Prove that there do not exist positive integers $x$ and $y$ with $x^2 - y^2 = n$ I'm working on a homework problem that is as follows:

Suppose that $n$ is a positive even integer with $n/2$ odd. Prove that there do not exist positive integers $x$ and $y$ with $x^2 - y^2 = n$.

It looked like a good candidate for proof by contradiction. So I know that I still assume the argument "$n$ is a positive even integer with $n/2$ odd" but will try to show that there exists positive integers $x$ and $y$ with $x^2 - y^2 = n$, and if this reaches a contradiction then I have proven the original conjecture.
So I started with the $n/2$ is odd and rewrote it as $n/2 = 2k+1$, for some $k$ in the integers.
Then I knew that $n = 4k+2$, and then tried so equate that with $x^2 - y^2 = 4k+2$.
EDIT: I then recognized that $x^2 - y^2$ is equivalent to $(x + y)(x - y)$ but that doesn't seem to be very helpful, because if you divide one or the other out you get a term on the RHS in terms of k and x and y.
I'm going to keep playing with it but I don't really have any good strategies going forward. Any help is appreciated!
 A: HINT


*

*$x^2 - y^2 = (x-y)(x+y)$

*If $x-y$ is even, then so is $x+y$

A: For every integer $x$, $x^2=0$ or $1\pmod{4}$. Hence, for every integers $x$ and $y$, $x^2-y^2=1$ or $0$ or $-1\pmod{4}$. On the other hand, if $n=2(2k+1)$ for some integer $k$, $n=2\pmod{4}$. 
Hence, $x^2-y^2=n\pmod{4}$ with $n$ twice an odd integer is impossible.
A: You look at $x^2-y^2=(x+y)(x-y)$, and this factoring can be used to complete the proof. 
Since $n$ is even, at least one of $(x+y),(x-y)$ must be even, but then since $n/2$ is odd, the other must be odd. That would mean $(x+y)+(x-y)=2x$ is also odd, which contradicts the existence of integer solutions.
A: A more or less mechanical approach is to work modulo $4$. Note that for any integer $k$, $k^2\equiv 0 \pmod{4}$ or $k^2\equiv 1\pmod{4}$.
So, modulo $4$, $x^2-y^2$ can only take on the values $0$, $1$, and $-1$.

More basic, and more useful, is to suppose that $x^2-y^2=n$. Then $(x-y)(x+y)=n$. Note that for any integers $x$ and $y$, the numbers $x-y$ and $x+y$ are both even or both odd. (If we want a proof, their difference $2y$ is even.) 
In neither case is $(x-y)(x+y)$ twice an odd integer.  You started along these lines. Note that you were one step from the end.
Remark: The reason the second idea is more useful is that when it comes to solving $x^2-y^2=n$, we express $n$ as a product $st$ of integers of the same parity, and solve the system $x-y=s$, $x+y=t$. The solution is $x=\frac{s+t}{2}$, $y=\frac{s-t}{2}$. If $n$ is twice an odd integer, then this process breaks down, because one of $s$ and $t$ will be odd and the other even, so we do not get integers $x$ and $y$. 
A: $\rm\begin{eqnarray}{\bf Hint\!\!:}\rm\ \  if\ prime\,\ p\:|\:a\!-\!b\,\ then\,\ p\:|\:ab\:&\Rightarrow&\:\rm p^2\:|\:ab,\ \,by\ \ a\equiv b\!\!\!\pmod{\!p} \\
\rm Therefore\ we\ deduce\ that\,\ 2\:|\:x^2\!-\!y^2&\Rightarrow&\:\rm\ \ 4\:|\:x^2\!-\!y^2\ \ for\ \ p=2,\,\ a,b = x\pm y\end{eqnarray}$ 
A: If $x^2-y^2= N$ and $N/2$ is odd this diophantine equation has no solutions. In fact you can rewrite this equation: $(x-y)(x+y)=N$, but $N$ is even and it is divisibile for $2$ then only one factor is even. So solving this equation with this formula: $x=(p+q)/2$ and $y=(p-q)/2$, where $pq=N$ you obtain two rational solutions.
A: Since for any 
$$a^2-b^2=(a+b)(a-b)=d_1 \cdot d_2=N$$
We have
$$\begin{align} \begin{cases}  d_1&=a+b \\ d_2&=a-b  \end{cases} \ \ \ \implies \ \ \ \begin{cases}   a&=\frac{d_1+d_2}{2}\\ b&=\frac{d_1-d_2}{2}   \end{cases}  \end{align} \ \ \implies \text{the identity}$$
$$\left(\frac{d_1+d_2}{2}\right)^2-\left(\frac{d_1-d_2}{2}\right)^2=d_1\cdot d_2$$
Meaning that $N$ needs to meet the requirement that it can be factored into two divisors of the same parity.  This can be done for any odd number, even prime, as $d_1=\text{prime}$ and $d_2=1$.  But not necessarily every even.  If the power of 2 in the prime factorization of N is greater than 1, then you have enough 2's to distribute to both divisors, $d_1$ and $d_2$. But if the power of 2 in the prime factorization of N is equal to 1 (as it would be if $\frac{n}{2}$ is odd), then that 2 will have to go in one of the divisors leaving the other odd, resulting in an odd sum or difference, resulting in a non-integer.  
A: Your equation is
$$
x^2-y^2=n\textrm{, }n>0.\tag 1
$$
I will show that 
THEOREM 1.
1) The number of solutions of (1) for any positive integer $n$ is
$$
r(n)=\sum_{d|n}\textrm{abs}\left((-1)^d+(-1)^{n/d}\right).\tag 2
$$
2) If $n=2P$, $(2,P)=1$, then (1) is imposible and the oposite.
PROOF.
1) For this assume that $r(n)$ is the number of representations of any positive integer $n$ in the form (1), with $x,y\in\textbf{Z}$. We can write $x^2-y^2=(x-y)(x+y)$ and setting $A=x-y$, $B=x+y$, we have such representations iff $x=\frac{A+B}{2}$ and $y=\frac{A-B}{2}$ i.e that is iff $AB=n$ and $A,B$ both even or both odd. Hence we can write that, for any positive integer $n\neq 0$, the number of solutions of (2) is
$$
r(n)=2\sum_{
\begin{array}{cc}
  1\leq A,B\leq n\\
 AB=n\\
 \frac{A+B}{2}=\textrm{integer}\\
 \frac{A-B}{2}=\textrm{integer}
\end{array}
}1=2\sum_{k|n}\left|\frac{(-1)^k+(-1)^{n/k}}{2}\right|.
$$
Hence finaly
$$
r(n)=\sum_{d|n}\textrm{abs}\left((-1)^d+(-1)^{n/d}\right).
$$
2) Moreover about the second state of the theorem we have: 
When $n$ is even with the prime decomposition  $n=2p_1^{a_1}p_2^{a_2}\ldots p_s^{a_s}$, $p_1<p_2<\ldots<p_s$, $p_i=$primes>2, $i=1,2,\ldots,s$, then in each $d|n$, we have  $d=$even and $n/d=$odd or $d=$odd and $n/d=$even. Hence all $\textrm{abs}\left((-1)^d+(-1)^{n/d}\right)=0$ and $r(n)=0$. QED
