Diophantine equation, help needed

Question

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

Answer 1

• 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.