Just curious about this pattern
$$x^2 = (x-1)(x+1) +1$$
So:
$$\begin{align}
1^2 &= \phantom{1}0\cdot\phantom{1}2+1 = 1 \\
2^2 &= \phantom{1}1\cdot\phantom{1}3+1 = 4 \\
3^2 &= \phantom{1}2\cdot\phantom{1}4+1 = 9 \\
4^2 &= \phantom{1}3\cdot\phantom{1}5+1 = 16 \\
9^2 &= \phantom{1}8\cdot10+1 = 81 \\
15^2 &= 14\cdot16+1 = 225
\end{align}$$
and so on.
Then, to know any number raised to the power of $2$, you can multiply the previous number $(x-1)$ by the next one $(x+1)$, and add $1$.
So, to know the squared root of a number like $64$, you have to substract $1$ ($63$) and look for two numbers multiplied are $63$ and subtracted are $2$:
$$x \cdot y = 63 \qquad x – y = 2$$
Solving the equation you get $9$ and $7$ (or $-7$ and $-9$). The number between these is the square root ($8$).
I don't know if this apply for power of $3$.
Does this fact/theorem/relation has a name or something?
Best Answer
The equation in your pattern is $x^2 = (x-1)(x+1) + 1$.
Both sides of the equation evaluate to $x^2$. It's just written differently on the right side. Expand the right side (using FOIL) and you get
$$(x-1)(x+1) + 1$$ $$= x^2 - x + x -1 + 1$$ $$= x^2 $$
This is basic algebra. No special theorems are involved.