$x^2 = (x-1)(x+1)$.
Does anybody know if this is true for any numbers (specifically the larger the better)?
Because: $256^2 = 255 \cdot 257 + 1$
It's very very close, just an interesting thing I noticed when dealing with 2-byte integers in computer programming.
By the distributive property, $$(x-1)(x+1)=(x-1)\cdot x+(x-1)\cdot1=x^2-x+x-1=x^2-1,$$ which means that it is false for all integers $x$.
In general, we have $$(x-y)(x+y)=x^2-y^2.$$ Yours is the case $y=1$.
This identity, $(x-1)(x+1) = x^2 - 1$, or more generally, $(x-a)(x+a)=x^2 - a^2$, occurs a lot in algebra. For example
\begin{align} 199 \times 201 = (200-1)(200+1) = 200^2 - 1^2 = 40000 - 1 &= 39999 \\ (\sqrt 7 - \sqrt 2)(\sqrt 7 + \sqrt 2) = 7 - 4 &= 3 \end{align}
You can also use it to solve the "sum-product problem". For example, to solve the system of equations: $$\text{$xy = -55 \quad$ and $\quad x+y = 6$.}$$
If $x+y = 6$, then $\frac 12(x+y) = 3$ is half-way between $x$ and $y$. That means there is some number, say $z$, such that $x = 3-z$ and $y = 3 + z$. We can now reason
\begin{align} xy &= -55 \\ (3-z)(3+z) &= -55 \\ 9 - z^2 &= -55 \\ -z^2 &= -64 \\ z^2 &= 64 \\ z &= 8 &\{\text{You could also have said $z = -8$.}\} \\ x = 3-8 &=-5 \\ y = 3+8 &= 11 \end{align}
