Math Notes 2025

I enjoy tinkering with math. Unfortunately, I often forget the little things I discover. On this page you will find all the tidbits I found interesting enough to record.

I don't strive for academic perfection, however I also do not want to spread blatant errors. If you notice any errors in these notes, please email me.

See all math notes

Fibonacci Moduli

April 8th 2025 Single Page

For the Fibonacci sequence, defined by $F_{1} = 1,\> F_{2} = 1,\quad F_{n} = F_{n-1} + F_{n-2}$

I believe the following is true (Haven’t found any counterexamples)

(F_{n} \operatorname{mod} n) \ne 0\quad \text{when}\quad (n \operatorname{mod} 6) \in \{2,3,4\}

Complete Factorization

April 9th 2025 Single Page

Quick! what are all the factors of $720$ ?

Instead of giving you an answer, here’s an algorithm to find it.

First, by repeated division, find the prime factors.

\begin{align*} 720 / 2 &= 360 \\ 360 / 2 &= 180 \\ 180 / 2 &= 90 \\ 90 / 2 &= 45 \\ 45 / 3 &= 15 \\ 15 / 3 &= 5 \\ 5 / 5 &= 1 \\ \end{align*}

F_{0} = [2,\,2,\,2,\,2,\,3,\,3,\,5]

Find the all unique factors and their cartesian products in $F_{0}$ to create $P_{0}$ .

P_{0} = \{2,\,3,\,5,\,6,\,10,\,15,\,30\}

S_{0} = P_{0}

Remove one of each unique factor from $F_{0}$ to make $F_{1}$ .

F_{1} = [\,\cancel{2},\,2,\,2,\,2,\,\cancel{3},\,3,\,\cancel{5}]

Find the all unique factors and their cartesian products in $F_{1}$ to create $P_{1}$ .

P_{1} = \{\, 2,\,3,\,6 \,\}

Multiply each number in $P_{1}$ by each number in $S_{0}$ to find $S_{1}$ Remove any duplicate values and remove any values which had been found in previous $S_{n}$ .

\begin{align*} S_{1} = \{\,& 4,\,\cancel{6},\,10,\,12,\,20,\,\cancel{30},\,60\\ & \cancel{6},\,9,\,\cancel{15},\,18,\,\cancel{30},\,45,\,90\\ & \cancel{12},\,\cancel{18},\,\cancel{30},\,36,\,\cancel{60},\,\cancel{90},\,180 \}\\ S_{1} = \{\,& 4,\,10,\,12,\,20,\,60,\,9,\,18,\,45,\,90,\,36,\,180\} \end{align*}

Remove one of each unique factor from $F_{1}$ to make $F_{2}$ .

F_{2} = [\,\cancel{2},\,\cancel{2},\,2,\,2,\,\cancel{3},\,\cancel{3},\,\cancel{5}]

Find the all unique factors and their cartesian products in $F_{2}$ to create $P_{2}$ .

P_{2} = \{\, 2\, \}

Multiply each number in $P_{2}$ by each number in $S_{1}$ to find $S_{2}$ Remove any duplicate values and remove any values which had been found in previous $S_{n}$ .

\begin{align*} S_{2} &=\{\,8,\, \cancel{20},\, 24,\,40,\, 120,\, \cancel{18},\, \cancel{36},\, \cancel{90},\, \cancel{180},\, 72,\, 360\,\}\\ S_{2} &=\{\,8,\, 24,\,40,\, 120,\, 72,\, 360\,\}\\ \end{align*}

Remove one of each unique factor from $F_{2}$ to make $F_{3}$ .

F_{3} = [\,\cancel{2},\,\cancel{2},\,\cancel{2},\,2,\,\cancel{3},\,\cancel{3},\,\cancel{5}]

Find the all unique factors and their cartesian products in $F_{3}$ to create $P_{3}$ .

P_{3} = \{\, 2\, \}

Multiply each number in $P_{3}$ by each number in $S_{2}$ to find $S_{3}$ Remove any duplicate values and remove any values which had been found in previous $S_{n}$ .

S_{3} = \{\,16,\,48,\,80,\,240,\,144,\,720 \,\}

Remove one of each unique factor from $F_{3}$ to make $F_{4}$ .

F_{4} = [\,\cancel{2},\,\cancel{2},\,\cancel{2},\,\cancel{2},\,\cancel{3},\,\cancel{3},\,\cancel{5}]

Seeing as there are no more prime factors left in $F$ , Halt. The factors of $720$ are found in the union $S_0,\,S_1,\,S_2,\, S_3$ and of course $1$ .

\begin{align*} S &= \{1\}\cup S_0 \cup S_1 \cup S_2 \cup S_3 \\ S &=\{\\ &\phantom{=\{}1,\,2,\,3,\,4,\,5,\,6,\,8,\,9,\,10,\,12,\,15,\,16,\,18,\,20,\,24,\,30,\,36,\\ &\phantom{=\{}40,\,45,\, 48,\,60,\,72,\,80,\,90,\,120,\,144,\,180,\,240,\,360,\,720 \\ &\phantom{=\{}\} \end{align*}

I don’t know the name of this algorithm. I figured it out myself, but the simplicity of the algorithm leads me to believe it has been known for quite some time. There are likely more efficient algorithms, but compared to the naive algorithm of dividing by all numbers less than, it is appreciably efficient. An advantage of this algorithm is that it can be done on pen-and-paper and also can be done using a computer.

Example Python Program

Squares mod 100

April 17th 2025 Single Page

Two neat little identities for squares mod 100.

n^2\equiv (n+50)^2 \mod 100

And

n^2\equiv (50-n)^2 \mod 100

These can be algebraically shown to be true

n^2 \equiv (n^2 + 2500 + 100n) \mod 100

What about the generic case of a modulus $k$ ? What is the specific $m$ to be added to $n$ to make the squares congruent?

n^2 \equiv (n+m)^2 \mod k

Expand

n^2 \equiv (n^2 + m^2 + 2mn) \mod k

The trivial case is for $m^2 \equiv 0 \mod k$ and $2mn \equiv 0 \mod k$ . Let’s just ignore the possibility that both of those aren’t true and yet still $(m^2 + 2mn) \equiv 0 \mod k$

The smallest natural number $m$ such that $2mn \equiv 0 \mod k$ for any integer $n$ is clearly $k/2$ if $k$ is even, and $k$ if $k$ is odd.

The smallest natural number $m$ such that $m^2\equiv 0 \mod k$ is a bit trickier.

$m^2 = k^2$ Obviously works, but the case of $m^2 = (k/2)^2$ only works if $k$ is divisible by $4$

m = \frac{k}{2}

m^2 = \frac{k^2}{2^2}

\frac{m^2}{k} = \frac{k}{4}

So for

n^2 \equiv (n+m)^2 \mod k

If $k$ is divisible by $4$

m = k/2

Otherwise:

m = k

Calculus of Inverses

May 10th 2025 Single Page

Let $f(x), g(x)$ be functions that are defined and differentiable over some interval. Further, let

\begin{aligned} f(g(x)) &= x\\ g(f(x)) &= x\\ \end{aligned}

Such that $g,f$ are inverses.

Differentiate by chain rule.

f'(g(x))g'(x) = 1

g'(x) = \frac{1}{f'(g(x))}

This already has some neat results

\begin{aligned} f(x) &= \sin(x)\\ g(x) &= \sin^{-1}(x)\\ \frac{d}{dx} \sin^{-1}(x) &= \frac{1}{\cos(\sin^{-1}(x))} \end{aligned}

\begin{aligned} f(x) &= \cos(x)\\ g(x) &= \cos^{-1}(x)\\ \frac{d}{dx} \cos^{-1}(x) &= -\frac{1}{\sin(\cos^{-1}(x))} \end{aligned}

\begin{aligned} f(x) &= \tan(x)\\ g(x) &= \tan^{-1}(x)\\ \frac{d}{dx} \tan^{-1}(x) &= \frac{1}{\sec^2(\tan^{-1}(x))} \end{aligned}

These aren’t the standard forms you’re familiar with, but they are equivalent.

By this we can create a very annoying looking integral (for fun of course).

\int \frac{1}{\sec^2(\tan^{-1}(x))\tan^{-1}(x)^2} \,dx

Inverse Inverse Trigonometry

May 10th 2025 Single Page

In the section on derivatives of inverses, these came up

\begin{aligned} \cos(\arcsin(x))\\ \sin(\arccos(x))\\ \sec^2(\arctan(x))\\ \end{aligned}

These forms can be simplified to remove any notion of trigonometry.

Draw a right triangle with angle $\theta$ , adjacent side length $\sqrt{1-x^2}$ , opposite side length $x$ , and hypotenuse side length $1$ .

Of course, $\sin(\theta) = x$ , so $\arcsin(x) = \theta$ within bounds of course.

And by definition $\cos(\theta) = \sqrt{1-x^2}$ , so

\cos(\arcsin(x)) = \sqrt{1-x^2},\quad\text{where}\quad -1\leq x\leq 1

Some more identities found by similar method

\begin{aligned} \cos(\arcsin(x)) &= \sqrt{1-x^2},\quad&\text{where}\quad &-1\leq x\leq 1\\ \sin(\arccos(x)) &= \sqrt{1-x^2},\quad&\text{where}\quad &-1\leq x\leq 1\\ \tan(\arcsin(x)) &= \frac{x}{\sqrt{1-x^2}},\quad&\text{where}\quad &-1\leq x\leq 1\\ \tan(\arccos(x)) &= \frac{\sqrt{1-x^2}}{x},\quad&\text{where}\quad &-1\leq x\leq 1\\ \end{aligned}

Arctangent is a bit more complex to deal with. Let’s do $\sec^2(\arctan(x))$

Draw a right triangle with angle $\theta$ , adjacent side length $A$ , opposite side length of $O$ , and hypotenuse side length $1$ .

We know that $\tan(\theta) = \frac{O}{A}$ so set $x =\frac{O}{A}$ . Also known is $O^2+A^2 = 1$

\begin{aligned} A^2+x^2A^2&=1\\ A^2(1+x^2)&=1\\ A^2 &= \frac{1}{1+x^2}\\ A &= \frac{1}{\sqrt{1+x^2}}\\ Ax &= O\\ O &= \frac{x}{\sqrt{1+x^2}} \end{aligned}

And so, the triangle has been completed.

Finally, $\csc^2(\theta) = \frac{1}{A^2}$

This leads to another set of identities

\begin{aligned} \sec^2(\arctan(x)) &= 1+x^2\\ \cos(\arctan(x)) &= \frac{1}{\sqrt{1+x^2}}\\ \sin(\arctan(x)) &= \frac{x}{\sqrt{1+x^2}} \end{aligned}

As a sidenote, it seems weird to me that

\begin{aligned} f(x)&= \sin(\arctan(x))\\ f^{-1}(x)&= \tan(\arcsin(x)) \end{aligned}

Textbook Derivatives

May 10th 2025 Single Page

I know, it’s in the textbook (any textbook), but I never really learned the derivative of $f(x) = \log_n(x)$

It’s not too complicated though using the logarithm change-of-base identity.

\frac{d}{dx} \log_n(x) = \frac{d}{dx} \frac{\ln(x)}{\ln(n)} = \frac{1}{x\ln(n)}

The exponential form isn’t that bad either.

\begin{aligned} g(x) &= n^x\\ \ln(g(x)) &= x\ln(n)\\ \frac{d}{dx} \ln(g(x)) &= \frac{d}{dx} x\ln(n)\\ \frac{1}{g(x)}g'(x) &= \ln(n)\\ \frac{d}{dx}n^x &= \ln(n)n^x \end{aligned}

Of course, given that $n^x$ and $\log_n(x)$ are inverses, we can also find the exponential through the identity found in the previous section.

\begin{aligned} f(x) &= \log_n(x)\\ g(x) &= n^x\\ \frac{d}{dx} n^x &= \frac{1}{\frac{1}{ \ln(n)n^x}} = \ln(n)n^x \end{aligned}

Partial sums of the Geometric Series+

May 26th 2025 Single Page

So, the geometric series: if you’ve taken some Calculus, you know of it.

r = 1 + x + x^2+x^3+ \dots

Solve for $r$

r = 1 + rx

r-rx = 1

r(1-x) = 1

r = \frac{1}{1-x}

And there’s the formula the textbook gave ya'

\sum_{j=0}^{\infty} x^j = \frac{1}{1-x}

But what about the partial sums?

\sum_{j=0}^{n} x^j = ?

Decompose into difference of sums

\sum_{j=0}^{n} x^j = \sum_{j=0}^{\infty} x^j - \sum_{j=n+1}^{\infty} x^j

We know the value of the left sum, and the right sum can be found similarly.

h = x^{n+1} + x^{n+2} + x^{n+3}+ \dots

h = x^{n+1} + x(x^{n+1} + x^{n+2} + \dots)

h = x^{n+1} + x(h)

h = \frac{x^{n+1}}{1-x}

So the partial sum can be found

\sum_{j=0}^{n} x^j = \frac{1}{1-x} - \frac{x^{n+1}}{1-x} = \frac{1-x^{n+1}}{1-x}

Some alternate forms are common

\sum_{j=0}^{n-1} x^j = \frac{1-x^{n}}{1-x} = \frac{x^n-1}{x-1}

Also note that, now that the upper bound of the sum is not infinity, the sum is defined for values $\lvert x \rvert \ge 1$

Partial Extended-Geometric Sums

This also applies to series like

x^1 + x^4 + x^7 + x^{10} + \dots

or in general terms where $a > b \ge 0$

r = x^{a+b} + x^{2a+b} + x^{3a+b} + \dots

Finding a closed form for this is not too much more difficult. First find the closed for to the simpler

h = x^{a} + x^{2a} + x^{3a} +\dots

h = x^{a}(1 + x^{a} + x^{2a} +\dots)

h = x^{a} + x^a h

h =\frac{ x^{a}}{1-x^a}

Now the more complex form can be found

r = x^{a+b} + x^{2a+b} + x^{3a+b} + \dots

r = x^{a+b} + x^{2a+b} + x^{3a+b} + \dots

r = x^b (x^{a} + x^{2a} + x^{3a} + \dots)

r= x^bh = \frac{x^{a+b}}{1-x^a}

\sum_{j=1}^{\infty} x^{aj+b} = \frac{x^{a+b}}{1-x^a}

Following similar reasoning as the regular geometric series, the partial form is found.

m = x^{an+b} + x^{a(n+1)+b} + x^{a(n+2)+b} + \dots

m = x^b( x^{an} + x^{a(n+1)} + x^{a(n+2)} + \dots)

m = x^b ( x^{an} + x^{a(n+1)} + x^{a(n+2)} + \dots)

Pause solving $m$ , introduce new equation, $t$

t = x^{an} + x^{a(n+1)} + x^{a(n+2)} + \dots

t = x^{an} + x^{an+1a} + x^{an+2a} + \dots

t = x^{an}(1 + x^{1a} + x^{2a} + \dots)

Pause solving $t$ , introduce new equation $o$

o = x^{a} + x^{2a} + x^{3a} + \dots

o = x^{a}(1 + o)

o = \frac{x^a}{1-x^a}

Resume solving $t$

t = x^{an}(1 + x^{1a} + x^{2a} + \dots)

t = x^{an}\left(1 + \frac{x^a}{1-x^a}\right)

t = x^{an}\frac{1}{1-x^a}

Resume solving $m$

m = x^b ( x^{an} + x^{a(n+1)} + x^{a(n+2)} + \dots)

m = \frac{x^{an+b}}{1-x^a}

Thus

\sum_{j=n}^{\infty} x^{aj+b} = \frac{x^{an+b}}{1-x^a}

\sum_{j=1}^{n-1} x^{aj+b} = \sum_{j=1}^{\infty} x^{aj+b} - \sum_{j=n}^{\infty} x^{aj+b}

\sum_{j=1}^{n-1} x^{aj+b} = \frac{x^{a+b}}{1-x^a} - \frac{x^{an+b}}{1-x^a} = \frac{x^{a+b} - x^{an+b}}{1-x^a} = \frac{x^{a+b}\left(1-x^{a(n-1)} \right)}{1-x^a}

For convenience, increment the value of $n$ by one.

\sum_{j=1}^{n} x^{aj+b} = \frac{x^{a+b}\left(1-x^{an} \right)}{1-x^a}

When $0 \le b < a$

I’m not sure if any of that made any sense, but from my bit of verification, I believe it’s true.

As far as verification goes, it wouldn’t hurt to do so through Induction.

First the base case of $n=1$

\sum_{j=1}^{1} x^{aj+b} = x^{a+b}

\frac{x^{a+b}\cancel{\left(1 - x^{a}\right)}}{\cancel{1-x^a}}

The base case is satisfied

The equation is valid if the following holds

\frac{x^{a+b}\left(1-x^{an} \right)}{1-x^a} + x^{a(n+1)+b} = \frac{x^{a+b}\left(1-x^{a(n+1)} \right)}{1-x^a}

x^{a+b}\left(1-x^{an} \right) + x^{a(n+1)+b}({1-x^a}) = x^{a+b}\left(1-x^{a(n+1)} \right)

x^{a+b} -x^{a(n+1)+b} + x^{a(n+1)+b}({1-x^a}) = x^{a+b}-x^{a(n+2) + b}

x^{a+b} -x^{a(n+1)+b} + x^{a(n+1)+b} - x^{a(n+2)+b} = x^{a+b}-x^{a(n+2) + b}

x^{a+b} \cancel{-x^{a(n+1)+b} + x^{a(n+1)+b}} - x^{a(n+2)+b} = x^{a+b}-x^{a(n+2) + b}

And so it has been proven.

Complex Exponentials and Trigonometric functions

August 26th 2025 Single Page

The value of the exponential function with an imaginary input is of interest.

e^{ix} = ?

Using the well known series expansion of the exponential function:

e^{x} = \sum_{n=0}^{\infty} \frac{x^{n}}{n!}

Substitute in imaginary constant

e^{ix} = \sum_{n=0}^{\infty} \frac{(i)^{n}x^n}{n!}

There is no direct way to simplify this expression as $(i)^n$ doesn’t directly work out to anything nice. This expression is equivalent to a few other, more-useful ones.

\begin{aligned} e^{ix} &= \sum_{n=0}^{\infty} \frac{(i)^{n}x^n}{n!}\\ &= \sum_{n=0}^{\infty} \frac{(i)^{2n}x^{2n}}{(2n)!} + \frac{(i)^{2n+1}x^{2n+1}}{(2n+1)!}\\ &= \sum_{n=0}^{\infty} \frac{(i)^{4n}x^{4n}}{(4n)!} + \frac{(i)^{4n+1}x^{4n+1}}{(4n+1)!} + \frac{(i)^{4n+2}x^{4n+2}}{(4n+2)!} + \frac{(i)^{4n+3}x^{4n+3}}{(4n+3)!} \end{aligned}

Now the powers of $i$ can be simplified

\begin{aligned} i^{4n} &= i^{4^n} = 1\\ i^{4n+1} &= i^{4n}i = i\\ i^{4n+2} &= i^{4n}i^2 = -1\\ i^{4n+3} &= i^{4n}i^3 = -i \end{aligned}

Back-substituting these into the above sum

e^{ix} = \sum_{n=0}^{\infty}\frac{x^{4n}}{(4n)!}+i\frac{x^{4n+1}}{(4n+1)!}-\frac{x^{4n+2}}{(4n+2)!}-i\frac{x^{4n+3}}{(4n+3)!}

Remember the power series for sine and cosine

\begin{aligned} \sin(x) &=\sum_{n=0}^{\infty} \frac{x^{2n+1}(-1)^{n}}{(2n+1)!}\\ \cos(x) &=\sum_{n=0}^{\infty} \frac{x^{2n}(-1)^{n}}{(2n)!} \end{aligned}

Reorder series

e^{ix} = \sum_{n=0}^{\infty}\frac{x^{4n}}{(4n)!} - \frac{x^{4n+2}}{(4n+2)!} + i \frac{x^{4n+1}}{(4n+1)!} - i\frac{x^{4n+3}}{(4n+3)!}

Split sums based on evens, and odds

\begin{aligned} e^{ix} &= \left[{\sum_{n=0}^{\infty}\frac{x^{4n}}{(4n)!} - \frac{x^{4n+2}}{(4n+2)!}}\right]\\ &+ \left[i\sum_{n=0}^{\infty} \frac{x^{4n+1}}{(4n+1)!} - \frac{x^{4n+3}}{(4n+3)!}\right] \end{aligned}

Notice the first sum is simply iterating over the even numbers, and the second is iterating over the odds. Simplify

\begin{aligned} e^{ix} &= \left[{\sum_{n=0}^{\infty}\frac{x^{2n}(-1)^n}{(2n)!}}\right]\\ &+ \left[i\sum_{n=0}^{\infty} \frac{x^{2n+1}(-1)^{n}}{(2n+1)!} \right] \end{aligned}

Notice

e^{ix} = \cos{x} + i\sin{x}

With Negatives

Recall these two properties of sine and cosine

\begin{aligned} \sin(-x) &= -\sin(x)\\ \cos(-x) &= \cos(x)\\ \end{aligned}

With that, use the exponential function with a negative imaginary number

e^{-ix} = \cos{x} - i\sin{x}

We can now solve for cosine and sine using the exponential function.

e^{-ix} + e^{ix} = 2\cos{x}

\cos{x} = \frac{e^{ix} + e^{-ix}}{2}

e^{ix} - e^{-ix} = 2\sin{x}

\sin{x} = \frac{e^{ix} - e^{-ix}}{2}

Proof by Multiplicative Inverse

August 26th 2025 Single Page

We have functions $f(x), g(x), h(x)$ .

Given $h(x) f(x) = 1$ . (they are multiplicative inverses)

We would like to determine whether $f(x) = g(x)$ .

Naturally, $h(x)g(x) = 1\ \implies\ f(x) = g(x)$

Application

I discovered this trick when searching for a proof to Euler’s Formula:

e^{it} = \cos{t} + i\sin{t}

We define

\begin{aligned} f(x) &= e^{it}\\ g(x) &= \left(\cos{t} + i\sin{t}\right) \end{aligned}

Naturally, $e^{it}h(x) = 1$ , so

h(x) = e^{-it}

So, if we can prove that $h(x)g(x) = 1$ , then it is clear that $f(x)=g(x)$ . Introduce $j(x)$ for brevity.

h(x)g(x) = j(x) = e^{-it}\left(\cos{t} + i\sin{t}\right)

Differentiate $j(x)$

\begin {aligned} j'(x) &= e^{-it}\left(-\sin{t} + i\cos{t}\right) - i e^{-it}\left(\cos{t} + i\sin{t}\right)\\ &= e^{-it}\left(-\sin{t} + i\cos{t}\right) + e^{-it}\left(-i\cos{t} + \sin{t}\right)\\ &= e^{-it}\left(i\cos{t} -\sin{t} + \sin{t} -i\cos{t} \right)\\ &= 0 \end{aligned}

That is, $j'(x) =0$ for all $x$ , which means that $j(x)$ is constant for all $x$ .

Find value of $j(x)$

j(0) = e^{0}\left(\cos{0} + i\sin{0}\right) = 1

So, $j(x) = 1$ for all $x$ , which shows that $g(x)$ is the multiplicative inverse of $h(x)$ . Given that $g(x)$ is also the multiplicative inverse of $h(x)$ :

f(x) = g(x)

Source: https://math.stackexchange.com/a/8612

Clever proof of solutions to linear diophantine equation

November 4th 2025 Single Page

A simple result in elementary number theory relates to the linear diophantine equation

ax + by = c

a,x,b,y,c \in \mathbb{Z}

A property of this equation (which is not proven here) is that it has infinitely many $(x,y)$ solutions if $c$ is a multiple of $\gcd(a,b)$ , otherwise it has none.

The clever part is, after finding the first solution we can find the rest. Suppose there exists a solution $(x,y)$ , and that $(x',y')$ is another solution:

c = ax+by = ax'+by'

Rearrange equation

a(x-x') = -b(y-y')

Let $d = \gcd(a,b)$ . Naturally, $a$ and $b$ are both divisible by $d$ . Divide by $d$

\frac{a}{d}(x-x') = -\frac{b}{d}(y-y')

Now, $a/d$ and $b/d$ share no common factors, therefor $a/d$ does not divide $b/d$ and $b/d$ does not divide $a/b$ . Note however that $b/d$ does divide the right side of the equation, and therefor must divide the left side of the equation. Because we already determined that $b/d$ does not divide $a/d$ , $b/d$ must divide $(x-x')$ , thus $(x-x')$ must be a multiple of $b/d$ (likewise $(y-y')$ is a multiple of $a/d$ ).

State that $(x-x')$ is a multiple of $b/d$ in a new equation:

\frac{bn}{d} = x - x'

Likewise of $y$

\frac{an}{d} = y - y'

Thus, given the initial solutions of $x,y$ , the other solutions are found as:

\frac{bn}{d}+x = x'

\frac{an}{d}+y = y'

This proof is not complete but highlights the clever trick of extracting an equation based on divisibility. It comes down to the (often subtle) fact that when $x$ divides $y$ , $y=xq$ .

Complex Difference of Squares

November 19th 2025 Single Page

Sometimes the solution to math puzzles lies in the difference off squares identity, $x^2-y^2 = (x+y)(x-y)$ . I’ve often thought that it’s a shame how there is no similar identity for $x^2+y^2$ , but… there is!

x^2+y^2 = (x+iy)(x-iy)

Where $i$ is the imaginary constant.