Prove that there are no $x,y ∈ \mathbb N$ for which $x^2-y^2 = 10$ I began by factoring and got $(x+y)(x-y) = 10$
Then I tried cases and was able to prove the ones where $x$ and $y$ are equal-> because the equation will result to zero.
and also where $x < y$, because the answer will be negative. 
How can I prove when $x > y$... or is there an easier way to do it? I think this is a proof by contrapositive?
 A: Definitely $x + y \ge x - y$. So you can consider following cases: $x + y = 10$ and $x - y = 1$ or $x + y = 5$ and $x - y = 2$. It is easy to see that in both cases $x$ and $y$ are not integer. 
A: $$
10 = (x+y)(x-y).
$$
There are only so many ways to factor $10$:
$$
1\cdot10,\qquad 2\cdot5.
$$
Either $x+y=10$ and $x-y=1$ (in which case $x=5.5$) or $x+y=5$ and $x-y=2$ (in which case $x=3.5$).
A: Note that $x+y$ and $x-y$ must be of the same parity. If both are even, then $4 \mid (x^2-y^2)$. If both are odd, then $(x^2-y^2)$ is also odd. $10$ is neither divisible by $4$ nor odd.
A: (I would normally have made this argument. But there are many ways to proceed. For example, the idea here is that squares get farther and farther apart, so when $x$ and $y$ are large, the difference $x^2-y^2$ is too big to be 10.  After eliminating these, only a finite number of cases are left.) 
Clearly $x>y$, so $k = x-y$ is positive.  Then we want $(y+k)^2  - y^2 = 10$, so $k^2 + 2yk= 10 $.  This is an increasing function of $k$ and $y$, and it is already too big whenever $y\ge5$ or  $k\ge3$.  So we have at most 8 cases to examine, consisting of $y\in\{1,2,3,4\}$ and $k\in\{1,2\}$.  These give, respectively:
$$\begin{array}{c|rr}
  & 1 & 2 \\ \hline
1 & 3 & 8 \\
2 & 5 & 12 \\
3 & 7 & 16 \\
4 & 9 & 20 \\
\end{array}$$
So there is not.
(The elements in the table are precisely the possible values of $x^2-y^2$ for small $x$ and $y$; for example $16 = 5^2 - 3^2$. The argument shows that for larger $x$ or $y$, the differences $x^2-y^2$ are all larger than 10.)
A: In general, a number $n\in\mathbb N$ is a difference of two squares, $n=x^2-y^2$, iff $n$ is odd or a multiple of $4$. This can be proved as follows, starting with $n=x^2-y^2=(x-y)(x+y)$:
If $n$ is odd, then one can solve $x-y=1$ and $x+y=n$ and get $x=\dfrac{n+1}{2}$ and $y=\dfrac{n-1}{2}$.
If $n$ is a multiple of $4$, then one can solve $x-y=2$ and $x+y=\dfrac{n}{2}$ and get $x=\dfrac{n}{4}+1$ and $y=\dfrac{n}{4}-1$.
Since $10$ is neither odd nor a multiple of $4$, it cannot be written as a difference of two squares.
A: Different approach:
If x <= y then it x*2 - y*2 < 0.
Hence x > y. x = y+z where z is of N.
x*2 - y*2 = y*2 +2yz +z*2 -y*2 = 2yz +z*2.
We are looking for 2 natural numbers y, z where z(2y +z) = 10.
There are 4 options z = 1, 2, 5, 10.
For any of these values there is no natural y that satisfies the equation
