Prove that $x^2-y^2=10$ has no integer solution I am trying to provide a proof by contradiction that $$x^2-y^2=10$$ has no integer solution. This is what I did, I am not sure it is correct, since it appears so simple...
Suppose an integer solution exists.
It follows that $$(x-y)(x+y)=10$$
At least one of these has to be even for 10 to be even. Without loss of generality I assume that $$(x+y)$$ is even.
It follows that $$x+y=2c$$ with c an integer
$$2c(x-y)=10$$
$$c(x-y)=5$$
Then both c and $(x-y)$ have to odd, but this contradicts the claim that $(x+y)$ is even.
Am I making an incorrect assumption in one of these steps?
 A: Here are some different ways for doing essentially what you did:

*

*We have $$(x - y)(x + y) = 10.$$ Note that $x + y = (x - y) + \color{red}{2}y$. Thus, $x + y$ and $x - y$ have the same parity. Since $10$ is even, one of them is divisible by $2$ and thus, so is the other. But this means that $10$ is divisible by $4$.


*Let us look at the possible remainders of a square modulo $4$. Since $$(n + 4)^2 \equiv n^2 \mod 4,$$ it suffices to only consider $n = 0, \ldots, 4$. Moreover, since $$n^2 = (-n)^2 \equiv (4 - n)^2 \mod 4,$$
it suffices to only consider $n = 0, 1$.
The possible remainders of squares are now seen to be $0^2 \equiv 0$ and $1^2 \equiv 1$. But $10$ has remainder $2$ modulo $4$. Thus, the difference of two squares being $10$ is impossible.
In this case, the second seems more cumbersome than the first but the method is useful in other situations as well by considering remainders modulo other $n$ instead of $4$.
As an example, $$x^2 - 5y^{1003} = 4343434343$$ has no integer solution as you can verify by going modulo $5$.
A: As DonAntonio said, for any integer $n$,  $n^2\equiv 0\bmod4$ or $n^2\equiv 1\bmod4$
LHS:
$$x^2\equiv0\text{ or }1\bmod4$$
$$y^2\equiv0\text{ or }1\bmod4$$
$$\implies x^2-y^2\equiv(0-0)\text{ or }(0-1)\text{ or }(1-0)\bmod4$$
(i.e) The possible remainders when $x^2-y^2$ is divided by $4$ are $\{-1,0,1\}\\$
RHS:
$$10\equiv2\bmod4$$
Since $2\notin\{-1,0,1\}$, $x^2-y^2\neq10$ when $x$ and $y$ are integers.

Ok, Why does $n^2\equiv 0\bmod4$ or $n^2\equiv 1\bmod4\ $ (where $n\in\mathbb{z}$)?
Case 1: $n$ is even:
$$n=2k \text{ (where } k\in\mathbb{z})$$
$$n^2=4k^2\text{ and } 4k^2\equiv0\bmod4$$
$$\implies n^2\equiv0\bmod4 \text{ when } n \text{ is even}$$
Case 2: $n$ is odd:
$$n=2k+1 \text{ (where } k\in\mathbb{z})$$
$$n^2=(2k+1)^2$$
$$=4k^2+4k+1$$
$$=4(k^2+k)+1$$
$$\implies n^2\equiv1\bmod4 \text{ when } n \text{ is odd}$$
Hence $n^2$ ($n\in\mathbb{z}$) leaves either $0$ or $1$ as a remainder when divided by $4$
A: An integer $n$ is said to be perpetual if $|n|=2m$ for some odd natural number $m$. We are going to prove the following lemma: If $n$ is a perpetual integer then the equation $x²-y²=n$ has no integer solutions. First we note that $x²-y²=n$ has integer solutions iff $y²-x²=-n$ has integer solutions. So, without loss of generality, we assume that $n\in \mathbb{N}$ is perpetual. Now, suppose there exists an integer solution $(r,s)$ of $x²-y²=n$. Then, we get $r²-s²=n=2m$ (where $n=2m$, and $m\in\mathbb{N}$ is odd). This implies $(r+s)(r-s)=2m$. Now, $(r+s)=(r-s)+2s$ and so $r+s$ and $r-s$ must both have the same parity. However, both can't be odd since then the product $2m$ will be odd, a contradiction. So, $r+s$ and $r-s$ are both even. Put $r+s=2k$ and $r-s=2l$. Then, we get $4kl=2m$ which is equivalent to $2kl=m$. Therefore, $2|m$, contradicting that $m$ is odd. Hence, $x²-y²=n$ has no integer solutions. Now for your problem, put $m=5$ and conclude.
A: Other answers have exploited parity and divisibility properties, I'll show you a completely different approach that might be useful to tackle other problems.
In fact squares are growing quite quickly, so the difference between two squares can be small only when $x,y$ are not to far apart.
Considering $x^2-y^2=10$ WLOG we can assume $x,y\ge 0$ (since $(\pm x,\pm y)$ also solution).
Also since $x^2=10+y^2>y^2$ then $x>y$ and we can set $x=y+n$ with $n>0$.
Reporting in the equation gives: $(y+n)^2-y^2=n^2+2ny=10\implies n^2=10-\underbrace{2ny}_{\ge 0}\le 10$
Therefore $n$ can only be $1,2$ or $3$ but none of $(1+2y=10),(4+4y=10),(9+6y=10)$ has integer solutions, so the overall problem has no solutions.
