The difference of two squares.

The difference of two squares is easily factored: We have,

$$a^2-b^2=a^2-ab+ab-b^2=a(a-b)+b(a-b)=(a+b)(a-b)=(a-b)(a+b).$$

Claim 11   Every odd positive integer $N>1$ is the difference of two positive integer squares.

Proof. Since $N$ is greater than 1, $N=n_1n_2$ for some odd positive integers $n_1$ and $n_2$ , $n_1>n_2$ . So,

$$\left(\frac{n_1+n_2}{2}\right)^2-\left(\frac{n_1-n_2}{2}\right)^2=n_1n_2=N.$$

Since $n_1$ and $n_2$ are both odd, $\frac{n_1+n_2}{2}$ and $\frac{n_1-n_2}{2}$ are integers. $\qedsymbol$

Note that $1<N=N\cdot 1,$ hence we can always choose $n_1=N$ and $n_2=1.$ Then

$$\left(\frac{N+1}{2}\right)^2-\left(\frac{N-1}{2}\right)^2=N\cdot 1=N.$$

Claim 12   If $N$ is an even positive integer then $N$ is the difference of two positive integer squares if and only if $N=4k$ for some integer $k>1$ .

Proof. $(\Rightarrow)$ Let $N=a^2-b^2=(a+b)(a-b)$ where $a$ and $b$ are positive integers, $a>b.$ Set $n_1=a+b$ and $n_2=a-b.$ So, $n_1n_2=N.$ Then

$$a=\frac{n_1+n_2}{2}$$   and$$\quad b=\frac{n_1-n_2}{2}.$$

Since $a$ and $b$ are integers then $n_1$ and $n_2$ are both odd or they are both even. They can not both be odd since $N$ is even. Hence for some integers $k_1$ and $k_2$ , $k_1>k_2$ , $n_1=2k_1$ and $n_2=2k_2$ . Therefore $N=n_1n_2=4k_1k_2=4k$ , $k>1$ .

$(\Leftarrow)$ If $N=4k$ , $k>1$ , then $k=k_1k_2$ for some positive integers $k_1$ and $k_2$ , $k_1>k_2$ . Therefore

$$N=4k=4k_1k_2=(k_1+k_2)^2-(k_1-k_2)^2.$$

$\qedsymbol$

Example: $N=15=15\cdot 1 =5\cdot 3.$ Therefore

$$15=\left(\frac{15+1}{2}\right)^2-\left(\frac{15-1}{2}\right)^2=\b... ...2} =\left(\frac{5+3}{2}\right)^2-\left(\frac{5-3}{2}\right)^2=\boxed{4^2-1^2}.$$

Example: $N=48$ . So $48/4=12=12\cdot 1=6\cdot 2=4\cdot 3.$ Therefore

$$48=(12+1)^2-(12-1)^2=\boxed{13^2-11^2}=(6+2)^2-(6-2)^2=\boxed{8^2-4^2} =(4+3)^2-(4-3)^2=\boxed{7^2-1^2}.$$

The only positive integers that can not be written as the difference of two positive integer squares are exactly $1, 4$ , and those integers $N$ such that
$4$ divides $N+2$ (that is, $2, 6, 10, \cdots$ ).

Theorem 9   The set of the differences of two squares of integers is closed under multiplication. In fact

$$\left(a^2-b^2\right)\left(c^2-d^2\right)=(ac+bd)^2-(bc+ad)^2=(ac-bd)^2-(bc-ad)^2.$$ (45)

Proof. Equation (45) is due to Diophantus and is a special case of Brahmagupta's identity

$$\left(-na^2+b^2\right)\left(-nc^2+d^2\right)=\left(nac+bd\right)^2-n\left(bc+ad\right)^2= \left(nac-bd\right)^2-n\left(bc-ad\right)^2$$ (46)

where $n=1$ . The Diophantus identity is easily derived straightforwardly.

$$\left(a^2-b^2\right)\left(c^2-d^2\right) =(a+b)(a-b)(c+d)(c-d)$$

$$=\Bigl((a+b)(c+d)\Bigr) \Bigl((a-b)(c-d)\Bigr)$$

$$=\Bigl((ac+bd)+(bc+ad)\Bigr)\Bigl((ac+bd)-(bc+ad)\Bigr)$$

$$=\boxed{(ac+bd)^2-(bc+ad)^2}$$

$$=\Bigl((a+b)(c-d)\Bigr)\Bigl((a-b)(c+d)\Bigr)$$

$$=\Bigl((ac-bd)+(bc-ad)\Bigr)\Bigl((ac-bd)-(bc-ad)\Bigr)$$

$$=\boxed{(ac-bd)^2-(bc-ad)^2} .$$

$\qedsymbol$

Example:

$$15=3\cdot 5=\left(2^2-1^2\right)\left(3^2-2^2\right) =\left(2\cdot 3+1\cdot 2\right)^2-\left(2\cdot2 +1\cdot 3\right)^2=\boxed{8^2-7^2}.$$

$$=\left(2\cdot 3-1\cdot 2\right)^2-\left(2\cdot2 -1\cdot 3\right)^2=\boxed{4^2-1^2}.$$

Example:    Solving the equation $x^2+y^2=z^2$ in relatively prime positive integers.

Solution:    We know that if $x$ and $y$ are relatively prime and $x^2+y^2=z^2$ then one of $x$ and $y$ is odd. Without loss of generality let $x>1$ be an odd positive integer. Then there exists relative prime odd positive integers $n_1$ and $n_2$ , $n_1>n_2$ , such that

$$n_1n_2=x=\left(\frac{n_1+n_2}{2}\right)^2-\left(\frac{n_1-n_2}{2}\right)^2.$$

$n_1=x$ and $n_2=1$ is one such pair.

Set

$$m=\left(\frac{n_1+n_2}{2}\right)$$   and$$\quad n=\left(\frac{n_1-n_2}{2}\right).$$

Thus, from equation (45),

$$x^2=\left(m^2-n^2\right)\left(m^2-n^2\right)=(mm+nn)^2-(mn+nm)^2 =\left(m^2+n^2\right)^2-(2mn)^2.$$

From whence

$$x=m^2-n^2,\quad y=2mn,$$   and$$\quad z=m^2+n^2$$

where $m$ and $n$ are relative prime positive integers of opposite parity, $m > n$ . This implies that positive integers $x, y,$ and $z$ are pair-wise relatively prime.

Supplemental

From equation (45) we have the inequality,

$$\left(a^2-b^2\right)\left(c^2-d^2\right)\le(ac-bd)^2.$$ (47)

And equality holds if $a=br$ and $c=dr$ for some integer $r$ .

Compare equation (45) to the Fibonacci identity ( another special case of Brahmagupta's identity. Set $n=-1$ in equation (46) and note that $\left(-x\right)^2=x^2$ ),

$$\left(a^2+b^2\right)\left(c^2+d^2\right)=(ac- bd)^2+(bc+ ad)^2=(ac+ bd)^2+(bc- ad)^2.$$ (48)

Similarly Fibonacci's identity can easily be derived straightforwardly. Set $i^2=-1.$ That is, we will factor each sum of two squares into Gaussian numbers. Hence

$$\left(a^2+b^2\right)\left(c^2+d^2\right) =(a+bi)(a-bi)(c+di)(c-di)$$

$$=\Bigl((a+bi)(c+di)\Bigr) \Bigl((a-bi)(c-di)\Bigr)$$

$$=\Bigl((ac-bd)+(bc+ad)i\Bigr)\Bigl((ac-bd)-(bc+ad)i\Bigr)$$

$$=\boxed{(ac-bd)^2+(bc+ad)^2}$$

$$=\Bigl((a+bi)(c-di)\Bigr)\Bigl((a-bi)(c+di)\Bigr)$$

$$=\Bigl((ac+bd)+(bc-ad)i\Bigr)\Bigl((ac+bd)-(bc-ad)i\Bigr)$$

$$=\boxed{(ac+bd)^2+(bc-ad)^2} .$$

This identity implies Cauchy's inequality for reals in two dimensions,

$$(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2.$$

Example 1.

In the Fibonacci identity, set

$$a=\cos \theta_1,\quad b=\sin \theta_1,$$

$$c=\cos \theta_2,\quad d=\sin \theta_2.$$

Then

$$\left(a^2+b^2\right)\left(c^2+d^2\right) =\left(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2\right)^2 +\left(\sin\theta_1\cos\theta_2+\sin\theta_2\cos\theta_1\right)^2$$

$$=\cos^2(\theta_1+\theta_2)+\sin^2(\theta_1+\theta_2)=1.$$

Similarly, the Diophantus identity (45), set

$$a=\cosh \theta_1,\quad b=\sinh \theta_1,$$

$$c=\cosh \theta_2,\quad d=\sinh \theta_2.$$

Then

$$\left(a^2-b^2\right)\left(c^2-d^2\right) =\left(\cosh\theta_1\cosh\theta_2+\sinh\theta_1\sinh\theta_2\right)^2 -\left(\sinh\theta_1\cosh\theta_2+\sinh\theta_2\cosh\theta_1\right)^2$$

$$=\cosh^2(\theta_1+\theta_2)-\sinh^2(\theta_1+\theta_2)=1.$$

Example 2.

See Multiplying Pythagorean triples.

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