Prove that $1+x+y$ not equal to $xy$, when $x$ and $y$ are positive odd numbers I was engaging in this equation and with the help of desmos.com I noticed it is not possible. So how can we prove? In which way it is easier, should I leave $x$ or $y$ alone, or is there any method different?
 A: First of all we can define odd numbers: 
$x=2m-1 \ \  \forall \ m \in \mathbb N \ \backslash \{0\}$ and $y=2n-1 \ \  \forall \ n \in \mathbb N \ \backslash \{0\}$
Then we have $1+x+y=x\cdot y\Rightarrow -1+2m+2n=(2m-1)(2n-1)$
$2m+2n-1=4mn-2m-2n+1$
$4m+4n=4mn+2$
$2(m+n)=2mn+1$
The LHS is always even and the RHS is always odd for all $n,m \in \mathbb N \ \backslash \{0\}$
A: Let's say $x \geq 3$ and $y \geq 3$ also. Then $x = 2m + 1$ with $m > 0$ and $y = 2n + 1$ likewise with $n > 0$. Then $1 + x + y = 1 + (2m + 1) + (2n + 1) = 2m + 2n + 3$. But $xy = (2m + 1)(2n$ $+ 1) = 4mn + 2m + 2n + 1$.
This means that $xy > 1 + x + y$, since $xy - (1 + x + y) = 4mn - 2$. Since $m$ and $n$ are both positive, $4mn$ has to be at least 4, which means a minimum difference of 2 between $xy$ and $1 + x + y$.
The case with either $x$ or $y$ equal to 1 should not present any special difficulty.
A: Just for fun, here is a somewhat convoluted proof, starting with a mod $6$ argument.
It is straightforward to see that the only way that $1+x+y\equiv xy$ mod $6$ with $x,y\in\{1,3,5\}$ (i.e., with $x$ and $y$ both odd) is if $\{x,y\}=\{3,5\}$.  Without loss of generality, let's suppose that $x=3u$ and $y=6v-1$, with $u$ odd. The equation now becomes
$$1+3u+(6v-1)=3u(6v-1)$$
which simplifies first to $u+2v=6uv-u$ and then to
$$u+v=3uv$$
But since $u\equiv1$ mod $2$ (i.e., since $u$ is odd), this implies $1+v\equiv v$ mod $2$, which is impossible. So there are no solutions to $1+x+y=xy$ with $x$ and $y$ both odd.
Remark: The argument here is related to the much simpler mod $4$ argument alluded to in dxiv's comment below the OP that the equation $1+x+y=xy$ can be rewritten as $(x-1)(y-1)=2$.
A: hint
$$1+x+y=xy \implies y=\frac{x+1}{x-1}$$
$$=1+\frac{2}{x-1}$$
thus
$$(x,y)=(2,3) \text{ or } (3,2)$$
