Does the fact that $x^2=(x-1)(x+1)+1$ have a name?

Question

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?

Answer 1

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.

Answer 2

If you were to subtract 1 from both sides, you would get $$ x^2 - 1 = (x + 1)(x - 1)$$ This is a simpler case of the more general factoring of the difference of squares: $$ x^2 - b^2 = (x + b)(x - b)$$

For more information see Wikipedia's Difference of two squares article.

Answer 3

Trickery;

$x^2= x^2 -1 +1 = (x^2-1) +1=$

$ (x+1)(x-1)+1$.

Used: $x^2-1=(x+1)(x-1)$.

P.S. Example :

$19^2= (20)(18)+1 = 360 +1=$

$ 361$.

Answer 4

A similar formula for $x^3$ is $$x^3=(x-1)(x^2+x+1)+1$$

You can extend this to higher powers

$$x^n= (x-1)(x^{n-1}+x^{n-2}+...+1)+1$$ Which is the basis for the sum of geometric series formula.

$$1+x+x^2+...+x^{n} = \frac {x^{n+1}-1}{x-1}$$