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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?

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up vote 13 down vote accepted

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.

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@user17762 how did you arrive that they cannot be divisible by 4? \ – Mac Oct 16 '13 at 0:27
@Mac Your question is unclear. All I said is that $10$ is not divisible by $4$, whereas if $x+y$ and $x-y$ are both even, then $x^2-y^2$ has to be divisible by $4$. – user17762 Oct 16 '13 at 0:29
@user17762 thank you, I understand now, I just misinterpreted your statement – Mac Oct 16 '13 at 0:30
@user17762, why do you say that 10 is not divisible by odd numbers? – Smylic Oct 16 '13 at 0:53
@Smylic All I said is $10$ is not odd. – user17762 Oct 16 '13 at 0:54

$$ 10 = (x+y)(x-y). $$

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 $x=3.5$).

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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.

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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.

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(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.)

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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

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