Difference of two squares

Usage

Factorization of polynomials and simplification of expressions

The formula for the difference of two squares can be used for factoring polynomials that contain the square of a first quantity minus the square of a second quantity. For example, the polynomial ${\displaystyle x^4-1}$ can be factored as follows:

$${\displaystyle x^4-1=(x^2+1)(x^2-1)=(x^2+1)(x+1)(x-1).}$$

As a second example, the first two terms of ${\displaystyle x^2-y^2+x-y}$ can be factored as ${\displaystyle (x+y)(x-y)}$, so we have: $${\displaystyle x^2-y^2+x-y=(x+y)(x-y)+x-y=(x-y)(x+y+1).}$$ Moreover, this formula can also be used for simplifying expressions: $${\displaystyle (a+b{)}^2-(a-b{)}^2=(a+b+a-b)(a+b-a+b)=(2a)(2b)=4ab.}$$

Complex number case: sum of two squares

The difference of two squares can be used to find the linear factors of the sum of two squares, using complex number coefficients.

For example, the complex roots of ${\displaystyle z^2+4}$ can be found by introducing a factor of ${\displaystyle -i^2=1}$, and using the difference of two squares:

$${\displaystyle \begin{array}{cc}{z}^2+4& =z^2-4i^2\\ & =z^2-(2i{)}^2\\ & =(z+2i)(z-2i).\end{array}}$$

Therefore, the linear factors are ${\displaystyle (z+2i)}$ and ${\displaystyle (z-2i)}$.

Since the two factors found by this method are complex conjugates, we can use this in reverse as a method of multiplying a complex number to get a real number. This is used to get real denominators in complex fractions.

Rationalising denominators

The difference of two squares can also be used, in reverse, in the rationalising of irrational denominators. This is a method for removing surds from expressions (or at least moving them), applying to division by some combinations involving square roots.

For example, the denominator of ${\displaystyle \frac{5}{4+\sqrt{3}}}$ can be rationalised as follows:

$${\displaystyle \begin{array}{cc}{\displaystyle \frac{5}{4+\sqrt{3}}}& ={\displaystyle \frac{5}{4+\sqrt{3}}}\times {\displaystyle \frac{4-\sqrt{3}}{4-\sqrt{3}}}\\ & ={\displaystyle \frac{5(4-\sqrt{3})}{4^2-{\sqrt{3}}^2}}={\displaystyle \frac{5(4-\sqrt{3})}{16-3}}=\frac{5(4-\sqrt{3})}{13}.\end{array}}$$

Here, the irrational denominator ${\displaystyle 4+\sqrt{3}}$ has been rationalised to ${\displaystyle 13}$.

Mental arithmetic

The difference of two squares can also be used as an arithmetical shortcut. If two numbers have an easily squared average, their product can be rewritten as the difference of two squares. For example: $${\displaystyle 27\times 33=(30-3)(30+3)={30}^2-3^2=891.}$$

Difference of two consecutive perfect squares

The difference of two consecutive perfect squares is the sum of the two bases n and n + 1. This can be seen as follows:

$${\displaystyle \begin{array}{cc}(n+1{)}^2-n^2& =((n+1)+n)((n+1)-n)\\ & =2n+1.\end{array}}$$

Therefore, the difference of two consecutive perfect squares is an odd number. Similarly, the difference of two arbitrary perfect squares is calculated as follows:

$${\displaystyle \begin{array}{cc}(n+k{)}^2-n^2& =((n+k)+n)((n+k)-n)\\ & =k(2n+k).\end{array}}$$

Therefore, the difference of two even perfect squares is a multiple of 4 and the difference of two odd perfect squares is a multiple of 8.

Galileo's law of odd numbers

A ramification of the difference of consecutive squares, Galileo's law of odd numbers states that the distance covered by an object falling without resistance in uniform gravity in successive equal time intervals is linearly proportional to the odd numbers. That is, if a body falling from rest covers a certain distance during an arbitrary time interval, it will cover 3, 5, 7, etc. times that distance in the subsequent time intervals of the same length.

From the equation for uniform linear acceleration, the distance covered $${\displaystyle s=ut+\frac{1}{2}a{t}^2}$$ for initial speed ${\displaystyle u=0,}$ constant acceleration ${\displaystyle a}$ (acceleration due to gravity without air resistance), and time elapsed ${\displaystyle t,}$ it follows that the distance ${\displaystyle s}$ is proportional to ${\displaystyle t^2}$ (in symbols, ${\displaystyle s\propto t^2}$), thus the distance from the starting point are consecutive squares for integer values of time elapsed.

Factorization of integers

Several algorithms in number theory and cryptography use differences of squares to find factors of integers and detect composite numbers. A simple example is the Fermat factorization method, which considers the sequence of numbers ${\displaystyle x_i:=a_i^2-N}$, for ${\displaystyle a_i:=\lceil \sqrt{N}\rceil +i}$. If one of the ${\displaystyle x_i}$ equals a perfect square ${\displaystyle b^2}$, then ${\displaystyle N=a_i^2-b^2=(a_i+b)(a_i-b)}$ is a (potentially non-trivial) factorization of ${\displaystyle N}$.

This trick can be generalized as follows. If ${\displaystyle a^2\equiv b^2}$ mod ${\displaystyle N}$ and ${\displaystyle a\not\equiv \pm b}$ mod ${\displaystyle N}$, then ${\displaystyle N}$ is composite with non-trivial factors ${\displaystyle gcd(a-b,N)}$ and ${\displaystyle gcd(a+b,N)}$. This forms the basis of several factorization algorithms (such as the quadratic sieve) and can be combined with the Fermat primality test to give the stronger Miller–Rabin primality test.

Generalizations

The identity also holds in inner product spaces over the field of real numbers, such as for dot product of Euclidean vectors:

${\displaystyle \mathbf{a}\cdot \mathbf{a}-\mathbf{b}\cdot \mathbf{b}=(\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})}$

The proof is identical. For the special case that a and b have equal norms (which means that their dot squares are equal), this demonstrates analytically the fact that two diagonals of a rhombus are perpendicular. This follows from the left side of the equation being equal to zero, requiring the right side to equal zero as well, and so the vector sum of a + b (the long diagonal of the rhombus) dotted with the vector difference a − b (the short diagonal of the rhombus) must equal zero, which indicates the diagonals are perpendicular.

Difference of two nth powers

If a and b are two elements of a commutative ring, then

$${\displaystyle a^n-b^n=(a-b)(\sum _{k=0}^{n-1}{a}^{n-1-k}{b}^k).}$$

The second factor looks similar to the binomial expansion of ${\displaystyle (a+b{)}^{n-1}}$, except that it does not include the binomial coefficients ⁠${\displaystyle (\genfrac{}{}{0ex}{}{n-1}{k})}$⁠.

Difference of two nth powers of matrices

In the case of two square matrices A and B, the upper identity does generally not hold true since the multiplication of matrices is not commutative. A similar result still holds:

$${\displaystyle {\mathbf{A}}^n-{\mathbf{B}}^n=\sum _{k=1}^n{\mathbf{A}}^{k-1}(\mathbf{A}-\mathbf{B}){\mathbf{B}}^{n-k}.}$$

Proof via induction: For ${\displaystyle n=1}$, the statement is true. Assuming the equation holds for an ${\displaystyle n\ge 1}$, then the induction step is

$${\displaystyle \begin{array}{cc}{\mathbf{A}}^{n+1}-{\mathbf{B}}^{n+1}& =\mathbf{A}{\mathbf{A}}^n-\mathbf{A}{\mathbf{B}}^n+\mathbf{A}{\mathbf{B}}^n-\mathbf{B}{\mathbf{B}}^n\\ & =\mathbf{A}({\mathbf{A}}^n-{\mathbf{B}}^n)+(\mathbf{A}-\mathbf{B}){\mathbf{B}}^n\\ & =\mathbf{A}\sum _{k=1}^n{\mathbf{A}}^{k-1}(\mathbf{A}-\mathbf{B}){\mathbf{B}}^{n-k}+(\mathbf{A}-\mathbf{B}){\mathbf{B}}^n\\ & =\sum _{k=0}^n{\mathbf{A}}^k(\mathbf{A}-\mathbf{B}){\mathbf{B}}^{n-k}\\ & =\sum _{k=1}^{n+1}{\mathbf{A}}^{k-1}(\mathbf{A}-\mathbf{B}){\mathbf{B}}^{(n+1)-k}.\end{array}}$$ That is, the statement holds true for ${\displaystyle n+1}$ and the result follows by the principle of induction.