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I am currently working on a problem with Diophantine equations

I need to prove (if it is true, which seems to be the case) the following :

Let $x,y \in \mathbb{N}$ be natural numbers, and $n \in \mathbb{N}$.

Let E be the following equation $$x^2-y^2=n \;(E)$$

then the couple $(x,y)=(\frac{n+1}{2};\frac{n-1}{2})$ is the unique solution iff $n$ is both an odd and prime number

What I managed to prove so far :

E has at least one solution iff $n=pq$ with $p \& q $ both odd or both even.

From there, I also proved that if $n$ is odd, then E has at least one solution.

Is it possible to prove what I need from this point ? Any help will be appreciated !

T.D

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    $\begingroup$ What is the relevance of $n$ being prime? Surely there are solutions with $n$ an odd composite, eg $(\frac{15+1}{2})^2-(\frac{15-1}{2})^2=8^2-7^2=15$? $\endgroup$ – Adam Bailey Oct 28 '19 at 22:41
  • $\begingroup$ @AdamBailey I have a program that uses odd prime numbers to do some deciphering. This could be of use to check the key provided by the user. $\endgroup$ – T.D. Oct 29 '19 at 7:41
  • $\begingroup$ @AdamBailey But the solution is not unique in this case, we can also write $15=4^2-1^2$. $\endgroup$ – Peter Nov 4 '19 at 10:55
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  • Suppose , $\ n\ $ is an odd prime number. $$x^2-y^2=n$$ means $$(x-y)(x+y)=n$$

Since $\ x-y\le x+y\ $ holds, we can conclude that $$x-y=1$$ $$x+y=n$$ implying $\ x=\frac{n+1}{2}\ $ , $\ y=\frac{n-1}{2}\ $, which is the unique solution

  • Suppose , we have the given solution. If we define $\ k:=\frac{n-1}{2}\ $, we have the solution $(k,k+1)$. Since $\ (k+1)^2-k^2=2k+1\ $ is odd , we can conclude that $\ n\ $ is odd. If $n$ is odd and composite, there are odd numbers $(a,b)$ with $\ 1<a\le b $ and $\ ab=n\ $. Solving the system $$x+y=b$$ $$x-y=a$$ gives then a solution different from the given one, namely $$x=\frac{b+a}{2}$$ $$y=\frac{b-a}{2}$$ which differs from the given solution because $\ 1<a\le b<n\ $ implies $\ b-a<n-1\ $.

Hence the solution is not unique.

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  • $\begingroup$ Note that for the criterion it is necessary that $0$ belongs to the natural numbers, otherwise $9$ would be a counterexample. $\endgroup$ – Peter Nov 4 '19 at 11:16
  • $\begingroup$ Thanks I understand most of it , however why would the last system give a solution different from the given one? $\endgroup$ – T.D. Nov 4 '19 at 18:29
  • $\begingroup$ Please add it to your answer ;) $\endgroup$ – T.D. Nov 5 '19 at 9:15

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