, we have $x = 1, y = 0$ and $x = -1, y = 0$ as solutions (more than one solution? Doesn't uniqueness imply only one?).

How do I show the other direction?: The solution for the Diophantine equation is unique $\Rightarrow$ $\mid N \mid$ or $\frac{\mid N \mid}{4}$ is $1$ or prime.

St Vincent
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  • The other direction is not quite true. Multiplying a number by $4$ does not change the number of representations, for the simple reason that both $x,y$ must then be even. Less clear, multiplying by $2$ does not change the number of representations, you go from some $m = x^2 + y^2$ to $2m = (x-y)^2 + (x+y)^2.$ Since you talk of uniqueness, you would be using absolute values in this and ordering them as well. – Will Jagy Jan 26 '15 at 01:50
  • "Unique" needs to be defined, since even if we do not allow negatives, we can interchange $x$ and $y$ and get, most of the time, a "new" solution. But even if we do that, there are other cases of non-uniqueness. For example $x^2+y^2=45$ has a "unique" solution. Or more boringly $x^2+y^2=9$. – André Nicolas Jan 26 '15 at 02:17

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It appears from the work in your question that the exercise is supposed to be about $x^2-y^2$, not $x^2+y^2$. As has been pointed out, the result is false for $x^2+y^2$, even with a generous interpretation of unique, so I’m going to assume that you meant $x^2-y^2$. In that case the statement is true, provided that one interprets unique rather generously by not counting solutions different only in the sign(s) of $x$ or $y$ as different. In particular, we may choose to represent any solution by its non-negative representative, so that $x,y\ge 0$. Moreover, $y^2-x^2=-(x^2-y^2)$, so there is a unique solution for $-N$ if and only if there is one for $N$. Thus, we may as well assume that $N\ge 0$ and $x\ge y\ge 0$.

Suppose that $a^2-b^2=N$ and $c^2-d^2=N$, where $a,b,c,d\ge 0$, where $\langle a,b\rangle\ne\langle c,d\rangle$. Then $(a-b)(a+b)$ and $(c-d)(c+d)$ are distinct factorizations of $N$, which therefore cannot be prime. We still have to show, however, that $N/4$ cannot be prime, either.

If $N/4$ were prime, it would be an integer, and $N$ would be even. Show that if $x^2-y^2=N$, where $N$ is even, then $x$ and $y$ have the same parity: both are odd, or both are even. Conclude that $x+y$ and $x-y$ are both even. Thus,

$$\frac{a-b}2\cdot\frac{a+b}2=\frac{N}4=\frac{c-d}2\cdot\frac{c+d}2\;,$$

and you should have no trouble completing the argument.

Brian M. Scott
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  • Thanks, this is very helpful. What does the $\langle \rangle$ notation mean explicitly? In $\langle a,b \rangle \ne \langle c, d \rangle$. – St Vincent Jan 26 '15 at 10:35
  • @StVincent: You’re welcome; glad it helped. $\langle a,b\rangle$ is another notation for the ordered pair that you probably write $(a,b)$. Thus, $\langle a,b\rangle\ne\langle c,d\rangle$ says that either $a\ne c$, or $b\ne d$, or both. – Brian M. Scott Jan 26 '15 at 18:24